Post by: guest39538 on 18/06/2015 19:12:40
y={1/∞}/t
X=player seat
t=time
P=probabilities
∅=random
P(x)=P(X∩y)/∅t
Post by: chiralSPO on 18/06/2015 19:23:18
I assume it has something to do with cards (1/52) and probability.
1/∞ is not defined itself, and so isn't a good part of a deffinition.
what does "player seat" mean?
what does "∅ = random" mean? is it a random number (generated how/chosen from what set with what probability function) is ∅ a randomizing function?
Post by: guest39538 on 18/06/2015 19:40:24
I have no idea what you're trying to derive here...
I assume it has something to do with cards (1/52) and probability.
1/∞ is not defined itself, and so isn't a good part of a deffinition.
what does "player seat" mean?
what does "∅ = random" mean? is it a random number (generated how/chosen from what set with what probability function) is ∅ a randomizing function?
I have just realised that ''y'' is wrong, i think the set is {∞/∞} , in this situation y axis has an infinite number of a single x, x being any one of 52 variants, a deck of cards.
52 in a x axis
then a sequence on top of that of 52 and so on for infinite times y axis,
123456789
123456789
123456789
123456789
Like that......1-52 being x axis and any x value being infinite y axis.
x is finite but y is infinite ,
Player seat is the receiver of x from x axis, but by random timing of events, we skip rows and receive a random row of x from y.
so player X, intercepts Y and between there i space time, but the gap is random ,
∅ just means random symbol, that the distribution of y is by random timing of x
P(x/t)≠P(y/∅t)
P(x/t)=221
P(y,x/∅t)=var(X)
x~=1/52
corr(X,Y)={1/∞}/var(X)~t
{x=y}≠{y/∅t}
Post by: PmbPhy on 19/06/2015 06:30:00
x={1/52}/tIt's all meaningless. You clearly ignored that i said about stating the meaning of the math you're writing. Without that it's all meaningless.
y={1/∞}/t
X=player seat
t=time
P=probabilities
∅=random
P(x)=P(X∩y)/∅t
Post by: Colin2B on 19/06/2015 07:36:49
It's all meaningless. You clearly ignored that i said about stating the meaning of the math you're writing. Without that it's all meaningless.To be honest even with explanation this is meaningless. He is just stringing together half understood terms in a totally illogical way.
I suppose it's easier than learning real maths and logic!
Post by: PmbPhy on 19/06/2015 10:18:25
Quite right my friend. And as ChiralSPO mentioned, 1/∞ is total nonsense. Anybody who knows what ∞ means knows better than to treat it like a number and thus make any attempt to try to divide one by it. Something like this rooted in complete ignorance.It's all meaningless. You clearly ignored that i said about stating the meaning of the math you're writing. Without that it's all meaningless.To be honest even with explanation this is meaningless. He is just stringing together half understood terms in a totally illogical way.
I suppose it's easier than learning real maths and logic!
Post by: guest39538 on 19/06/2015 17:40:39
Quite right my friend. And as ChiralSPO mentioned, 1/∞ is total nonsense. Anybody who knows what ∞ means knows better than to treat it like a number and thus make any attempt to try to divide one by it. Something like is rooted in complete ignorance.It's all meaningless. You clearly ignored that i said about stating the meaning of the math you're writing. Without that it's all meaningless.To be honest even with explanation this is meaningless. He is just stringing together half understood terms in a totally illogical way.
I suppose it's easier than learning real maths and logic!
I know very well that ∞ represents infinite, I have a Y axis that is infinite and contains an infinite number of variant x's, where my x axis only contains 52 variants of x.
I am trying to associate maths, and represent 52 variants of x, and 52 variants stacked for infinite time.
52
52
52
52
like that, I need the for Y, to conclude my formula.
Post by: alancalverd on 19/06/2015 17:49:48
Post by: guest39538 on 19/06/2015 18:31:43
1/x → 0 as x → ∞, so why not just write 0?
y=0 ? that would not make any sense?
Probability of X from x is 1/52
probability of X from y is infinite
added model
http://www.badscience.net/forum/viewtopic.php?f=3&t=36878&p=1382194#p1382194
p580
Post by: PmbPhy on 19/06/2015 23:53:38
Quote from: Thebox
y=0 ? that would not make any sense?It's an abuse of infinity because you're treating it as a number and not using it in the proper sense, i.e. in terms of limits. Infinity is not a number and division is only defined for numbers.
Post by: Colin2B on 20/06/2015 00:19:42
y=0 ? that would not make any sense?Correct, it does not make any sense, but that is what you wrote!
Nothing else makes sense either.
probability of X from y is infiniteProbability can only take values from 0 to 1. Infinite probability is meaningless.
Why are you using time as a value. To paraphrase Tina Turner "What's time got to do with it?"??
Post by: guest39538 on 20/06/2015 09:31:26
y=0 ? that would not make any sense?Correct, it does not make any sense, but that is what you wrote!
Nothing else makes sense either.probability of X from y is infiniteProbability can only take values from 0 to 1. Infinite probability is meaningless.
Why are you using time as a value. To paraphrase Tina Turner "What's time got to do with it?"??
I am using time, because time decides what point of Y you receive. Intersection of x,y by random timing of a tables hand.
But I think I am just going to quit science now and not bother any more , It is like talking to a brick wall at times.
x ⊇ ∞y
every element of x is an infinite element of y.
[x : y]=∞
Post by: Colin2B on 20/06/2015 10:51:10
I am using time, because time decides what point of Y you receive. Intersection of x,y by random timing of a tables hand.It doesn't change the probability. Elements are not events.
every element of x is an infinite element of y."Infinite element" is meaningless unless you explain what you mean by it.
But I think I am just going to quit science now and not bother any more , It is like talking to a brick wall at times.This is a brick wall of your own building. You seem committed to building it as thick and high as you can.
Please stop to think before you post.
And do try to learn basic maths. Probability is not a good place to start as your misunderstandings show. You are also misusing set notation. Don't pick symbols and think you understand their meaning without studying their maths theory first.
People are willing to help you, but you do need to put some effort in, it's not easy but I'm convinced you can do it.
Post by: PmbPhy on 20/06/2015 12:24:28
Quote from: Thebox
I know very well that ∞ represents infinite, ..You misunderstood what I said. I wrote
Quote
Anybody who knows what ∞ means knows better than to treat it like a number and thus make any attempt to try to divide one by it.Did you see me say or imply that you don't know what ∞ means? No. You didn't. Your problem with it has always been that you're treating infinity as if it was a number, and it's not.
Quote from: Thebox
I have a Y axis that is infinite ...This statement is meaningful because the y-axis in a Cartesian coordinate system is unbounded and never ends and that's the meaning of infinite. So you got that part correct. Then you went downhill from there and said
Quote from: Thebox
and contains an infinite number of variant x's, where my x axis only contains 52 variants of x.when you never defined what "variant x's" means. In the context that you used it, it's not a mathematical term so you must have defined it yourself. Perhaps its just poor English on your part.
Quote from: Thebox
I am trying to associate maths, and represent 52 variants of x, and 52 variants stacked for infinite time.But what you've ended up doing is posting things which are completely meaningless. Nobody besides yourself knows what the hell you're talking about.
Quote from: Thebox
But I think I am just going to quit science now and not bother any more , It is like talking to a brick wall at times.That's up to you but you never gave science a shot. All you've done is made erroneous claims as well as claim that physics is wrong when you couldn't understand it. You've never taken the time to learn it. I gave you the opportunity to do so. But instead of reading a physics text you used what little time that you claim to have posting all this nonsense. Any reason that you give for not reading is bogus. Everyone who's serious about learning physics can most certainly take 15 minutes out of their day to read a few pages in a text. You claimed you couldn't but I believe that you've been lying to me. That's probably why you refuse to send me a message every day telling me what you could accomplish that day, even if it was nothing. What's your excuse for not doing that? I've asked you that almost every single day and you ignored my question every single day. That's why you were suspended from my forum. I'll give you two weeks to start doing that after which you'll be permanently banned.
The worst part about your attitude in this particular thread is that you asked us if the math was correct and when we explain that it isn't you claim that we're wrong.
Post by: guest39538 on 21/06/2015 06:27:36
The worst part about your attitude in this particular thread is that you asked us if the math was correct and when we explain that it isn't you claim that we're wrong.
I assume you know what it means without explanation. X is any one of the 52 variants of x axis. X in the y axis is any one of the 52 variants of x*∞. There is an infinite amount of rows of 52, y axis.
So lets say row 10 , column 1, there is an X with the value of being an ace.
By random timing this could be intercepted of the distribution. Do you agree?
12345
23145
53241
12345
In the above I have a 2/4 chance of receiving a 1 if my go is first of the distribution.
So if we increased the rows, lets say to 500 rows, and lets say in column one there was 150 ones spread out over y, I have 150/500 chance of receiving a 1.
Now do you see?
shuffle a deck of cards, there is a 4/52 chance that the top card is an ace.
shuffle another deck of cards, there is also a 4/52 chance the top card is an ace.
have a choice of either decks,
that is two chances that the top card is an ace, 4/52^2 or 4/52*2 or something else.
I have a 1/52 chance of any card being in my distribution place, so consider that I also have like my example,
also I have a 150/500 chance from Y being inline,
added - eureka moment
I have an easy explanation now, it came to me.
shuffle a deck of cards, the odds of the top card being an ace is 4/52 every single shuffle.
shuffle 100 decks of cards, the odds of any of the singular decks top card is 4/52
This is the problem, lets say for example purposes, that in 100 decks of pre-shuffled cards , 15 of the top cards out of the 100 decks, was an ace.
You then have a choice of any deck, your odds are 15/100 of getting an ace and not 4/52.
Now this is as simple explanation as it gets, surely you all understand this?
Post by: alancalverd on 21/06/2015 10:54:59
Quote
shuffle a deck of cards, the odds of the top card being an ace is 4/52 every single shuffle.
shuffle 100 decks of cards, the odds of any of the singular decks top card is 4/52
This is the problem, lets say for example purposes, that in 100 decks of pre-shuffled cards , 15 of the top cards out of the 100 decks, was an ace.
and there's your problem. If you know that 15/100 are aces, then you have good reason to believe that the cards were not randomly shuffled. If they were properly shuffled the most probable number of aces would be between 7 and 8. You can use a Gauss curve to determine the probability of 15 aces in 100 random shuffles: I don't have the time to do right now it but my guess is way less than 5%.
Post by: PmbPhy on 21/06/2015 13:15:58
Quote from: Thebox
I assume you know what it means without explanation.Therein lies your problem. You can't make assumptions of what people know in either math or physics. It's just bad juju.
Quote from: Thebox
X is any one of the 52 variants of x axis.Here's a good example of what you're doing wrong. The term variant means either
Quote
1: manifesting variety, deviation, or disagreementBut you haven't said what it is a variant of. The number "52" reminds us of poker but you didn't mention poker so you're forcing us to make assumptions and that's a very poor thing for someone to do in either physics or math. And if its 52 different things what are those different things? Are the letters? Numbers? If they're numbers are they real or complex? Are they vectors, tensors, 1-forms, etc?
2: varying usually slightly from the standard form
Quote from: Thebox
X in the y axis is any one of the 52 variants of x*∞.A mathematician wouldn't even talk to you if you didn't stop writing things like this nonsense "x*∞." And the statement " X in the y axis" is meaningless.
Quote from: Thebox
There is an infinite amount of rows of 52, y axis.Also meaningless.
Quote from: Thebox
So lets say row 10 , column 1, there is an X with the value of being an ace.Nobody can agree with something that's so poorly stated as to have little meaning. I.e. the phrase this could be intercepted of the distribution is meaningless, just plain nonsense.
By random timing this could be intercepted of the distribution. Do you agree?
What you've posted here is so poorly phrased that you won't find anybody who will be able to understand it. Your problem is that you want to talk about a more difficult area of math, i.e. probability, when you have no understanding of algebra yet,
Post by: guest39538 on 23/06/2015 07:09:05
xxx
xxx
xxx
xxx
each row has 3 units 1-3 that are randomly shuffled, the odds of any row having a 1 in position one (the left) is 1/3.
What are the odds of choosing a row from the five rows, that number one fell into that position one?
Post by: alancalverd on 23/06/2015 07:48:03
Post by: guest39538 on 23/06/2015 18:56:53
1/3, obviously. Since the rows are independent, it doesn't matter which you choose.
consider the random sequence is set, you are choosing from the Y axis, how can it be 1/3 when there is clearly 5 choices?
........................y
........................y
x axis 123456789
........................y
........................y
........................y
x ≠ y
P(x)=1/9
p(y)={1/9}+{1/5}=?
x∩y
Post by: alancalverd on 23/06/2015 19:07:15
Consider a simpler example. I have tossed a coin 1,000,000 000 times. What are the odds that it will come down heads next time?
Now I roll a die 1,000,000,000 times. What are the odds of getting a six on the next roll?
Post by: guest39538 on 23/06/2015 19:17:34
You only make one choice. None of the groups knows anything about any of the other groups.
Consider a simpler example. I have tossed a coin 1,000,000 000 times. What are the odds that it will come down heads next time?
Now I roll a die 1,000,000,000 times. What are the odds of getting a six on the next roll?
I understand what you are saying , but you are not considering after the role or toss.
I toss a coin 1,000,000 times, and record the results, Heads is a win and tails is a loss.
The sequence of the first ten throws are
hhththttth
This is all good, based on only you playing, now lets consider that there is 2 players, I and you, except by randomness, the already results, are distributed to us both.
you get the already tossed 3rd toss, and the 5th toss, and the ninth toss.
You receive 3 tails in a row.
can you understand that?
added - 52²
the rows are different to the columns. The Y deck contains repeat variants.
see model here, surely you understand this one page 581
http://badscience.net/forum/viewtopic.php?f=3&t=36878&p=1383135#p1383135
Post by: alancalverd on 23/06/2015 23:03:21
Post by: guest39538 on 24/06/2015 06:20:49
Why don't I get the 1st and 7th?
You may well get the 1st and 7th because it is random, but that does not matter, what matters is that x is not equal to Y.
you do not get 1/52 from the Y axis, you would get ?/52 if 52² and 52 is x axis, and 52 high Y axis.
123
231
123
132
all x axis would be 1/3 where y axis you can clearly observe is different.
randomly mix the x axis as much as you like it will remain 1/3 always. where Y is subject to constant change .
Y axis is not 52 variants, it is ? variants by alignment.
there could be 50 variants, having 6 aces in the Y axis.
Post by: alancalverd on 24/06/2015 07:58:24
What is the probability that you get 4 heads and 1 tail?*
What is the probability that you get 40000 heads and 10000 tails?
You can't discern or prove randomness from a small number of trials. 4:1 is unlikely but would not suggest a biassed coin, whilst 40000:10000 looks like a bent penny.
If you study any table of random numbers you will find a few short sequences, but the mere presence of sequences is not proof of predictability. Indeed the absence of sequences is evidence that the numbers are not random.
*these numbers are not chosen at random! Test cricket fans keep a record of captains winning the toss, and anyone who wins more than 3 in a 5-match series is honoured in the record books even if the team performs disastrously.
Post by: Colin2B on 24/06/2015 11:59:26
Read carefully what Alan has written and try to understand that probability only tells you about the likelihood of random events over a large number of trials. You seem to be confusing known and unknown events with probability.
Example:
If you draw a card from the top of a pack there is 4/52 chance of an ace. Let's say it is the ace of clubs. If you replace the ace in the centre of the pack and do not shuffle you know the next card is not the ace of clubs, so there are only 3 chances the next card is an ace. But is the probability of an ace now 3/52 or 3/51? What do you think?
123
231
123
132
all x axis would be 1/3 where y axis you can clearly observe is different.
But you don't know that this is the sequence. If you know then you are dealing with ceertainty, not probability. The sequences could also be:
132
123
132
321
you just don't know.
The sequence of the first ten throws are
hhththttth
This is all good, based on only you playing, now lets consider that there is 2 players, I and you, except by randomness, the already results, are distributed to us both.
you get the already tossed 3rd toss, and the 5th toss, and the ninth toss.
You receive 3 tails in a row.
can you understand that?
No, we can't understand because your scenario still isn't clear and you are still confusing random with known. If one of you knows the sequence then they are betting different odds.
As Alan said, why doesn't he get 1st, 3rd, 5th etc; your scenarios are confusing.
To make this clear, let me go back towards the beginning.
Well, without explanation we are confused because in post #1 you say that X=player seat, but above you provide 2 different definitions of X. Confused we are.
I assume you know what it means without explanation. X is any one of the 52 variants of x axis. X in the y axis is any one of the 52 variants of x*∞. There is an infinite amount of rows of 52, y axis.
So lets say row 10 , column 1, there is an X with the value of being an ace.No, because the phrase "By random timing this could be intercepted of the distribution" is incomprehensible.
By random timing this could be intercepted of the distribution. Do you agree?
What has timing got to do with it? if you select a card it doesn't matter whether you do it today or tomorrow.
What distribution?
12345
23145
53241
12345
In the above I have a 2/4 chance of receiving a 1 if my go is first of the distribution.
OK, now this is becoming understandable, but is very different from anything you have said before.
This sounds like:
I have 4 packs containing 5 shuffled cards numbered 1 to 5. If I draw the first card from each pack what is the chance of drawing a 1. However, the probability you have quoted is not for a random shuffle, but only works for the known sequence you have given.
This question however, bears no resemblance to you original questions which talk about y=∞. If you had an infinite number of packs you would spend a very long time taking the 1st card from each pack!!
As you can see, it is important to specify the scenario before you try to introduce maths expressions.
Newton didn't just quote formulae to everyone, he spent a great deal of time writing down and giving lectures on his ideas so everyone could understand what the formulae meant.
EDIT - PS the link you provide is to an equally incomprehensible diagram. You seem to be posting the same thing on a number of sites, giving the same responses (using the same words!!) and getting the same degree of misunderstanding.
Post by: guest39538 on 24/06/2015 19:10:34
In this situation we have 5 sets of 3 variants. Each set is labelled in alphabet form.
a
b
c
d
e
Each set contains the numbers 1,2 and 3 that are randomly shuffled, set, and then sit in an unknown position
a)xxx
b)xxx
c)xxx
d)xxx
e)xxx
We will make this a game of 2 players that the players have a 1/3 chance of receiving a number 1 as the first card from any row, we will call the rows the X axis, and players on the left receive from the right to the left the cards in order of turn.
player 1................b)xxx
player 2................a)xxx
c)xxx
d)xxx
e)xxx
Both players are playing 1/3 odds and playing 1,2,3, that are random shuffled and in an unknown position.
In this scenario the dealer tells player 2 that they have to move seat, and sit south of the decks,
player 1,,,,,,abcde
.................player 2
player 2 is also told they now have to randomly pick 1 of the decks each turn.
a.xxx
b.xxx
c.xxx
d.xxx
e.xxx
We will call this the Y axis of abcde v's the X axis of 123
Player 2 becomes confused, he can clearly see that they are no longer playing 123 randomly distributed over the X axis. They consider they are playing 123 randomly displaced by abcde, player 2 knows that y could have possible repeat values.
5 is not the same as 3 so therefore y≠x , x=1/3, where y=1/3+?/5
the X axis can only produce a combination of 1,2,3, while the Y axis can produce combination of 2,2,2,2,2 therefore having a possible 0/5 chance of receiving a 1.
The X deck is different to the Y deck, 5² and 1-5 being randomised rows leaves an uncertainty of Y the columns, repeat values unknown quantities.
which is simple to show if we reveal and do an experiment with 5 decks of cards.
123
123
123
123
123
each X axis as 123 where Y is constant to a value or values
added- it is infinite because when deck is gone a new one arrives, it is constant
p.s you can flip it and same result
11111∞
22222∞
33333∞
where y axis is now 1/3
and x axis becomes 0 to ∞ odds/t
because the decks are distributed by timing of the table hands which are random time based on players action time.
you are player 5 and will receive the 5th card you have always a 1/5 chance of getting a J,
play the Y axis you have 100% chance of getting a J.
AKAQj
KAQAJ
QKAkJ
player 3 , x axis 1/5 chance of an ace
y axis , 2/3
cards in internet poker should be 1/52 and not 0-∞ by random timing choice of decks.
Post by: alancalverd on 25/06/2015 00:39:56
Post by: guest39538 on 25/06/2015 04:41:39
Your last shuffle and deal was a bad one. The probability of C5=J five times out of five is very small (1/55 = 0.00032) if the shuffles are truly random.
What about if we used 1,000,000 decks, and the same scenario, how many hands out of 1,000,000 decks would contain a J in the 5th position?
The shuffles are truly random, the rng works fine that was never the problem. There is many players who claim online poker is rigged, I know it is not rigged but has a distribution process problem which hopefully I have shown.
Can you confer my scenario is the correct logic and maths?
added -consider spacing and random time of deck distribution
1.akj
.
.time
.
34.akj.
.
.time
.
107.akj
By good luck or bad luck of table timing, would you agree that a player could receive better or lesser than the expected odds of 1/52. The time dots represent decks that have gone elsewhere. ?
Would you agree that the player could intercept values by chance and that the maths is player (a) intercepts x,y over time
P(a)∩(x,y)=0-∞/t
and x≠y/t
added - it just came to me
reel 1
akj
you have to land on an ace to win
reel 1 and 2 both have akj, two aces have to align for you to win.
consider a fruit machine like this , on its side.
Post by: Colin2B on 25/06/2015 21:02:34
Can you confer my scenario is the correct logic and maths?I don't know about the logic because you haven't fully described the scenario.
........
Would you agree that the player could intercept values by chance and that the maths is player (a) intercepts x,y over time
P(a)∩(x,y)=0-∞/t
and x≠y/t
The maths is wrong. You are saying the probability of player (a), what does that mean? If you have a player (a) then probability value is 1, if you don't have a player it is 0. So P(a) is confusing me!
(x,y) what does that mean? You have cards in x,y. So it is really not valid to put these together with the probability of having a player!
What does 0-∞ mean? It has been pointed out before that using ∞ in equations is not valid, to be honest the closest you could say for this is that it is -∞ , what does that mean?
Then you divide -∞ by t. This doesn't make sense either, is -∞/60 seconds really any different from -∞/100seconds??
Time is not an issue here. You can't divide a probability by time.
It would be better if you took the trouble to learn maths rather than stringing random maths symbols together! In most other forums, you would be ridiculed for these equations.
As far as the scenario goes, you will need to describe it in more detail. But let me make some suggestions.
Your game obviously involves cards. Let's say, to follow one of your examples, that you have 5 packs of cards each shuffled.
If each player takes 5 cards from separate packs, clearly they both have equal probability of any sequence of 5 cards. If they both move on to new packs for the next hand, or they return their cards to the packs and shuffle, then again on the next draw they both have equal probability. You don't need to assume an infinite number of packs.
The only reason there might be a difference between the players is if one is expected to make a selection which has a different probability to the other.
Say player 1 draws 5 cards from his pack, player 2 then draws 5 from his own pack. On the next hand player 1 draws 5 cards from his pack without replacing the 1st 5. But player 2 gets a new pack. In this case the probabilities will be different for each player.
Unless you can clearly explain your scenario, without the false maths, it is impossible to know whether you are right. For example, this sequence is not clear:
added -consider spacing and random time of deck distribution
1.akj
.
.time
.
34.akj.
.
.time
.
107.akj
It seems to have something to do with selection of decks?
Post by: guest39538 on 26/06/2015 05:28:38
Can you confer my scenario is the correct logic and maths?I don't know about the logic because you haven't fully described the scenario.
........
Would you agree that the player could intercept values by chance and that the maths is player (a) intercepts x,y over time
P(a)∩(x,y)=0-∞/t
and x≠y/t
The maths is wrong. You are saying the probability of player (a), what does that mean? If you have a player (a) then probability value is 1, if you don't have a player it is 0. So P(a) is confusing me!
(x,y) what does that mean? You have cards in x,y. So it is really not valid to put these together with the probability of having a player!
What does 0-∞ mean? It has been pointed out before that using ∞ in equations is not valid, to be honest the closest you could say for this is that it is -∞ , what does that mean?
Then you divide -∞ by t. This doesn't make sense either, is -∞/60 seconds really any different from -∞/100seconds??
Time is not an issue here. You can't divide a probability by time.
It would be better if you took the trouble to learn maths rather than stringing random maths symbols together! In most other forums, you would be ridiculed for these equations.
As far as the scenario goes, you will need to describe it in more detail. But let me make some suggestions.
Your game obviously involves cards. Let's say, to follow one of your examples, that you have 5 packs of cards each shuffled.
If each player takes 5 cards from separate packs, clearly they both have equal probability of any sequence of 5 cards. If they both move on to new packs for the next hand, or they return their cards to the packs and shuffle, then again on the next draw they both have equal probability. You don't need to assume an infinite number of packs.
The only reason there might be a difference between the players is if one is expected to make a selection which has a different probability to the other.
Say player 1 draws 5 cards from his pack, player 2 then draws 5 from his own pack. On the next hand player 1 draws 5 cards from his pack without replacing the 1st 5. But player 2 gets a new pack. In this case the probabilities will be different for each player.
Unless you can clearly explain your scenario, without the false maths, it is impossible to know whether you are right. For example, this sequence is not clear:
added -consider spacing and random time of deck distribution
1.akj
.
.time
.
34.akj.
.
.time
.
107.akj
It seems to have something to do with selection of decks?
Colin my friend you have just said it,
''Say player 1 draws 5 cards from his pack, player 2 then draws 5 from his own pack. On the next hand player 1 draws 5 cards from his pack without replacing the 1st 5. But player 2 gets a new pack. In this case the probabilities will be different for each player.''
x is not equal to y
the x axis of 52 individual variants of a deck of cards, finite,
the y axis as an unlimited number of decks,
Every table in a game gets a new deck every hand,
picked randomly by timing of table hands.
123
123
123
x is 1/3 always , y is not. Y is not the same as x, x as 1-52. picking from y gives you a different deck because the cards are in the y axis in formation by the shuffle of the x axis aligning cards to your seat different than 52 individual variants, the Y has repeat values and cards missing compared to the x axis.
Consider what you just said. You have it.
and y is 0 to infinite not 0-infinite, - represented to.
akj
akj
jka
play the x axis has 3 decks, or play the Y axis as 3 decks, see the difference?
Post by: alancalverd on 26/06/2015 06:25:07
Quote
What about if we used 1,000,000 decks, and the same scenario, how many hands out of 1,000,000 decks would contain a J in the 5th position?
4,000,000/52, i.e. 4/52 (there being 4 jacks in a pack) of the total number of trials. This is called "regression to the mean" and is a good test of randomness.
Post by: Colin2B on 26/06/2015 15:37:47
what you have shown here is not the same as the scenario I gave above. Let me give some examples to show how different - the scenario is everything.
akj
akj
jka
play the x axis has 3 decks, or play the Y axis as 3 decks, see the difference?
Before I do, let me check that you really understood what Alan was saying when you asked the probability that the 5th card drawn is a J. Do you understand that it is 4/52?
OK, scenarios based on looking back at descriptions you have given.
Let's say 2 players Alan and Box play as follows:
Game 1
there are an infinite number of decks 1,2,3,4,5,etc of 52 cards.
Alan takes pack 1 and draws 5 cards, Box takes pack 2 and draws 5 cards. Equal probability for both players, OK?
Next hand Alan takes Pack 3, but Box is told not to take pack 4, but choose at random from the rest of the packs, he choses pack 9. Probability for both players is still equal OK?
Game 2.
Alan is given a pack of 52 cards and draws 5. From the x axis as you call it.
Box is given 5 packs of 52:
52
52
52
52
52
and is told to take the 1st card from each pack, in other words from the y axis as you call it.
This is not an equal game, the probabilities are different for each player, but it is not the same game or probabilities as the example in my previous post:
Say player 1 draws 5 cards from his pack, player 2 then draws 5 from his own pack. On the next hand player 1 draws 5 cards from his pack without replacing the 1st 5. But player 2 gets a new pack. In this case the probabilities will be different for each player.
Do you understand why the scenarios are so important and even small differences can make a big difference to the probabilities?
I'm saying this because I don't understand how your comments below affect the scenarios you are talking about:
picked randomly by timing of table hands.
or:
123
123
123
x is 1/3 always , y is not. Y is not the same as x, x as 1-52. picking from y gives you a different deck because the cards are in the y axis in formation by the shuffle of the x axis aligning cards to your seat different than 52 individual variants, the Y has repeat values and cards missing compared to the x axis.
I hope by accident I have managed to cover your scenario.
PS, learn maths!
Post by: guest39538 on 26/06/2015 16:22:36
I,you and Alan sit down at a table , we will be distributed two cards each, Alan sits in the small blind position and will receive the first card, the odds of an ace are 4/52.
You sit in the big blind position so will receive the second card. Again 4/51odds of receiving a ace.
I am on the dealer chip so will receive the 3 rd card my odds are also 4/50
We each in turn then receive another card each
We play the hand,
The hand finishes
You are then asked to choose a random deck from several decks that are all ready shuffled
Now consider your comment and scenario ,this is not an equal game
X is 52
Y is repeats etc
X is not equal to y
One deck is always 52 different cards , so imagine card two is an ace, then imagine several decks that card 2 was also an ace, partitioning these decks is several other decks that card 2 is not an ace, you then pick random decks , by luck you pick every deck that gives you an ace, this is not normal to a game of poker.
If we labelled the winning decks red and the losing decks blue and randomly shuffled the decks ,a sort of roulette would happen if distribution was based on random timing of the wheel
Post by: Colin2B on 26/06/2015 17:32:47
the first card, the odds of an ace are 4/52.These probabilities are only valid if the first 2 players do not receive an ace. Ok with that?
the second card. Again 4/51odds of receiving a ace.
the 3 rd card my odds are also 4/50
Note, these are probabilities not odds, odds are different.
You are then asked to choose a random deck from several decks that are all ready shuffledSo how does play proceed? I need to see the full scenario. Do I receive cards from this new deck and you and Alan from the original deck? If so, then yes unequal game. If we all play from the new deck in the same way as described at the beginning, then it is an equal game.
X is 52What you are describing here does not change the probability of receiving an ace from the 1st or 2nd cards in the pack chosen. It doesn't matter which pack you choose.
Y is repeats etc
X is not equal to y
One deck is always 52 different cards , so imagine card two is an ace, then imagine several decks that card 2 was also an ace, partitioning these decks is several other decks that card 2 is not an ace, you then pick random decks , by luck you pick every deck that gives you an ace, this is not normal to a game of poker.
If we labelled the winning decks red and the losing decks blue and randomly shuffled the decks ,a sort of roulette would happen if distribution was based on random timing of the wheel
It is only an issue if I am playing from the new pack and you and Alan continue playing from the original pack. If no aces have been received from this original pack then you and Alan have a greater chance of receiving an ace than I do.
How would a 'normal' game be played? You would have to shuffle or change the deck at some point.
Post by: guest39538 on 26/06/2015 18:10:08
You have gone of track because you then considered the x axis again and not the y axis random formation of variants that were in place by the shuffling of the x axis. Consider an x axis. It can only have 52 variants, where y can have multiple unknown values.
Imagine a single deck of cards face up around a wheel, that was lying flat, like a roulette wheel, on t?op of this wheel was 51 more wheels lying the same way, like a !fruit machine lying on its side,
Spin all the wheels randomly , consider how many cards line up with a single point or a full table 9 points
Added DJ turn tables in a y axis or x axis playing the same record, strobe timing is needed for unison.
Post by: Colin2B on 26/06/2015 22:39:00
Hi Colin you are getting colder and away from the thinkingOh no I'm not, I'm getting warmer if not hot, because I think I can see where the problem is.
Let me propose a game. Alan has gone home, so just the 2 of us.
We will each have a deck of cards in front of us and we will turn over the top card and highest wins (aces high), best out of three.
I take a new pack and shuffle it, put it down in front of me.
In front of you are 10 shuffled packs, and you choose one at random, place it in front of you.
We play. I then return my card and shuffle my pack. You discard your pack and choose another from the remaining 9.
We play, I shuffle, you pick from the remaining 8.
We play, we have a winner.
This is an equal, fair game. We both have equal chance of winning. If we played 100 games we would, most likely, both win around 50.
Do you agree?
I suspect not, because you will say what about the other 7 decks, they might have held better hands for you.
That is irrelevant to probability.
Probability only deals with likely outcomes over a large number of games.
Probability theory says that all the decks were well shuffled, all had 52 cards and all had equal probability of a winning hand. Forget the other 7 decks, they are irrelevant.
Post by: guest39538 on 27/06/2015 06:46:33
That would not be the same game as I am talking about.Hi Colin you are getting colder and away from the thinkingOh no I'm not, I'm getting warmer if not hot, because I think I can see where the problem is.
Let me propose a game. Alan has gone home, so just the 2 of us.
We will each have a deck of cards in front of us and we will turn over the top card and highest wins (aces high), best out of three.
I take a new pack and shuffle it, put it down in front of me.
In front of you are 10 shuffled packs, and you choose one at random, place it in front of you.
We play. I then return my card and shuffle my pack. You discard your pack and choose another from the remaining 9.
We play, I shuffle, you pick from the remaining 8.
We play, we have a winner.
This is an equal, fair game. We both have equal chance of winning. If we played 100 games we would, most likely, both win around 50.
Do you agree?
I suspect not, because you will say what about the other 7 decks, they might have held better hands for you.
That is irrelevant to probability.
Probability only deals with likely outcomes over a large number of games.
Probability theory says that all the decks were well shuffled, all had 52 cards and all had equal probability of a winning hand. Forget the other 7 decks, they are irrelevant.
OK imagine your game, you have a single deck and I have 100000000 decks and can randomly choose any deck, how many of those decks have an ace as the top card?
You have a 4/52 per every shuffle
A 1/52 chance of any particular card
I have a 1/52 chance of any card and also a 4/52 chance of an ace being in my seat alignment from any individual deck,
But what is my cross odds, what is the odds that a pick a deck that as already been shuffled that as an ace aligned to my seat?
Ajk
Kja
Kaj
Akj
Kaj
I am not relying on the shuffle, I am relying on deck choice.in this situation my cross odds are 2/5
With 1000000000 decks my cross odds are ?/1000000000
With an infinite distribution my cross odds are zero to infinite
I will try science , consider time and distance, an vector x is finite that contains 52 meters,
Each meter as its own value represented numerical one to fifty two.
Vector y is infinite and contains multiple variants of meters.y axis can contain several one meters where x axis contains only one.
123
312
123
321
321
123
123
312
312
X is not equal to y
10m is not equal to 5m over time , x is short term and y is long term I do not have two lifetimes,
Post by: Colin2B on 27/06/2015 09:38:14
OK imagine your game, you have a single deck and I have 100000000 decks and can randomly choose any deck, how many of those decks have an ace as the top card?To correctly work out the probability you also have to consider the decks that don't have an ace as the top card and there are far more of those.
You have a 4/52 per every shuffleThere you have it, we both have the same chance of winning for any particular game.
A 1/52 chance of any particular card
I have a 1/52 chance of any card and also a 4/52 chance of an ace being in my seat alignment from any individual deck,
But what is my cross odds,There is no such thing as cross odds
what is the odds that a pick a deck that as already been shuffled that as an ace aligned to my seat?You have chosen a particular set of 5 packs, but there are many more sets where you don't get an ace. In order to understand probability you have to consider not only the favourable sets, but also the unfavourable ones.
Ajk
Kja
Kaj
Akj
Kaj
I am not relying on the shuffle, I am relying on deck choice.in this situation my cross odds are 2/5
Probability isn't about individual games, it's about what happens over a large number of games. If we play your game or mine a large number of times we will both win an equal number of times, so the games are even.
I will try science , ...................I don't know what you think science is, but this is not it.
.............., x is short term and y is long term I do not have two lifetimes,
There is no probability in this, it is not relevant.
Probability is a very difficult subject for many people to understand. It is full of pitfalls like the one you have fallen into.
I don't blame you, we all have an intuitive view of probability which is often incorrect and we have to be very careful in our analysis of favourable cases vs unfavourable cases to really understand what is happening.
I'm going to leave this one now, because I'm not convinced that you will be convinced.
Please learn maths.
Post by: guest39538 on 27/06/2015 10:06:22
It is not me understanding, and I know this because you think I am confused. Probabilities is very simple.
I will show you I understand
a dice 1/6
a deck of cards 1/52
a roulette wheel 1/36
a coin 1/2
that is your odds of any of the above
if I rolled a number 1 with a dice, the next go my odds are 1/6^2 to roll another number 1.
I understand.
4/52 to get an ace is not the same as x/x, if we do not know how many decks there are, and we do not know how many aces have fell in the position of the top card, we can not say 4/52.
it would be would it not?
a={x+y}=
{4/52}²~
t
Post by: Colin2B on 27/06/2015 11:29:50
4/52 to get an ace is not the same as x/x, if we do not know how many decks there are, and we do not know how many aces have fell in the position of the top card, we can not say 4/52.No it wouldn't. Mathematical gibberish.
it would be would it not?
a={x+y}=
{4/52}²~
t
OK Steve, one last try:
Think of each time you shuffle a deck as being the equivalent of an infinite number of preshuffled decks. Each deck from the infinite stack has a 4/52 chance of a ace as 1st card so it doesn't matter which you choose.
You say "we do not know how many aces have fell in the position of the top card, we can not say 4/52". That is is whole point of probability, we don't know. Probability does not deal with known in this way. And yes we can and do say that the probability for each deck is 4/52.
I will show you I understandThis is wrong, if you have already thrown a 1, the odds of your second throw being a 1 is 1/6.
if I rolled a number 1 with a dice, the next go my odds are 1/6^2 to roll another number 1.
If you have not thrown any dice the odds of you getting two 1s for the 1st 2 throws are (1/6)^2
Ok, to see if you do understand, let me ask you another question based on the tossing of a coin 10 times.
Is any of the following sequences more likely to occur than the others?
A) HTHTTHTHHT
B) HTTHTTHHTH
B) HTTHTTTHTT
As I say, I'm unlikely to convince you, so I'm not going to continue to discuss this here, we will have to agree to disagree.
Post by: alancalverd on 27/06/2015 16:40:11
Post by: Colin2B on 27/06/2015 23:16:33
The third sequence contains an excess of tails so might draw the attention of an amateur statistician but a professional would tell him that the excess is not statistically significant - yet.I was wondering whether he knew enough to avoid that obvious trap. I suspected from his previous posts that he didn't.
Post by: alancalverd on 27/06/2015 23:38:26
if I rolled a number 1 with a dice, the next go my odds are 1/6^2 to roll another number 1.
I understand.
Apparently not. The odds of you rolling a 1 on the next throw, and indeed any throw, are 1/6, because the throws are independent.
So here's a little puzzle for you. The odds of throwing six successive 1's in 6 throws are clearly (1/6)6 but what are the odds of throwing (a) at least and (b) exactly one 1 in 6 throws?
I somehow think this is the problem you are trying to solve.
Post by: guest39538 on 28/06/2015 06:57:34
if I rolled a number 1 with a dice, the next go my odds are 1/6^2 to roll another number 1.
I understand.
Apparently not. The odds of you rolling a 1 on the next throw, and indeed any throw, are 1/6, because the throws are independent.
So here's a little puzzle for you. The odds of throwing six successive 1's in 6 throws are clearly (1/6)6 but what are the odds of throwing (a) at least and (b) exactly one 1 in 6 throws?
I somehow think this is the problem you are trying to solve.
Dice can not compare like this, the problem is I have 2 roads, one road is a small length of road and somewhere along the distance of the road , sits a £10 note . The second road is an infinite length, and along that road separated by space is ? quantity of £10 notes sitting in place.
You walked along the first road and randomly dropped the £10 note.
You then walked along the 2nd road and randomly placed or did not place £10 notes.
I am not even sure if there is a £10 note down the second road. I stand at the crossroads.
Post by: jccc on 28/06/2015 07:14:19
gone with the wind.
don't play online poker.
Post by: guest39538 on 28/06/2015 07:17:30
the wind took the note.
gone with the wind.
don't play online poker.
Very good Jccc,
d=..................£10...........................road one
d=....................................................................................∞ road two
£10 notes on road two ?
£10 notes blew away from road two ?
road 1 as the ten pound note stuck to the floor, it is always on road one and only ever one.
r1=.......£10..................................
r1=................£10.........................
r1=..........................£10...............
r1=......£10...................................
.............r2...................................
look vertical from r2, and consider picking a random road.
I call this cross odds.
Post by: alancalverd on 28/06/2015 09:16:55
Online gambling is an industry, not a charity.
Post by: guest39538 on 28/06/2015 09:51:55
If the second road is infinite, anything and everything can happen as you travel it, but the probability of any one thing happening before you die is negligible.
Online gambling is an industry, not a charity.
Poker is not suppose to be gambling, it is a +ev game over time, we play online poker thinking we are playing poker, where it is a very different game to the live game.
It is not negligible probability, it is unknown probabilities, the probability could be an infinite line of aces or an infinite line of no aces, it could also be very unequal, where poker suppose to be the same odds/ probability to us all.
Table one player 2 receives deck 1,8,11,56,72, luckily by timing receiving good starting hands.
Table two player 2 receives deck 2, 9,12,55,70, unluckily receiving poor starting hands.
Post by: Colin2B on 28/06/2015 10:58:58
what are the odds of throwing (a) at least and (b) exactly one 1 in 6 throws?I don't think you are going to get an answer to this, given his poor understanding of probability, unless someone tells him.
I somehow think this is the problem you are trying to solve.At times it is very difficult to understand what he is trying to solve, as some examples appear to be related to selecting from rows and columns.
However, I believe the crux of the problem is as follows:
If you and I are playing a face to face card game and we need to change the deck, we gather the cards together and shuffle the deck. Probability of an ace 1st card is 4/52,
In online poker at change of deck the player apparently selects, at random, from a group of previously shuffled decks. This appears to be where Box derives the infinite number of decks.
However, his concern is with the unselected decks. Let's say you select decks 2, 4 and 7 for your next 3 games. I would say that for each deck the probability of an ace 1st card is 4/52, but box does not. His view is that decks 1, 3, 5 and 6 might have given better hands, hence the probability should take account of this 'infinite vertical y axis' and so the probability for a particular deck is not 4/52 but a complex combination of infinity, time and the x, y axes.
This is more confused by some of the other scenarios he gives, and his poor understanding of maths.
See also his lack of understanding in this reply to you
Table one player 2 receives deck 1,8,11,56,72, luckily by timing receiving good starting hands.
Table two player 2 receives deck 2, 9,12,55,70, unluckily receiving poor starting hands.
I've given him my view, which is that it doesn't matter which decks you select.
I'm not going to keep hitting my head against a brick wall when the probability of success is so low, and particularly if jccc is joining the random fray!
Post by: guest39538 on 28/06/2015 15:34:22
At times it is very difficult to understand what he is trying to solve, as some examples appear to be related to selecting from rows and columns.
However, I believe the crux of the problem is as follows:
If you and I are playing a face to face card game and we need to change the deck, we gather the cards together and shuffle the deck. Probability of an ace 1st card is 4/52,
In online poker at change of deck the player apparently selects, at random, from a group of previously shuffled decks. This appears to be where Box derives the infinite number of decks.
However, his concern is with the unselected decks. Let's say you select decks 2, 4 and 7 for your next 3 games. I would say that for each deck the probability of an ace 1st card is 4/52, but box does not. His view is that decks 1, 3, 5 and 6 might have given better hands, hence the probability should take account of this 'infinite vertical y axis' and so the probability for a particular deck is not 4/52 but a complex combination of infinity, time and the x, y axes.
This is more confused by some of the other scenarios he gives, and his poor understanding of maths.
See also his lack of understanding in this reply to you
No. You are not understanding , we do not select the decks, when our table finishes a hand, we get a random new deck from the already shuffled decks.
So does every other table.
and yes
''should take account of this 'infinite vertical y axis' and so the probability for a particular deck is not 4/52 but a complex combination of infinity, time and the x, y axes.''
table one table 2 table 3 table 4 table 5
deck one deck 2 deck 3 deck 4 deck 5
........................deck 6......................
table 3 finishes their hand first so get the next deck.
Post by: PmbPhy on 28/06/2015 17:07:13
Post by: Colin2B on 28/06/2015 23:02:57
....we do not select the decks, when our table finishes a hand, we get a random new deck from the already shuffled decks.It doesn't matter who selects the new deck or in what order, whether it is the player, the dealer, an onlooker or a computer.
So does every other table.
and yesand no
''should take account of this 'infinite vertical y axis' and so the probability for a particular deck is not 4/52 but a complex combination of infinity, time and the x, y axes.''
should not take account of this 'infinite vertical y axis' and so the probability for a particular deck is not a complex combination of infinity, time and the x, y axes, but the probability of an ace 1st card is 4/52 (ie odds in favour 1:12) for any deck however selected.
table one table 2 table 3 table 4 table 5Doesn't matter.
deck one deck 2 deck 3 deck 4 deck 5
........................deck 6......................
table 3 finishes their hand first so get the next deck.
Please give up this pseudomaths and follow PmbPhy's advice.
Post by: guest39538 on 29/06/2015 18:18:46
should not take account of this 'infinite vertical y axis' and so the probability for a particular deck is not a complex combination of infinity, time and the x, y axes, but the probability of an ace 1st card is 4/52 (ie odds in favour 1:12) for any deck however selected.
Hi Colin , I still can see you are not understanding what I am saying exactly. You and I are talking about two entirely different things.
For some reason you keep resorting back to thinking x axis and disregarding the Y axis.
Do you not understand that I understand basic probabilities, and I am saying to you there is more with a 100% certainty that I know of and am trying to explain.
One deck of cards, we will call this (a), and in (a) there is 4 aces, which we will call an ace, (n).
A second deck of cards, any we will call this (b) , and in (b) is also 4 (n)'s.
Both (a) and (b) have a 4/52 chance of (n) being the top card after a shuffle.
If we were to (a)+(b)=104 then (n) would be 8/104 chance of (n) being the top card.
So if we have a choice of either deck, our first odds would be, 8/104, and then once we have made that choice, the odds return to 4/52.
This is to show you that I understand.
added to clarify, imagine looking at 2 decks of card, there is 8 aces in total in the two decks, and a total of 104 cards, there is an 8/104 chance that one of , or both of the decks, has an ace as the top card.
added -
machine (a) randomly displaces variants on a horizontal conveyor belt for 1 minute.
machine (b) randomly displaces variants on a horizontal conveyor belt for 1 minute.
machine (c) randomly displaces variants on a horizontal conveyor belt for 1 minute.
they stop, creating y axis.
machine (a) randomly variants on a displaces horizontal belt for 1 minute conveyor .
machine (b) displaces randomly variants on a horizontal conveyor belt for 1 minute.
machine (c) randomly horizontal displaces variants on a 1 minute conveyor belt for .
Post by: chiralSPO on 29/06/2015 20:38:56
If you shuffle two decks together, there is still the same chance that the top card is an ace (4/52 = 8/104 ≈ 7.7%) Same if you shuffled 1000 decks together (4000/52000 ≈ 7.7%)
However, if you have two sheffled decks side-by-side, each has a 4/52 chance of having an ace on top. The probability of BOTH decks having an ace on top is (4/52)*(4/52) ≈ 0.59%. The probability that only one of them has an ace on top (and the other has some card other than an ace on top) is 2*(4/52)*(48/52) ≈ 14.2% the probability that neither of them has an ace on top is (48/52)*(48/52) ≈ 85.2%. (quick reality check: the only options in this scenario are two aces on top, one ace on top, and no aces on top, so the probability of one of those three options occurring must be 100%. 0.59 + 14.2 + 85.2 = 99.9, which is within my rounding error of 100%)
If you have these two decks side by side, and you look at the top card of one deck, and it is an ace, you still don't know anything about the other deck, so it still has 4/52 chance of having an ace on top. But if you have the two decks shuffled together, there is an 8/104 chance of the top card being an ace, and if that first card is an ace, then you know the next card has a 7/103 chance of being an ace.
Does that make sense?
Post by: Colin2B on 29/06/2015 21:04:52
So if we have a choice of either deck, our first odds would be, 8/104, and then once we have made that choice, the odds return to 4/52.Unfortunately you don't understand as much as you think you do.
This is to show you that I understand.
This is conditional probability. You don't play with 2 decks, so you choose one and the probability for that game is 4/52
added to clarify, imagine looking at 2 decks of card, there is 8 aces in total in the two decks, and a total of 104 cards, there is an 8/104 chance that one of , or both of the decks, has an ace as the top card.Again you don't understand probability.
Two decks of cards A and B
Let's call probability (not odds, that's different) that the top card of deck A is an ace
, and probability that it's B 
. If it's not an ace we show it as 
or 
.If we put the 2 decks side by side turn over the 2 top cards there are 4 possible events:
Because these are mutually exclusive events there are 2 ways of handling this.
We can either use
which gives the probability of either 1 or 2 acesor more easily
either way the probability is around 14% as ChiralSPO said not 8% as you seem to think, you can follow the arithmetic in his post.
However, ChiralSPO also asked you if that made sense, and given your poor understanding of probability, I suspect not.
Edit: please stop writing probabilities and calling them odds, they are not the same.
The probability of tossing a coin is 1/2 or 50%, or 0.5
The odds are 1:1 or 50:50
Don't confuse the 2
Post by: Colin2B on 30/06/2015 07:33:33
Sorry, in my post above I forgot to answer this part of your question fully.So if we have a choice of either deck, our first odds would be, 8/104, and then once we have made that choice, the odds return to 4/52.Unfortunately you don't understand as much as you think you do.
This is to show you that I understand.
This is conditional probability. You don't play with 2 decks, so you choose one and the probability for that game is 4/52
If you have 2decks and are going to choose between them, then before you choose the probability is 1/2*4/52=4/108 not 8/108 as you think.
However, as I pointed out, you have already decided to choose one deck and never make a different decision, so this selection follows the rules of conditional probability. As you seem to want to use symbolic notation for these situations, it is shown as P(A|B)=4/52.
Hint, learn maths.
EDIT: Added
Steve
I'm addding this because I'm going to try and explain one last time.
Forget your roads, conveyer belts etc and concentrate on cards.
I do understand that you are trying to arrange your decks as 52 cards in the x axis and decks 1 through to infinity in the y axis. Let's forget infinity for the moment.
When we write that the probablity of an ace is 4/52 we should really treat this as a decimal fraction in order to compare different probabilities. In this case ≈0.077.
Probabilities range from 0 to 1 with 0 being never happens to 1 being a certainty. So if we toss a coin 1/2=0.5 mid way on the probability scale (odds 50:50). If we toss a coin the probability of either a head or a tail =1 certainty because you have to get one or the other. If you toss a coin the probability of 6=0 = will never happen.
So if you have 2 decks of cards probability of ace first card = 8/104 = 0.077, same as 4/52 in other words not very likely to happen.
For 10 decks 40/520=0.077
100 decks 400/5200=0.077
All the same all the way up to infinity.
No difference when you include the y axis, which is why I said it was irrelevant.
You however are playing a little game of retrospective probability = what if. What if the packs you didn't choose had a better hand.
A similar situation occurs every time there is an aircrash. Someone misses the plane and then asks "what if I had caught that plane"? Well, probability says it doesn't matter which one you caught, that one, the one before, the one after, the probability for that aircraft, that company that route is the same. We are not concerned about the single event, it's probability is the same as all other events.
But let's look at your missed decks, those better hands. If we consider an ace first card, probability tells us that for every favourable deck there 12 unfavourable, so unlikely you find a better one. Also probability as I explained above is not concerned with the outcome of specific games, but with what happens when you play a large number of games. In that case the probabilities for all decks are as described no matter how the decks are chosen. It really doesn't matter which deck you choose, 12 will be unfavourable for every 1 in your favour.
If you don't agree I suggest you write up your cross odds theory and post it in New Theories. I'll stay away and see what others think. OK
Post by: guest39538 on 30/06/2015 16:42:17
Time plays a key role, it important to the conclusion, I just need to find the correct wording and steal some of the provided maths.
added - it just came to me how I can explain, (n) is a winner, I need two players,
player 1 . please pick from the x axis,
player 2, please pick from the y axis,
a
c
b
n
e
f
g
n
h
n
i
k
l
n
abcdefghijklmnopqrstuvwxyz
m
n
o
p
q
r
n
s
t
u
z
of cause you can not see the values in reality,
player 1 or player 2 has the better chance of picking an (n)?
Post by: Colin2B on 30/06/2015 17:07:28
of cause you can not see the values in reality,Yes, the way you have 'rigged' it, player 2 has a better chance.
player 1 or player 2 has the better chance of picking an (n)?
But you have given y 6 times what would occur from a random distribution and as you play more and more games the sequence tends towards a random distribution.
Also, the way the decks are shuffled you will tend to get 4/52 in the y direction as well as x, so in the long run ( which is what probability is all about) you will get the same result.
PS, I think you know, but I forgot to say that probability is often shown as % ie 1=100%=certainty etc
Have fun with your new theory
Post by: guest39538 on 30/06/2015 17:59:26
of cause you can not see the values in reality,Yes, the way you have 'rigged' it, player 2 has a better chance.
player 1 or player 2 has the better chance of picking an (n)?
But you have given y 6 times what would occur from a random distribution and as you play more and more games the sequence tends towards a random distribution.
Also, the way the decks are shuffled you will tend to get 4/52 in the y direction as well as x, so in the long run ( which is what probability is all about) you will get the same result.
PS, I think you know, but I forgot to say that probability is often shown as % ie 1=100%=certainty etc
Have fun with your new theory
x axis 1-52
y axis ?
52²
1-52 is randomly shuffled along the x axis. When the shuffle ends, the unknown sequence is set of each variant in position within the square,
the y axis contains 1-52 variants in rows where the x axis contains columns of unknown values
stupid me , I had it back to front, sort of.
It does not matter if I un-rig it becuase x is not equal to y
x
x
x
x
x
x
xxxxxxxxxxxxxxxx
x
x
x
x
we know the vertical or Y axis as repeat values over a large quantity.
lets do this properly 52²
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
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....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<x axis
....................................................<52 players
What is the chance of x axis shuffling randomly its 52 variants of each row and aligning 1-52 to each players column?
Post by: Colin2B on 30/06/2015 23:55:10
What is the chance of x axis shuffling randomly its 52 variants of each row and aligning 1-52 to each players column?So eyewateringly small that it is impossible.
The number of ways of arranging a deck of 52 cards is 52!=1068 so the probability of dealing a particular ordering of 52 cards is too small to consider possible. So don't even consider doing it for 52 packs! You won't live to see it even if you deal 24 hrs a day for your lifetime.
Post by: guest39538 on 01/07/2015 05:55:29
What is the chance of x axis shuffling randomly its 52 variants of each row and aligning 1-52 to each players column?So eyewateringly small that it is impossible.
The number of ways of arranging a deck of 52 cards is 52!=1068 so the probability of dealing a particular ordering of 52 cards is too small to consider possible. So don't even consider doing it for 52 packs! You won't live to see it even if you deal 24 hrs a day for your lifetime.
Thank you Colin for your help.
on a ratio of any given x, they individually have 4/52 of producing one of a set of 4 variants as the first card in the rows. What is the chance of there being more or less than 4/52 in the columns aligned to players?
Post by: Colin2B on 01/07/2015 10:29:35
on a ratio of any given x, they individually have 4/52 of producing one of a set of 4 variants as the first card in the rows. What is the chance of there being more or less than 4/52 in the columns aligned to players?
Before I answer this let me check what you are trying to do otherwise the answer will mislead you.
player 1 . please pick from the x axis,
player 2, please pick from the y axis,
For simplicity lets take a example you have given of 4 decks a b c d
a 132
b 123
c 132
d 321
Each player will receive 2 cards
Player 1 is going to use the x direction and is given deck a and draws 13
Player 2 is going to use the y direction chooses 2 decks b&c and hence draws 11
This seems a funny sort of game and I can't see how it relates to your real life game!
Note, these numbers do not relate to reality and player 2 could also have lost with a different distribution
EDIT:
Just to make sure I understand what you mean by using the y axis.
If we have 4 decks of cards:
y4 d d x
y3 d d x
y2 d d x
y1 d d x
x1 x2 x3 x4 x5 x6 etc ............ x52
If you want to use the y axis to deal 4 cards to a player, from x3 as shown, you would have to take decks y1.....y4 discard the top 2 cards from these packs =d and deal the cards x to the player.
Post by: guest39538 on 01/07/2015 16:19:32
on a ratio of any given x, they individually have 4/52 of producing one of a set of 4 variants as the first card in the rows. What is the chance of there being more or less than 4/52 in the columns aligned to players?
Before I answer this let me check what you are trying to do otherwise the answer will mislead you.player 1 . please pick from the x axis,
player 2, please pick from the y axis,
For simplicity lets take a example you have given of 4 decks a b c d
a 132
b 123
c 132
d 321
Each player will receive 2 cards
Player 1 is going to use the x direction and is given deck a and draws 13
Player 2 is going to use the y direction chooses 2 decks b&c and hence draws 11
This seems a funny sort of game and I can't see how it relates to your real life game!
Note, these numbers do not relate to reality and player 2 could also have lost with a different distribution
EDIT:
Just to make sure I understand what you mean by using the y axis.
If we have 4 decks of cards:
y4 d d x
y3 d d x
y2 d d x
y1 d d x
x1 x2 x3 x4 x5 x6 etc ............ x52
If you want to use the y axis to deal 4 cards to a player, from x3 as shown, you would have to take decks y1.....y4 discard the top 2 cards from these packs =d and deal the cards x to the player.
Ok,
yes Y represents the individual sets of decks,
y1=deck 1
y2=deck 2
and so on.
expanding this
1 to 52 along the x axis.
123456789 and so on.
so in Y and x form it looks like this
123
123
123
I am sure you understand this part from your post.
Now we will add some players for our game, 3 players on the Y axis and 3 players along the x axis
y
p3 {1. 2. 3}
p2 {1 .2. 3}
p1 {1. 2. 3}
....p1 p2 p3
p1y gets to chooses a variant out of 3 random shuffled variants, number 2 is a winning number.
they have a 1/3 chance of guessing where the 2 is using the correct axis of x.
p1x also have to pick a variant, from the Y axis, they have 0/3 chance of guessing where the two is in this random formation of x making y.
So now consider poker using one deck,
and then consider the consequence of picking a random deck from the Y axis as my scenario clearly shows.
My chance of an ace suppose to be 4/52 following an x axis of a singular deck, my chance of an ace using the y axis is 0 to infinite. Every time a deck is gone, a new one appears. It is infinite for as long as you play.
X and Y alignment is not the same as just x.
In saying pick a deck, the deck is randomly given to your table, after every hand, each table gets a new deck, timing of table hands deciding what deck you get out of the system.
y1
y2
y3
y4
would be equal to x if y1,y2,y3,y4, arrived at the same table. But by random distribution by time , it is possible to receive y1 then y4 , which just happen give you an ace, while some other poor unlucky player receives y2 and y3 which is a kick in the teethe in this instant.
Imagine a tournament
time=...
Now imagine a player having luck by timing
time=aaa..........aaa.........a.a...a....a....a................a...................a..........a..........a.a.a..a
a=ace
now imagine a player having bad timing luck
time=......................................................I stop here because they ran out of blinds and were forced to fold every hand because they was bad.
Post by: guest39538 on 01/07/2015 16:41:06
''Where a deck of cards is shuffled has absolutely no bearing as to
whether it is shuffled randomly or not. As we have explained to
your repeatedly, the cards are shuffled randomly. They *are*
dealt to your table alone, in order, without being altered in any
way. As I write this, there are more than 25,000 tables in
action, and each of them requires a new, shuffled deck of cards
foe every hand it deals. Each table requires a deck more than
once per minute, on average. It makes no sense to have the cards
shuffled at each table. That would require tens of thousands of
individual shuffle servers, which is just silly.
Instead, there are servers that do nothing but shuffle decks of
cards. They have no idea how those cards will be used, they just
shuffle decks and line them up. When your table needs a deck, it
pings the server, and the next deck in line is delivered to that
table. The cards are then dealt in order. There is no such
thing as a "cool deck"; every deck is shuffled randomly, so it is
a fair deck. It cannot *possibly* favor any particular player.
NOTHING is ever done to alter the decks -- they are simply dealt
in order, fairly.
The "user input" we refer to is not done at your table and does
not result in "control" of the hands. It is a steady stream of
data which is used as one part of the shuffling process. No
*individual* alters the dealing. NOBODY can alter the dealing.
The user input is the entire mass of data coming in from every
single ''edited out name of company'' user in the world, all at once. Understand
that your ''edited out name of company'' table is a representation. As far as the
client and our server are concerned, your table is just a grid of
numbers. When you move your mouse or click a button, it is
transmitting pairs of numbers to our server which represent the
location you are clicking on. For example, a "Fold" might be
something like 28,604x58,789 -- two *large* numbers which
represent a location. Right now, as I am writing this, there are
more than 160,000 players on our site, and *every one of them* is
transmitting a string of these pairs of numbers to our server --
that's how it knows what you want to do.
That *MASSIVE* amount of information is turned into an even more
*MASSIVE* strings of binary data, which is *one* of the sources
of data which our shuffle servers use to shuffle decks of cards.
That string of data has no *meaning* whatsoever. If you move
your mouse left or right, or click a slightly different spot,
that string of data does *change*, but it doesn't influence the
dealing in any particular way. There is no specific control.
Even if you *wanted* to somehow influence the dealing, you would
need to know what every single poker player on ''edited out name of company'', in
every part of the world, is doing at that moment, *AND* you would
have to know how every single bit of that data will be used,
which is *impossible* to know.
Furthermore, that string of binary data is combined with another
string of complete random binary data which comes from the
Quantis Random Number Generator, as described to you already and
linked to on our website. Between those two completely
unpredictable sources, a *massive* number of randomly shuffled
decks of cards are generated every single day -- many tens of
millions of them.
Yes, it could be argued that you moving your mouse a millimeter
to the left somehow *changed* something. But there is absolutely
no way to know *HOW* it changed anything. You don't know the
algorithms used to shuffle, or how the two sources of data are
intertwined, or anything else about that change. You don't even
know the pairs of numbers *you* are generating. It's all
completely impenetrable. That's the most basic quality of
randomness -- it's unpredictable.
The truth is, the shuffling on''edited out name of company'' is likely *MORE* random
than the shuffling of a deck of cards at a live poker table.
That's because the dealers at a live table have a good shuffle
method (scramble, shuffle, cut, shuffle, box, shuffle, cut), but
it's not perfect. It's *POSSIBLE* to track visible cards through
that shuffle. It's *POSSIBLE* for a shuffling expert to
manipulate that shufflie. It's *POSSIBLE* that the dealer will
flash a couple of cards when he's shuffling or cutting. Or it
will "waffle" while it's being pitched across the table. It's
*POSSIBLE* that some player will get an advantage via such dealer
error. It's *POSSIBLE* that the cards are marked. *Usually*
nobody knows what's coming, or any of the cards, but I don't know
any casino player who has not seen a flashed card, or experienced
a mis-deal. On PokerStars, that *NEVER* happens. You *NEVER*
know what's coming, and cards are *NEVER* flashed. It really is
a completely random deal.''
User input is not a "terrible flaw" in our dealing, it's an
*extra* bit of random, unpredictable information. This is used
*along with* another random source. This methodology has been
inspected and approved by Cigital, a company whose job is it to
analyze such things. The *results* of this method of shuffling
and dealing have also been analyzed by countless people for more
than *eleven years* now, and not one person has ever found
anything wrong. The math in our games *is* correct. That's not
a matter of opinion, it's a *fact* and it's completely
verifiable. That's the whole power of hand histories -- you
don't have to have beliefs about the dealing, you can actually
check the results.
As to your claim that "the math is not working out" for you, you
talk about checking your stats. Have you actually done this
math? How did you gather your raw data? What stats have you
checked? How did you do this calculation? What software did you
use? How big is your sample? What method did you use to measure
the significance of your findings? Please send us the details of
your findings.
Or is this all just observations? Are you just frustrated with
losing, and venting that frustration?
Yes, you take bad beats when you play poker. *Every* person who
plays poker takes bad beats. But bad beats are *NOT* why you are
losing. Even *if* it were true that you always bust out on bad
beats (which is *not* true) That wouldn't prove anything. That's
how poker tournaments work -- you play and play and play until
you either make a mistake or take a bad beat. It's the same for
everyone. You just have to lose *ONE* hand and you're busted.
To win a big poker tournament you have to survive a dozen or more
all-in hands. Even if you get your money in ahead every single
time, you're *still* a favorite to lose one of them. For
example, if you get your money in as an 80% favorite (like pair
over pair) just *four* times, your odds of winning all four are
.8x.8x.8x.8 = .4096 -- you're a 2:3 *underdog* to win all of
them. Yes, you remember the time you lose, but you *do* win
those contests 80% of the time. And like I said, you have to win
*many* more than four to win an MTT. Plus, sometimes you don't
get your money in as an 80% favorite -- sometimes it's a flip,
sometimes it's 60%, sometimes you're even an underdog. You *DO*
win most of the time with your good hands, and when you do, you
just keep on playing. *Every* all-in hand is an *opportunity* to
bust out. Complaining about the ones you lose is just silly. OF
COURSE you lost those hands -- that's why you busted out, and why
you remember it.
The last MTT you played here a few days ago had 4,501 players in
it. If you're an average player, your odds of winning a
tournament that big are 4,500:1. Those are *very* long odds.
You did *NOT* bust out of that tournament on a bad beat -- you
busted out with a good hand, AKs, but you lost to a better hand
-- Aces.
In your entire history here, you've only played 643 MTTs with a
cash buy-in. The *average* player count in those events is
3,996. Yet you're upset that you haven't won one yet? Even the
best poker player in the world would not be a favorite to win 1
of 643 tries in a tournament with almost 4,000 players. You
cashed in 90 of them, which is 14% -- that's a normal amount,
since we usually pay 12-15% of the field. Your results are
perfectly normal.
You don't lose every time you're ahead. You win *and* lose
hands, just like everyone else does. You've won a bunch of Sit &
Go's, which of course are easier to win because they have fewer
players in them. When you win them, it's because you *don't*
take a bad beat on an all-in hand, or because *you* got lucky.
The last heads-up S&G you won, just a few days ago, you won when
your opponent pushed all-in with King High, and you called with
Jacks and he *didn't* hit a King. Your good hand *won*. In the
one after that, you lost with KJ to AJ -- *not* a bad beat. In
the 180-player S&G you played just before that, you busted out
when you called *two* pre-flop all-ins with 8-7 (Eight High!),
and lost to pocket Aces. Obviously that's not a bad beat,
either. The truth is, you only bust out of tournaments on bad
beats *sometimes*. *Most* of the time, you bust out when you get
your money in behind. You *don't* "lose to the bad beat always".
Attached to this email is a chart of *every* all-in hand you've
been involved in for the last month in tournaments. Go through
it carefully. Count the times you were ahead and the times you
were behind. Count the wins *and* the losses. What you will
find is that when you have the better hand, you *do* win most of
the time. When you have the worse hand, you lose most of the
time. You take some bad beats and you give some bad beats. It's
all just normal poker. You only have to glance at the "Was
Allin" column and the "Won" column to see the All-In hands where
you lost. It's *obvious* that most of the time you did *not*
bust out on a bad beat. Starting from the top: AK vs AA, KJ vs
AJ, A9 vs 66, QJ va AT, 78 vs AA, Q9 vs A7, J7 va A5, T7 vs 89
(you started ahead but got the money in behind), AQ vs QT...
Finally a true bad beat. Like I said, *most* of the time you do
*not* bust out on a bad beat. You are *not* losing because of
bad beats or some flaw in our dealing. You're losing because
you're getting your money in as an underdog too often.
That's what real poker analysis is about -- checking the
percentages that hands win in similar situations. It's not about
the painful beats. Winning at poker is about recognizing the
mistakes that you make and correcting them. It's *not* about
blaming the dealer because you took a few beats when you had a
good hand.
We are happy for you to make your "findings" public as long as
they are mathematically sound. Your opinion that there is a flaw
in our dealing methodology is incorrect, and your assertions
about how you lose are inaccurate. If you would like to do a
complete, unbiased analysis of your hands, we welcome it. We
have 243,557 of your hands on file, and we will send them all to
you if you ask. There are many software packages available
online which you can use to do a full analysis. It is 100%
certain that if you go through *all* of your hands, you will find
that you have been treated completely fairly. You have won and
lost according to the normal poker math percentages. Here are
some of the hand analysis tools which are available online:
Post by: alancalverd on 01/07/2015 18:23:42
2. In poker, you are playing against a whole lot of other people.
3. In online poker you don't know who they are.
4. The casino is not a charity
5. Each hand is winner-takes-all
6. Very good players can win money in the long term
7. Therefore anyone who is less than very good is likely to lose money.
See? No complicated maths required!
Post by: guest39538 on 01/07/2015 19:51:29
1. A bad workman blames his tools.timing is not a tool
2. In poker, you are playing against a whole lot of other people.yes
3. In online poker you don't know who they are.yes
4. The casino is not a charitypoker is not the same as a casino game,
5. Each hand is winner-takes-allnot really ,
6. Very good players can win money in the long termcrap players win money online
7. Therefore anyone who is less than very good is likely to lose money.not true
See? No complicated maths required!untrue
The relevant part.
''As I write this, there are more than 25,000 tables in
action, and each of them requires a new, shuffled deck of cards
foe every hand it deals. Each table requires a deck more than
once per minute, on average. It makes no sense to have the cards
shuffled at each table. That would require tens of thousands of
individual shuffle servers, which is just silly.
Instead, there are servers that do nothing but shuffle decks of
cards. They have no idea how those cards will be used, they just
shuffle decks and line them up. When your table needs a deck, it
pings the server, and the next deck in line is delivered to that
table. ''
y≠x
123
231
213
P(n)={4/52}/x
...............t
P(n)=?/y
...........t
Post by: chiralSPO on 01/07/2015 20:12:31
These guys spent a lot of time making sure their algorithms are fair, and you are not going to prove otherwise with your current level of mathematical prowess.
Post by: guest39538 on 01/07/2015 20:32:15
Box, it sounds like you have played some online poker and lost. Then it sounds like you complained to the administrators, who sent you a very long, very detailed, and very awesome message. I would not recommend that you continue your argument with them (I might also recommend you stop playing online poker, but that is entirely up to you [:)])
These guys spent a lot of time making sure their algorithms are fair, and you are not going to prove otherwise with your current level of mathematical prowess.
I will give you an idea of my poker ability,
45 man tournaments, 8-9 final tables in a row, on several occasions, hat tricks of first places on several occasions, 180 man 1st places, huge amount of players, cashes. including 3rd places.
HENL Just Lika $1.10 3659 /4 $201
30 Apr 10 No Limit Hold'em HENL Zwane69 $1.10 3802 /5 $162
I am a skilled player, want to see?
http://www.boomplayer.com/poker-hands?search=holdemace486
By the way the company never did send my hand audit, and if you type my player name in google you will see some crazy posts, you will also see if they still exist posts where I am experimenting with online poker.
What he says in his post to me , was when I was on a rant and trying to time a win.
Post by: Colin2B on 01/07/2015 21:57:38
Now we will add some players for our game, 3 players on the Y axis and 3 players along the x axisIn order to make this work you would have to lay out all the cards from the y decks into the xy matrix you show. Players would then have to choose a row or column. This is not how card games are played.
y
p3 {1. 2. 3}
p2 {1 .2. 3}
p1 {1. 2. 3}
....p1 p2 p3
p1y gets to chooses a variant out of 3 random shuffled variants, number 2 is a winning number.
they have a 1/3 chance of guessing where the 2 is using the correct axis of x.
p1x also have to pick a variant, from the Y axis, they have 0/3 chance of guessing where the two is in this random formation of x making y.
Many people are confused when told that the probability of an ace 1st card is 4/52. What they don't realise is that this is the same probability for a 2, a 3, a 9, Q, J etc. In other words all the cross points of your matrix have the same probability of being an ace (or any other card).
So now consider poker using one deck,No, your scenario doesn't show this, it shows only a matrix which the players cannot access for a single deck.
and then consider the consequence of picking a random deck from the Y axis as my scenario clearly shows.
My chance of an ace suppose to be 4/52 following an x axis of a singular deck, my chance of an ace using the y axis is 0 to infinite.Again you cannot access the y axis for a single deck
would be equal to x if y1,y2,y3,y4, arrived at the same table. But by random distribution by time , it is possible to receive y1 then y4 , which just happen give you an ace, while some other poor unlucky player receives y2 and y3 which is a kick in the teethe in this instant.As I have said before, for every favourable deck there are 12 unfavourable so little point trying to change.
If the decks are distributed at random then each deck has the same probability of a winning hand, the order in which you receive them is irrelevant.
You are playing a game here of 'what if', which as I have said before is not what probability is about.
If you genuinely think that the choice of deck order influences the probability of winning I suggest you set up a Monte Carlo simulation for the 2 types of game and see what happens.
Post by: alancalverd on 02/07/2015 00:14:14
1. A bad workman blames his tools.timing is not a tool
2. In poker, you are playing against a whole lot of other people.yes
3. In online poker you don't know who they are.yes
4. The casino is not a charitypoker is not the same as a casino game,
5. Each hand is winner-takes-allnot really ,
6. Very good players can win money in the long termcrap players win money online
7. Therefore anyone who is less than very good is likely to lose money.not true
See? No complicated maths required!untrue
This is known as The Gambler's Delusion. It's remarkably common and the reason that very few people can make a living from playing poker.
Post by: guest39538 on 02/07/2015 05:47:53
I have no question about your poker ability, nor your understanding of how the games are played. I am however having difficulty understanding how your xy matrix relates to real card games. Let me explainNow we will add some players for our game, 3 players on the Y axis and 3 players along the x axisIn order to make this work you would have to lay out all the cards from the y decks into the xy matrix you show. Players would then have to choose a row or column. This is not how card games are played.
y
p3 {1. 2. 3}
p2 {1 .2. 3}
p1 {1. 2. 3}
....p1 p2 p3
p1y gets to chooses a variant out of 3 random shuffled variants, number 2 is a winning number.
they have a 1/3 chance of guessing where the 2 is using the correct axis of x.
p1x also have to pick a variant, from the Y axis, they have 0/3 chance of guessing where the two is in this random formation of x making y.
Many people are confused when told that the probability of an ace 1st card is 4/52. What they don't realise is that this is the same probability for a 2, a 3, a 9, Q, J etc. In other words all the cross points of your matrix have the same probability of being an ace (or any other card).So now consider poker using one deck,No, your scenario doesn't show this, it shows only a matrix which the players cannot access for a single deck.
and then consider the consequence of picking a random deck from the Y axis as my scenario clearly shows.My chance of an ace suppose to be 4/52 following an x axis of a singular deck, my chance of an ace using the y axis is 0 to infinite.Again you cannot access the y axis for a single deck
would be equal to x if y1,y2,y3,y4, arrived at the same table. But by random distribution by time , it is possible to receive y1 then y4 , which just happen give you an ace, while some other poor unlucky player receives y2 and y3 which is a kick in the teethe in this instant.As I have said before, for every favourable deck there are 12 unfavourable so little point trying to change.
If the decks are distributed at random then each deck has the same probability of a winning hand, the order in which you receive them is irrelevant.
You are playing a game here of 'what if', which as I have said before is not what probability is about.
If you genuinely think that the choice of deck order influences the probability of winning I suggest you set up a Monte Carlo simulation for the 2 types of game and see what happens.
Hi Colin, you do not pick columns, the columns are a set order and inline with each player always, the rows are distributed over random timing of hands, in simple terms there is gap spacing between the decks your table receives.
pocket aces are 1/221 on average, ow consider this if we start to make space gaps between hands, i.e play hand one, then do not play another hand until hand 35, a space of 33 hands. This is where time comes into it.
................................................. unbroken/t
. . . . . broken/t
Post by: Colin2B on 02/07/2015 10:07:48
pocket aces are 1/221 on average, ow consider this if we start to make space gaps between hands, i.e play hand one, then do not play another hand until hand 35, a space of 33 hands. This is where time comes into it.
................................................. unbroken/t
. . . . . broken/t
As I've said before the order of the packs or the gaps in between doesn't make a difference to the probability, if you believed that what about all the packs that pass by while you are sleeping?
To check you understand what I am saying:
Q1 shuffle a deck and turn over the top card, what is probability it is an ace?
Q2 shuffle the deck but do not look at the top card, what is probability of an ace?
Q3 still without looking, waste this pack, shuffle and turn over top card, probability of an ace?
Q3 shuffle, put the top card face down on table, no peeking. Turn over 2nd card what is probability it is an ace?
Post by: guest39538 on 02/07/2015 17:03:51
pocket aces are 1/221 on average, ow consider this if we start to make space gaps between hands, i.e play hand one, then do not play another hand until hand 35, a space of 33 hands. This is where time comes into it.
................................................. unbroken/t
. . . . . broken/t
As I've said before the order of the packs or the gaps in between doesn't make a difference to the probability, if you believed that what about all the packs that pass by while you are sleeping?
To check you understand what I am saying:
Q1 shuffle a deck and turn over the top card, what is probability it is an ace?4/52
Q2 shuffle the deck but do not look at the top card, what is probability of an ace?4/52
Q3 still without looking, waste this pack, shuffle and turn over top card, probability of an ace?4/52
Q3 shuffle, put the top card face down on table, no peeking. Turn over 2nd card what is probability it is an ace?4/51 if you discard the card you have separated or 4/52 if you include it.
I think Colin you are still unsure of what I am on about in full,
You are wiser than most and close to understanding in full I feel.
Lets us discuss the time issue,
Lets say I and you sit down at the same poker table in a game.
You have £1000 and I have £1000 in starting chips , we are playing a small tournament.
In a tourney we post blinds and antes, so we can say that £1000 is equal to so many blinds and antes
If we sat out and did not play a hand, we will both get blinded out. We can say and by using measurement,
that there would be an uncertainty of time it will take before we get blinded out, but we could put an approx time value to this, a range between ?hrs-?hrs based on the blind increase time, and an estimate means of how long a table takes to play a hand.
Ok so far?
Post by: Colin2B on 02/07/2015 17:47:41
Q3 shuffle, put the top card face down on table, no peeking. Turn over 2nd card what is probability it is an ace?4/51 if you discard the card you have separated or 4/52 if you include it.
No, the probability is 4/52 in both situations, as long as you don't know what the first card is.
It's important you understand this before thinking about timing because it is fundamental to your understanding of the y axis.
Remember I said that for all the points in your xy matrix -the sample space - there is an equal probability of it being any card from the pack, 1/52.
You have a concern that the 1st card in each deck (y diRection) can have repeat values eg 2 ace of diamonds, which would not occur in the x direction as you are limited to one deck.
Because of this you see the y direction as being different from the x. But, remember what I said, the probability of any card in any space x or y of your matrix - the sample space - is the same, 1/52. This is no different to a 52 sided dice! You would be very happy to accept that a Dice rolled many times might have multiple 6s occuring in succession, but after a large number of games it would be even. Cards in the y direction are the same because each card has a probability of 1/52, so there is no problem if you skip decks, the probability for any deck given to you is the same as any other.
There is no point talking timing, because there is nothing timing can influence.
Post by: guest39538 on 02/07/2015 18:02:23
Q3 shuffle, put the top card face down on table, no peeking. Turn over 2nd card what is probability it is an ace?4/51 if you discard the card you have separated or 4/52 if you include it.
No, the probability is 4/52 in both situations, as long as you don't know what the first card is.
It's important you understand this before thinking about timing because it is fundamental to your understanding of the y axis.
Remember I said that for all the points in your xy matrix -the sample space - there is an equal probability of it being any card from the pack, 1/52.
You have a concern that the 1st card in each deck (y diRection) can have repeat values eg 2 ace of diamonds, which would not occur in the x direction as you are limited to one deck.
Because of this you see the y direction as being different from the x. But, remember what I said, the probability of any card in any space x or y of your matrix - the sample space - is the same, 1/52. This is no different to a 52 sided dice! You would be very happy to accept that a Dice rolled many times might have multiple 6s occuring in succession, but after a large number of games it would be even. Cards in the y direction are the same because each card has a probability of 1/52, so there is no problem if you skip decks, the probability for any deck given to you is the same as any other.
There is no point talking timing, because there is nothing timing can influence.
Timing is what defines what deck you get out of the Y axis.
The Y axis columns are not 1/52, only x has 52 individual values of a set.
123
123
123
123
x axis 1/3
y axis 1/3 in 4
''You have a concern that the 1st card in each deck (y diRection) can have repeat values eg 2 ace of diamonds, which would not occur in the x direction as you are limited to one deck.
Because of this you see the y direction as being different from the x.''
You understand
And Y axis is obviously different to x, take any 3 variants of a set, add more sets of 123,
123
123
123
123
123
123
randomly shuffle all the sets,
x≠y
This is evidential , we can all see this to be true, an axiom. X≠Y is the correct formula, x does not equal y, we can all observe this.
added - add 3 players to the bottom of the columns
player 1y≠player 2y≠player 3y, all the probabilities of Y are differential. Where the probability of x is constant to all observers.
in reality player 1,2,3 is multiplied in the y axis
123
123
123
123
123
ppp
ppp
ppp
ppp
ppp
p=player
i did say it was complex and all this is divided by random time.
Post by: alancalverd on 03/07/2015 13:29:39
Quote
The Y axis columns are not 1/52, only x has 52 individual values of a set.
WRONG! Since each pack is randomly shuffled, each point on the y axis is random, therefore
A. in an infinite number of hands you will find each card value appears 1/13 in each position.
B. in a finite number of hands it would not be surprising (1/6.5) if a given card value appeared in the same position twice in succession
C. but 3 times in succession is pretty unlikely
D. and if you nominate the position (say the first card) then 3 in succession would suggest a fix.
Post by: guest39538 on 03/07/2015 15:29:13
QuoteThe Y axis columns are not 1/52, only x has 52 individual values of a set.
WRONG! Since each pack is randomly shuffled, each point on the y axis is random, therefore
A. in an infinite number of hands you will find each card value appears 1/13 in each position.
B. in a finite number of hands it would not be surprising (1/6.5) if a given card value appeared in the same position twice in succession
C. but 3 times in succession is pretty unlikely
D. and if you nominate the position (say the first card) then 3 in succession would suggest a fix.
Hi Alan, I know from your post that you are still not understanding and still playing the x axis.
In simple explanation I will say this.
In the x axis of infinite rows there can be a possibility of only 1 ace of diamonds per row.
In the Y axis of an infinite column, the x axis shuffle makes the Y axis, and the Y axis contains ? ace of diamonds.
The top card of each individual deck as a 1/52 chance of being an ace of diamonds considering rows.
so lets us put this into a reality situation
1/52
1/52
1/52
1/52
1/52
In the above there is a 1/52 chance of the x axis's top card being an ace of diamonds.
The Y axis obviously reads (1/52)*5
Post by: alancalverd on 03/07/2015 18:09:36
As I interpret your model, you lay a single deck along the x axis, then shuffle and lay another deck above it, thus building up the y axis.
Then there is a 1/13 chance of the card in any given position being the value you state, in each row.
Post by: Colin2B on 03/07/2015 18:18:30
The Y axis columns are not 1/52, only x has 52 individual values of a set.The 52 values in x are what define the probability in y.
You either have not read my post properly or have not understood. Similarly you are not understanding what Alan is saying, because he is saying the same as me.
I will explain again with a simpler example below with 123.
123This does not reflect what happens in reality.
123
123
123
x axis 1/3
y axis 1/3 in 4
x≠yNo we can't all observe this. The probability in y is the same as in x
This is evidential , we can all see this to be true, an axiom. X≠Y is the correct formula, x does not equal y, we can all observe this.
Let me explain again using 123.
If P(x) is the probability in the x axis then for x1, x2, x3 P(x1)=P(x2)=P(x3)=1/3
This is as you describe above.
However, x1 for any deck is the same as yn in that line - this you understand.
So P(x1)=P(y1)=P(y2)=P(y3)=P(y4)............P(y∞)
1/3 1/3 1/3
1/3 1/3 1/3
1/3 1/3 1/3
1/3 1/3 1/3
etc
So the probability for any position is 1/3 and so P(x)-P(y)=1/3
You do not compute the rows as (1/3)*3
Why do you compute the columns as (1/3)*4?
Because this is what you are doing in your response to Alan
If the probability in y is (1/52)*5 then probability in x is (1/52)*52 which is nonsense.
1/52
1/52
1/52
1/52
1/52
In the above there is a 1/52 chance of the x axis's top card being an ace of diamonds.
The Y axis obviously reads (1/52)*5
i did say it was complexNo, it's not complex it is very simple. This is very basic probability and if you cannot understand this you will never understand probability.
There is no difference in the probability in the x or y directions and until you read carefully what we have written and fully understand, there is little value in continuing this.
EDIT: I notice Alan has replied while I was typing. Again we are both saying the same thing. Do try to understand.
Post by: guest39538 on 03/07/2015 19:13:35
If the probability in y is (1/52)*5 then probability in x is (1/52)*52 which is nonsense.
Where are you getting (1/52)*52 from?
In my example I put 5 rows of 1/52
5*52=260
5/260 in the correct position
(1/52)*5
1/52
1/52
1/52
1/52
1/52
there is 5 chances of 1/52 that the top card is an ace of diamonds
in fact there is a 0-∞/t chance an ace of diamonds is the top card.
52x²=(1/52)*52=1
52x²=2704/13/4=52
Y=52a
x=52b
Post by: Colin2B on 04/07/2015 00:03:51
Where are you getting (1/52)*52 from?I'm getting it from the same place you get (1/52)*5, just to show how you are wrong in what you are doing. If you do that to the column you have to do it to the rows as well.
there is 5 chances of 1/52 that the top card is an ace of diamondsThese 5 events are not mutually exclusive so you can not say (1/52)*5, but must use (1/52)5, so the probability of 5 decks in a row all yielding a specific card, eg ace of diamonds, as first card is 1/380204032, in other words unlikely. It is worth noting that this probability is the same for any selected group of 5 decks, whether consecutive or spaced eg decks 2 3 4 5 6 or 2 4 5 8 15 both these sequences have the same probability.
However, this is not what you are doing in the games you play, you are only ever taking one deck at a time and therefore the probability is always 1/52. This again is independent of the order of the decks so y=x.
in fact there is a 0-∞/t chance an ace of diamonds is the top card.All of this is mathematical gibberish, and wrong.
52x²=(1/52)*52=1
52x²=2704/13/4=52
Y=52a
x=52b
Please look back at the posts by Alan, ChiralSPO and myself and try to understand where you are going wrong.
When you do understand, let me know and we can talk some more.
Edit: if you understand what we have written, you will then be able to understand why timing ie sequence is irrelevant.
Post by: alancalverd on 04/07/2015 06:43:05
It is also obvious that if you shuffled a very large number N of decks, the number of times you would draw AD as the top card should approach N/52 as N increases, but you still have no means of predicting which shuffle will achieve this.
Post by: guest39538 on 04/07/2015 09:02:31
Now Alan seem's close to understanding.
I suggest the problem is with communication, I am translating your posts and you are translating my posts differently to there meanings.
The messages intentions and meanings are being lost in translation, can I please suggest we start over and try to keep it simple, I understand in my head exactly what I mean, people I speak to I know understand me.
So in starting new, let us define some values and stick to the values.
One deck of cards, containing 52 individual variants we will call this {N} and put on set brackets to define a set.
A single deck of cards spread out in a row, we will call this the (x) axis.Brackets defining it is a scalar direction
more rows of cards, we will call this (y) axis.Brackets defining it is a scalar direction
t=time
r=radius
R=random
P=probabilities
p=player
Each deck has subsets of 4 values, i.e 4 aces, we will call this {_N}
For a single variant, we will call this {_n}
Can we agree that P{_n}=1/52 and the P{_N}=4/52 using a single deck?
If p1 receives the top card and it is an ace, the probability of the top card being an ace after the next shuffle is P{_N}=4/52^2?
P{_N} from any singular (x) from (y) is 4/52?
P{_n} from any singular (x) from (y) is 1/52?
We also need to add {_sn} , which represents a specific variant of the 52.
~=distribution of
#=not equal to
•=repeat
(1)(2)(3) ect= deck numbers from the (y) axis.
ok so far?
Ok first scenario,
we take the first 3 cards of each deck and leave them face down unseen.
(p1)nnn
(p2)nnn
(p3)nnn
(p4)nnn
(p5)nnn
You are player 3, can you confirm the probabilities of the first {_sn} you will receive?
P{_sn}=?
secondly ....we will take the first 5 {_n} of each {N} from (y)
nnnnn
nnnnn
nnnnn
nnnnn
nnnnn
ppppp
Can you please now confirm the P{_sn} from the (y) , you are still p3.
P{_sn} from (y)=
Post by: alancalverd on 04/07/2015 09:32:12
Quote
If p1 receives the top card and it is an ace, the probability of the top card being an ace after the next shuffle is P{_N}=4/52^2?
No. The shuffles are independent so P{_N} = 4/52 every time.
But the probability of drawing two aces in successive shuffles is obviously (1/13)^2
Hence "beginner's luck". You remember your first win, whether it was your first game or your 13th, but the probability of winning n games in a short sequence thereafter is the square, cube.... nth power of 1/13 so it looks as though your luck runs out even though it hasn't changed at all.
And of course it's more complicated in a real poker game because "first to draw an ace" is not the winning hand, and a good bluffer can win with an empty hand.
Post by: guest39538 on 04/07/2015 09:39:19
"scalar direction" is meaningless.QuoteIf p1 receives the top card and it is an ace, the probability of the top card being an ace after the next shuffle is P{_N}=4/52^2?
No. The shuffles are independent so P{_N} = 4/52 every time.
But the probability of drawing two aces in successive shuffles is obviously (1/13)^2
Hence "beginner's luck". You remember your first win, whether it was your first game or your 13th, but the probability of winning n games in a short sequence thereafter is the square, cube.... nth power of 1/13 so it looks as though your luck runs out even though it hasn't changed at all.
And of course it's more complicated in a real poker game because "first to draw an ace" is not the winning hand, and a good bluffer can win with an empty hand.
please answer
we take the first 3 cards of each deck and leave them face down unseen.
(p1)nnn
(p2)nnn
(p3)nnn
(p4)nnn
(p5)nnn
You are player 3, can you confirm the probabilities of the first {_sn} you will receive?
P{_sn}=?
secondly ....we will take the first 5 {_n} of each {N} from (y)
nnnnn
nnnnn
nnnnn
nnnnn
nnnnn
ppppp
Can you please now confirm the P{_sn} from the (y) , you are still p3.
P{_sn} from (y)=
added - ''Chaos theory is the field of study in mathematics that studies the behavior of dynamical systems that are highly sensitive to initial conditions—a response popularly referred to as the butterfly effect. Small differences in initial conditions (such as those due to rounding errors in numerical computation) yield widely diverging outcomes for such dynamical systems, rendering long-term prediction impossible in general.''
https://en.wikipedia.org/wiki/Chaos_theory#/media/File:Chaos_Sensitive_Dependence.svg
Post by: Colin2B on 04/07/2015 11:39:04
Hi, I am confused,I thought Colin was close to understanding and agreeing with me,No, there is a big difference between understanding and agreeing. I do understand the mistaken assumptions you are making, but I have never agreed with you about your random timing and your interpretation of probabilities in x&y. You frequently misunderstand what I am writing.
Now Alan seem's close to understanding.Alan also understands the mistakes you are making.
You are right about misunderstandings as many of your posts are very confusing, and it is very easy to misunderstand your questions.
If p1 receives the top card and it is an ace, the probability of the top card being an ace after the next shuffle is P{_N}=4/52^2?No, this has been explained before. It is 4/52.
It is only 4/52^2 if you play both decks together and turn over the top 2 cards.
Once you have played a deck and know the outcome the probability for the next shuffle is 4/52. This is called conditional probability
Please reread posts #56 and #57 where this is also explained.
Those posts also give you the method for determining the probability that at least one of the top cards of 5 packs is an ace.
As explained there, the probability for 2 packs that one and only one of the top cards is an ace is 2*(4/52)*(48/52)
For 5 decks the probability that one and only one of the top cards is an ace is 5*(4/52)*(48/52)4
Also as I have explained above, the order in which you select the decks - what you call random timing - does not affect the probabilities.
'Edit: Don't try to understand chaos theory until you thoroughly understand probability. Chaos theory is not appropriate here because we are not dealing with approximations and this example is a simple probability problem.'
If you are not reading and understanding my posts I see no point continuing.
So I'll leave it to Alan to see if his explanations can help you understand.
Hence "beginner's luck". ...Alan, even though you are armed with much more than beginner's luck, you are going to need more luck than I have had.
I wish you ..... luck?
Post by: alancalverd on 04/07/2015 14:55:10
2. Poker has nothing to do with chaos theory. The fall of the cards is pure linear statistics.
Post by: guest39538 on 05/07/2015 07:34:16
1. I have absolutely no idea what {_sn} means
2. Poker has nothing to do with chaos theory. The fall of the cards is pure linear statistics.
Huh? I explained what {_sn} means further up the page, it means a specific variant of a set, i.e the ace of diamonds.
P{_sn} from x is always 1/52
P{_sn} from Y is ?
It is unknown and can never be known, it is chaos.
I really do understand what I am on about.
take 123 and mix it randomly, take another 123 and mix it randomly, the mixing of 123 and 123 generates Y axis.
231
213
In this example x remains 1/3 where is Y is differential. Just consider rows and columns.
P(x)#P(y)
advanced model
http://www.badscience.net/forum/viewtopic.php?f=3&t=36878&p=1386649#p1386649
post#14608
page 585
P{_sn}/(x)
t
=1
P{_sn)/(y)
t
=0-∞
x=123
y=
1
1
1
or
0
0
0
Post by: Colin2B on 05/07/2015 09:41:42
Huh? I explained what {_sn} means further up the page, it means a specific variant of a set, i.e the ace of diamonds.No you didn't explain at all. Your use of set notation is confusing, better not to use it unless you are intending to manipulate sets.
The way you wrote your subset n as being sets of 4 cards means P(n)=1/4 - which is not what you intended.
P{_sn} from x is always 1/52Agreed
P{_sn} from Y is ?Rubbish. It is also 1/52
It is unknown and can never be known, it is chaos.
I really do understand what I am on about.No you don't.
However, I am beginning to see where you are confusing yourself. If I get time I will put together an explanation later today.
I'm not going to comment on the rest of your post as it is so confusing, both to you and others.
Post by: guest39538 on 05/07/2015 09:49:06
Huh? I explained what {_sn} means further up the page, it means a specific variant of a set, i.e the ace of diamonds.No you didn't explain at all. Your use of set notation is confusing, better not to use it unless you are intending to manipulate sets.
The way you wrote your subset n as being sets of 4 cards means P(n)=1/4 - which is not what you intended.P{_sn} from x is always 1/52AgreedP{_sn} from Y is ?Rubbish. It is also 1/52
It is unknown and can never be known, it is chaos.I really do understand what I am on about.No you don't.
However, I am beginning to see where you are confusing yourself. If I get time I will put together an explanation later today.
I'm not going to comment on the rest of your post as it is so confusing, both to you and others.
if Y has 5o rows it can not be 52, it would be /50 .
and it just came to me, players by random timing of table hands are quantum leaping through space-time and intercepting values by timing luck.
p.s you keep turning the visual around in your head, so you keep getting 1/52
left to right is x,
bottom to top is y
keep the perspective the same.
123
231
123
the rows are not the same as the columns.
P(1)/x=1/3
P(1)/y=?/3
x is a constant of 123 where as Y is not a constant of 123, x creates unknown y.
Post by: alancalverd on 05/07/2015 10:01:08
If you play any game of pure chance against n -1 other players, the probability of your winning is 1/n.
Poker cards are randomly shuflled for each hand so if all players are of zero skill and stake the same amount, your longterm return will be 1/n of everyone's stake - i.e, your stake (less the house commission, of course). And everyone else will receive exactly the same.
But there is a considerable element of skill in poker, so in real life your return will be x/n (where x represents your fraction of the total skill around the table) as long as everyone plays. But people drop out and the final head-to-head is effectively a winner-takes-all contest of skill, but because it involves a large element of chance it takes longer than a darts or boxing match to resolve.
No difficult maths involved. As with any game from chess to cricket, if you play a lot, against better players, you will learn a bit and lose a lot. At least in chess and cricket there are leagues and ratings tables so you can choose an opponent of your own level and have fun.
I prefer backgammon.
Post by: guest39538 on 05/07/2015 10:09:33
Let's get to the nub of this.
If you play any game of pure chance against n -1 other players, the probability of your winning is 1/n.
Poker cards are randomly shuflled for each hand so if all players are of zero skill and stake the same amount, your longterm return will be 1/n of everyone's stake - i.e, your stake (less the house commission, of course). And everyone else will receive exactly the same.
But there is a considerable element of skill in poker, so in real life your return will be x/n (where x represents your fraction of the total skill around the table) as long as everyone plays. But people drop out and the final head-to-head is effectively a winner-takes-all contest of skill, but because it involves a large element of chance it takes longer than a darts or boxing match to resolve.
No difficult maths involved. As with any game from chess to cricket, if you play a lot, against better players, you will learn a bit and lose a lot. At least in chess and cricket there are leagues and ratings tables so you can choose an opponent of your own level and have fun.
I prefer backgammon.
Yes Poker is a game of skill, and in the long term it is +ev for good players.
But all the skill in the world can not prevent you being dominated by ''quantum leaping'',
Timing is everything in poker, if you make a final table you hope you get some good hands, you do not want to be bad ''quantum leaping'' and landing bad hands.
I could explain consequence all day long, ''butterfly effect'',
x/t would be standard, x,y/t is completely random winners.
Post by: alancalverd on 05/07/2015 10:19:42
123
231
123
the rows are not the same as the columns.
P(1)/x=1/3
P(1)/y=?/3
Wrong.
Since the rows are independent, the probability of anything happening anywhere in the y axis is exactly the same as in the x axis. If it were not so, you would be asserting that choosing an element in row 1 affects the distribution of row 2, i.e. the rows are not independent.
In the case of online poker, I think the promoters have gone to great lengths to ensure that each shuffle is indeed independent.
Your problem of understanding is this:
By shuffling from a limited set, you obviously can't get two identical cards in sequence on the x axis in any one shuffle. Each card has only one degree of freedom in a shuffle.
But it is entirely possible that he same card will appear in the same position in two rows in the y axis PRECISELY because the shuffles are random.
The a priori probability of finding a given card in a given position is obviously 1/52 for each row, but less demonstrably 1/52 for each column because each card has one degree of freedom in EVERY shuffle so you need an infinite number of shuffles to prove it.
Now to add to your confusion!
1. If you find an ace at position p in one shuffle, what is the probability Pp of finding an ace at p in the previous or subsequent shuffle?
2. If you nominate position q, what is the probability Pq of finding an ace at q in two subsequent shuffles?
Post by: alancalverd on 05/07/2015 10:23:41
I could explain consequence all day long, ''butterfly effect'',
x/t would be standard, x,y/t is completely random winners.
The butterfly effect is the consequence of a trigger in an unstable system. In poker the system is stable but contains a random element. The art is to ride the wave you are given, but unlike surfing, you can't see it coming.
Post by: guest39538 on 05/07/2015 10:44:10
I could explain consequence all day long, ''butterfly effect'',
x/t would be standard, x,y/t is completely random winners.
The butterfly effect is the consequence of a trigger in an unstable system. In poker the system is stable but contains a random element. The art is to ride the wave you are given, but unlike surfing, you can't see it coming.
Unfortunately the waves of y are different to x, you are not considering random choice of Y and the set sequence.
are you really persisting your argument saying that x is equal to y?
123
321
123
X is obviously not equal to Y,
All observers looking left to right observe 1/3.
All observers looking up observe ?/3
column 1, 131 =player 1 the small blind
column 2, 222 =player 2 the big blind
column 3, 313 =player 3 under the gun
so how is x =y?
it obviously is not.
Post by: alancalverd on 05/07/2015 12:07:20
It's your money and your life. Gamblers Anonymous have more experience and patience than me in dealing with your problem.
Alfa Charlie out.
Post by: guest39538 on 05/07/2015 18:16:51
There is no set sequence in poker. You cannot usefully compare three carefully chosen groups wth an infinity of random ones. It's another part of the Gambler's Delusion.
It's your money and your life. Gamblers Anonymous have more experience and patience than me in dealing with your problem.
Alfa Charlie out.
It is called gamblers fallacy and not gamblers delusion,
and if you shuffle any deck, and stop, the sequence is set although unknown.
I will give up , although I know very well I am correct,
Post by: Colin2B on 05/07/2015 18:36:28
I will give up , although I know very well I am correct,You give up without having read and understood our posts. For that reason you will never understand why you are wrong, and you will never learn maths and probability.
A pity, because you could have.
Post by: guest39538 on 05/07/2015 18:48:13
I will give up , although I know very well I am correct,You give up without having read and understood our posts. For that reason you will never understand why you are wrong, and you will never learn maths and probability.
A pity, because you could have.
oh but I understand.
You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?
i bet you say 4/52 which is incorrect.
I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?
Post by: jccc on 05/07/2015 19:18:54
I will give up , although I know very well I am correct,You give up without having read and understood our posts. For that reason you will never understand why you are wrong, and you will never learn maths and probability.
A pity, because you could have.
oh but I understand.
You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?
i bet you say 4/52 which is incorrect.
I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?
4/53 is correct for every deck.
100 cards you get 100/13 aces.
i see your fishing trip is very successful, good job bro!
Post by: guest39538 on 05/07/2015 19:23:58
I will give up , although I know very well I am correct,You give up without having read and understood our posts. For that reason you will never understand why you are wrong, and you will never learn maths and probability.
A pity, because you could have.
oh but I understand.
You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?
i bet you say 4/52 which is incorrect.
I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?
4/53 is correct for every deck.
100 cards you get 100/13 aces.
i see your fishing trip is very successful, good job bro!
7.69230769231≠0.07692307692
Post by: chiralSPO on 06/07/2015 02:16:01
oh but I understand.
You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?
i bet you say 4/52 which is incorrect.
I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?
4/52 is correct for every deck.
100 cards you get 100/13 aces.
i see your fishing trip is very successful, good job bro!
jccc is absolutely correct here.
7.69230769231≠0.07692307692
No, it is exactly 100 times bigger (because you're drawing 100 cards, not 1)...
Post by: jccc on 06/07/2015 02:29:19
oh but I understand.
You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?
i bet you say 4/52 which is incorrect.
I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?
4/52 is correct for every deck.
100 cards you get 100/13 aces.
i see your fishing trip is very successful, good job bro!
jccc is absolutely correct here.
please save that line for my gravity/light theory.
Post by: guest39538 on 06/07/2015 06:15:31
oh but I understand.
You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?
i bet you say 4/52 which is incorrect.
I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?
4/52 is correct for every deck.
100 cards you get 100/13 aces.
i see your fishing trip is very successful, good job bro!
jccc is absolutely correct here.
7.69230769231≠0.07692307692
No, it is exactly 100 times bigger (because you're drawing 100 cards, not 1)...
No, you are only drawing one card from 100 cards, the same as choosing a deck from 100 pre-shuffled decks and taking the top card.
x is not equal to y. I believe I am correct in my idea?
Post by: Colin2B on 06/07/2015 09:18:31
Consider y
Toss a coin. 1/2 chance for any face value H or T. Toss 100 times still 1/2 each throw. Toss ∞ times still 1/2 each throw. yes you can get multiple heads in 100 throws but they even out over a large number of throws so in 100 throws you are more likely to get 50 H and 50 T.
Now throw a dice. 1/6 chance of any face value. Throw 100 times still 1/6 each throw. Throw ∞ times still 1/6 each throw. Again evens out over a large number of throws.
Now throw a 52 sided dice with a card value painted on each face. Better still deal a shuffled pack and choose the top card. 1/52.
As with the coin and 6 sided dice you can get repeats but they even out. The probabilities or odds are not unknown.
Probability of repeats?
4 H in a row 1/2*1/2*1/2*1/2=(1/2)4
4 ace of diamonds in a row (1/52)4= not very likely.
100 coin tosses you are more likely to get 50 H and 50 T.
Draw top card from 100 decks, then you are likely to have 100/13 aces.
In reality, you will get some slight variations from these, but over a large number eg >300 the actual numbers get closer to the calculated probabilities.
These are not unknown probabilities, they are known and understood.
Happy fishing by the way.
PS good one jccc you understand probability, we'll have you understanding QM soon. Probability of electron etc [;)]
EDIT:
I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?These are not the same question. On the 1st jccc is right.
.
.
No, you are only drawing one card from 100 cards, the same as choosing a deck from 100 pre-shuffled decks and taking the top card.
On the 2nd read what ChiralSPO and I have written above then consider:
If you toss a coin 99 times what is the probability of a H on the next toss? it is 1/2.
Now get someone to toss a coin 100 times and write down the outcomes. Then you choose one of them, if you choose #100 then the probability it is a H is still 1/2. You are not choosing from 100, you are choosing from 2 possibilities, either it is a H or it is a T = 1/2.
So if you choose a deck from 100 preshuffled decks the probability of an ace top card is 4/52. The 99 other decks are irrelevant, you are only choosing this one.
I think you are confusing probability of singular events with that of combinational events. Don't feel bad about it, this is a very common error and leads to people believing that if they have just tossed 5 H in a row, the next is more likely to be a T.
Post by: guest39538 on 06/07/2015 15:52:28
jccc is right
Consider y
Toss a coin. 1/2 chance for any face value H or T. Toss 100 times still 1/2 each throw. Toss ∞ times still 1/2 each throw. yes you can get multiple heads in 100 throws but they even out over a large number of throws so in 100 throws you are more likely to get 50 H and 50 T.
Now throw a dice. 1/6 chance of any face value. Throw 100 times still 1/6 each throw. Throw ∞ times still 1/6 each throw. Again evens out over a large number of throws.
Now throw a 52 sided dice with a card value painted on each face. Better still deal a shuffled pack and choose the top card. 1/52.
As with the coin and 6 sided dice you can get repeats but they even out. The probabilities or odds are not unknown.
Probability of repeats?
4 H in a row 1/2*1/2*1/2*1/2=(1/2)4
4 ace of diamonds in a row (1/52)4= not very likely.
100 coin tosses you are more likely to get 50 H and 50 T.
Draw top card from 100 decks, then you are likely to have 100/13 aces.
In reality, you will get some slight variations from these, but over a large number eg >300 the actual numbers get closer to the calculated probabilities.
These are not unknown probabilities, they are known and understood.
Happy fishing by the way.
PS good one jccc you understand probability, we'll have you understanding QM soon. Probability of electron etc [;)]
EDIT:I take the 100 unknown top cards of each deck, how many of the 100 cards are aces?These are not the same question. On the 1st jccc is right.
.
.
No, you are only drawing one card from 100 cards, the same as choosing a deck from 100 pre-shuffled decks and taking the top card.
On the 2nd read what ChiralSPO and I have written above then consider:
If you toss a coin 99 times what is the probability of a H on the next toss? it is 1/2.
Now get someone to toss a coin 100 times and write down the outcomes. Then you choose one of them, if you choose #100 then the probability it is a H is still 1/2. You are not choosing from 100, you are choosing from 2 possibilities, either it is a H or it is a T = 1/2.
So if you choose a deck from 100 preshuffled decks the probability of an ace top card is 4/52. The 99 other decks are irrelevant, you are only choosing this one.
I think you are confusing probability of singular events with that of combinational events. Don't feel bad about it, this is a very common error and leads to people believing that if they have just tossed 5 H in a row, the next is more likely to be a T.
I am not making no mistakes, my logic is accurate, but from this post I now know why you are looking always from the wrong angle.
You are not considering that the coin is already tossed 100 times.
You are not considering the already set unknown sequence of a deck of cards.
The top card is not random, it is a unknown value. Once the top card is set, it is set and nothing can change this.
You are making a mistake and still looking a the wrong angle and perspective.
Answer this,
Take two decks of cards for simplicity, try it at home, shuffle both decks individually,
both decks have a probability that an ace is the top card of 4/52.
I ask you to take the top card of each deck , and discard the other cards of the decks, at this stage there is still a 4/52 chance that any of the two cards is an ace.
I have a quick look, and can confirm that one of the cards is an ace.
Please state the probability of the two cards now?
added -
100*4/52 = 7.69, which is roughly 8.
a
x
x
x
x
a
x
x
x
x
a
a
a
x
x
x
quantum leap , skip some space time, get lucky and land on another a.
it is very simple
pick a card anywhere from in a deck. 1/52
pick a card from another deck, 1/52
and continue this 100 times.
x
x
x
x
x
x
x
x
x
and so on to 100.
now what is the probability of drawing an ace from the new made y axis?
You are player 1, you will get card 1, pick a deck
1-52
1-52
1-52
1-52
and so 100 times in total.
You are not picking from 1-52, you are picking from 1-100 of unknown variant quantities.
Post by: Colin2B on 07/07/2015 00:07:46
You are not considering that the coin is already tossed 100 times.It doesn't matter. Previous tosses do not affect future tosses. This is a common error, the coin has no memory. Probability is 1/2 every time
You are not considering the already set unknown sequence of a deck of cards.Probability isn't about a specific set sequence, it is the likelihood of an event occurring or having occurred and is based on lots of trials not just the one or two you have carefully selected to make your point.
The top card is not random, it is a unknown value.A card can be both random and unknown, in fact it usually is both until looked at. So I don't understand what you mean by this.
Once the top card is set, it is set and nothing can change this.No one is saying you can change it, but probability only deals with what is likely or unlikely not with what is actually there.
You are not picking from 1-52, you are picking from 1-100 of unknown variant quantities.This is wrong as already explained.
I ask you to take the top card of each deck , and discard the other cards of the decks, at this stage there is still a 4/52 chance that any of the two cards is an ace.No the probability is not 4/52, ChiralSPO and I explained this in posts #54 & 55.
You clearly haven't read these posts. I'm not going to repeat them again, please read them.
I have a quick look, and can confirm that one of the cards is an ace.Again, this was explained fully in those 2 posts. This is condition probability which I have also explained in other posts. It is a mistake to assume P(AB)=P(B|A) as I previously explained.
Please state the probability of the two cards now
If you are not going to bother reading and understanding our posts there really isn't any point continuing. I have persisted this far because I am aware that in a few years your son will be starting secondary school. His first exposure to probability in maths will be through these very basic first year examples using coins and cards. If you teach him your pseudomaths he will fail his coursework and be considered a fool by his friends and teachers. This will be unfair. You owe it to him to learn real maths, not myth.
Go back, read what Alan, ChiralSPO and I have written. If you have specific questions on understanding what we have written I will try to answer, but I am unable to deal with your delusion regarding the 'timing' of decks and it's effect on outcomes.
Whatever it is you think you are doing it isn't probability.
Like Alan, I'm out.
Post by: guest39538 on 07/07/2015 15:50:19
And like every other forum on the internet, again your versions are being forced onto me. The things you defend are not even your things, they were some ones from history things.
I ask where is your own thought and consideration for the talking point? without resorting back to present information.
Does nobody on this planet have the ability to think about what is actually said?
I wish to discuss x is not equal to y, I have shown axiom logic and models and maths. All you guys seem to do is say no, and resort back to present information, this is not really discussing my point or even trying to agree with my point.
You instantly cast something out because you do not understand it. I know from your posts that you can not see it or visualise it what I am actually referring to.
Post by: Colin2B on 07/07/2015 17:53:28
I ask where is your own thought and consideration for the talking point?These are my own thoughts.
I have looked at probability, I have done the experiments and the maths and understood.
I wish to discuss x is not equal to y, I have shown axiom logic and models and maths.And we have shown by logic, models and maths that we disagree with you
All you guys seem to do is say no, and resort back to present information, this is not really discussing my point or even trying to agree with my point.If you tell me there are no fish in the sea, I have to disagree with you.
If you tell me there is no ocean between UK and US, I have to disagree with you.
I have to be true to myself, I cannot agree with false 'facts', false logic or false maths.
You instantly cast something out because you do not understand it. I know from your posts that you can not see it or visualise it what I am actually referring to.No, I can visualise what you are trying to say and I understand it, but I also understand why you are seeing it incorrectly.
Again I cannot agree with falsehoods and it would not be respectful to you if I did.
Post by: guest39538 on 07/07/2015 19:01:05
I ask where is your own thought and consideration for the talking point?These are my own thoughts.
I have looked at probability, I have done the experiments and the maths and understood.I wish to discuss x is not equal to y, I have shown axiom logic and models and maths.And we have shown by logic, models and maths that we disagree with youAll you guys seem to do is say no, and resort back to present information, this is not really discussing my point or even trying to agree with my point.If you tell me there are no fish in the sea, I have to disagree with you.
If you tell me there is no ocean between UK and US, I have to disagree with you.
I have to be true to myself, I cannot agree with false 'facts', false logic or false maths.You instantly cast something out because you do not understand it. I know from your posts that you can not see it or visualise it what I am actually referring to.No, I can visualise what you are trying to say and I understand it, but I also understand why you are seeing it incorrectly.
Again I cannot agree with falsehoods and it would not be respectful to you if I did.
I am not telling you that there is no fish in the sea, I am telling you that there is too may fish in the sea.
100 decks of randomly shuffled cards will contain approx 8 aces as the top card of the 100 decks.
The small blind is in alignment with card 1 of each deck.
''There are 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000 arrangements of a set of 52 cards. 7.69% of them have an ace at the top.''
So if I have 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000 decks of cards pre-shuffled, column 1 to the small blind contains 7.69% worth of aces. compared to 4/52
it is possible that all 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000 decks have an ace as the top card.
Post by: Colin2B on 08/07/2015 04:39:03
the small blind contains 7.69% worth of aces. compared to 4/52My dear Box, at last we are in agreement.
Yes 7.69% does indeed = 4/52
Now you can see that the probability in the y direction is the same as in the x direction.
P(y)=P(x)
y=x
Now at last you understand what we are saying and we can begin to explore the ways in which x and y are different.
it is possible that all 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000 decks have an ace as the top card.Yes, I pointed this out to you many posts ago. However, as I also pointed out it is extremely unlikely to happen in anybody's lifetime.
There has never been any question that there are different sequences in the x and y directions, the problem has been your misinterpretation of the probabilities and the effect they have on the games you play. You asked "is this maths correct" and we have explained that it is not correct.
Post by: guest39538 on 08/07/2015 16:31:24
the small blind contains 7.69% worth of aces. compared to 4/52My dear Box, at last we are in agreement.
Yes 7.69% does indeed = 4/52
Now you can see that the probability in the y direction is the same as in the x direction.
P(y)=P(x)
y=x
Now at last you understand what we are saying and we can begin to explore the ways in which x and y are different.it is possible that all 80,658,175,170,943,878,571,660,636,856,403,766,975,289,505,440,883,277,824,000,000,000,000 decks have an ace as the top card.Yes, I pointed this out to you many posts ago. However, as I also pointed out it is extremely unlikely to happen in anybody's lifetime.
There has never been any question that there are different sequences in the x and y directions, the problem has been your misinterpretation of the probabilities and the effect they have on the games you play. You asked "is this maths correct" and we have explained that it is not correct.
Thank you Colin, interesting that probabilities may not be the answer I am looking for, although alternative scenarios must apply?
I do not why you think I am agreeing with the maths and that Y=X. Not for one second do I think Y=X.
Forget probabilities for a while and please answer me this, in the diagram below , does Y=X?
.
.
.
. . . .
4²
Post by: alancalverd on 08/07/2015 17:25:51
I think you are confusing probability of singular events with that of combinational events. Don't feel bad about it, this is a very common error and leads to people believing that if they have just tossed 5 H in a row, the next is more likely to be a T.
Good point! In fact if you have just tossed 5H, it's quite likely that the coin is biassed so the next toss is more likely to be H than T!
Post by: guest39538 on 08/07/2015 17:31:13
I think you are confusing probability of singular events with that of combinational events. Don't feel bad about it, this is a very common error and leads to people believing that if they have just tossed 5 H in a row, the next is more likely to be a T.
Good point! In fact if you have just tossed 5H, it's quite likely that the coin is biassed so the next toss is more likely to be H than T!
^5 reverse bet not fallacy, I would bet you could not throw another heads, I am not betting 1/2 of heads or tails
and I forgot science likes simple
x
x
x
a
b
c
x=1 or 2
x~abc random over time.
what is the chance that (a) will receive a 1.
Post by: alancalverd on 09/07/2015 00:04:51
, I would bet you could not throw another heads,
Gotcha! You are ignoring the evidence in favour of your preconception. The perfect mug.
Post by: guest39538 on 11/07/2015 09:33:52
, I would bet you could not throw another heads,
Gotcha! You are ignoring the evidence in favour of your preconception. The perfect mug.
You have not got me, I do not ignore solid evidence, I know the coin still has 1/2 chance. And this has nothing to do with my poker ideas.
I understand randomness, I understand sequence, and I certainly understand x and y. It is not my misconception, it is the entire human races perspective view of just about everything, from atoms to probability.
The human race looks at things in a 1 dimensional perspective, where everything in my space is see through.
Post by: guest39538 on 11/07/2015 10:30:18
It has come to my attention by playing online poker, and by my observations and knowledge of the distribution process that there is a flaw in the process. The flaw is a probability function, that at the start of any hand, all players have uneven probabilities, a flaw caused by a multitude of pre-shuffled decks, that are randomly issued to game tables each and every hand.
A time based distribution, based on when a table finishes it's hand, it gets a new deck from the system that is pre-waiting in storage.
This makes an anomaly, where standard random sequences over time from using a single deck, is altered by periodic ''jumping'' of spacial dividers.
We can assert that of a multitude of individual pre-shuffled decks, that the top cards of each deck will all have a 4/52 chance of it being an ace, we can with a certainty be assured that a single deck only contains 4 aces. However we can also also assert that the top card of each deck could all be an ace, or all could be a different card to an ace.
If we consider a diagram form and a variation of top cards,
ace
queen
four
ace
the four cards representing 4 individual decks and the top card of each deck, if a table received deck 1, being at the top , and deck 4 being the at the bottom, then that table card 1 is both times an ace. This is the spacial dividers over time ''jumping'' I refer to.
ace,queen,four,ace
ace........time........ace
it is not hard to consider an oblong format of random sequences of rows,
Obviously x axis does not equal y axis,
123
231
123
X makes y, and Y is different to x, x remains 1/3 whilst Y remains unknown ,
I am not saying ''that the probability of getting an ace decreases as the number of decks increases''.
I am saying there is possibly more or less aces in the Y axis, which has been agreed with, and a player could ''quantum leap'' through space-time to receive another ace, and on the reverse a player could quantum leap through space-time and receive no aces ever.
If we call 1, (n), P(n)/x=1/3 but P(n)/y=?, because we could never possibly know what values had fell in place of the Y axis columns, example
123
123
123
In this situation column one on the left, the P(1)/y=3/3 where as x always remains 1/3.
And if we hide the values, the result could still be the same.
nnn
nnn
nnn
we know left to right of each row is still 1/3, but we do not know the columns values.
Post by: alancalverd on 11/07/2015 19:45:03
Quote
I am saying there is possibly more or less aces in the Y axis, which has been agreed with, and a player could ''quantum leap'' through space-time to receive another ace, and on the reverse a player could quantum leap through space-time and receive no aces ever.
This is entirely due to the randomness of the distribution of aces. It is known as the luck of the draw, or the consequence of shuffling, and is what turns poker from an algorithm into a game.
However your misappropriation of the term "quantum leap" is deplorable in a science forum. What you are saying is that if you sit at a random table, there is a 1/13 chance that the first card you receive will be an ace, and the probability of it happening twice in two successive independent shuffles is 1/169.
Post by: guest39538 on 11/07/2015 22:18:50
QuoteI am saying there is possibly more or less aces in the Y axis, which has been agreed with, and a player could ''quantum leap'' through space-time to receive another ace, and on the reverse a player could quantum leap through space-time and receive no aces ever.
This is entirely due to the randomness of the distribution of aces. It is known as the luck of the draw, or the consequence of shuffling, and is what turns poker from an algorithm into a game.
However your misappropriation of the term "quantum leap" is deplorable in a science forum. What you are saying is that if you sit at a random table, there is a 1/13 chance that the first card you receive will be an ace, and the probability of it happening twice in two successive independent shuffles is 1/169.
I do not see how advancing forward is deplorable in a science forum?
I have no idea where you have 1/169 from?
or 1/13?
I have one deck of cards 4/52 chance of an ace following my x axis, you you can make the x a Y if you like and the result is the same.
I have the y axis, at ?/? .
12
21
x=y
21
21
x≠y
only 2 values are needed to show this.
12
21
xy≠xy
Post by: alancalverd on 12/07/2015 00:46:52
If P = probability of something happening in one trial, the probability of it happening n times in succession in independent trials = P^n.
So probability of the first card being an ace in two (and only two) independent trials = (1/13)^2 = 1/169
A posteriori analysis of a small number of trials does not give you the probability of an event. Choosing or constructing two or three possible sequences tells you nothing about the likely outcome of any actual trial, let alone the probability of a given outcome in sequential trials.
For the very last time: if the shuffles are independent (a) the probability of an outcome is exactly the same in each shuffle, by definition of independent, and (b) the statistics of actual outcomes only tend towards the calculated probability for a very large number of trials, by definition of probability.
Belief in anything else is the first step on the road to gambling bankruptcy.
Post by: guest39538 on 12/07/2015 10:41:10
4/52 = 1/13
If P = probability of something happening in one trial, the probability of it happening n times in succession in independent trials = P^n.
So probability of the first card being an ace in two (and only two) independent trials = (1/13)^2 = 1/169
A posteriori analysis of a small number of trials does not give you the probability of an event. Choosing or constructing two or three possible sequences tells you nothing about the likely outcome of any actual trial, let alone the probability of a given outcome in sequential trials.
For the very last time: if the shuffles are independent (a) the probability of an outcome is exactly the same in each shuffle, by definition of independent, and (b) the statistics of actual outcomes only tend towards the calculated probability for a very large number of trials, by definition of probability.
Belief in anything else is the first step on the road to gambling bankruptcy.
I know Alan you still do not understand my argument and what I am arguing about. My argument is not about the probabilities of independent decks.
I know each deck has a 4/52 chance of the top card being , but I also know this only applies using an x-axis scenario.
To say Y is equal to X when talking probability is absurd when it is evidentially not, no matter how many digits is in use.
xxxxx
xxxxx
xxxxx
xxxxx
xxxxx
5², each row as the values of 1 to 5, these values have been displaced of any order by a random shuffle, when the shuffle stops, in alignment with a Y axis and of 5 columns, each column also has 5 values, can you tell me what values are in the columns?
I can tell you that in every row there is 1 to 5 in no order.
Post by: Colin2B on 12/07/2015 17:38:01
I do not why you think I am agreeing with the mathsYour interpretation of the probabilities in the x&y directions has always been a sticking point in discussing you theory. In particular you think that the probability of an ace when picking from the top cards of 100 decks is different to that of picking from a single deck of 52. Well, in one post you said:
You have got 100 decks of shuffled cards in front of you, and you are asked to draw the top card from one of the decks, what is the chance it is an ace?However, later when talking about 100 decks you say:
i bet you say 4/52 which is incorrect.
..........column 1 to the small blind contains 7.69% worth of aces. compared to 4/52So, because 7.69%=4/52=8/104=0.769 you are agreeing that probability in columns = probability in rows
I forgot science likes simpleNo, science is comfortable with complex. What it does like is clarity. If you ask vague, woolly, unclear or confusing questions you can expect to be asked "what do you mean by...", "define ....", "what are you really asking". You have to be clear, something you are usually not.
However, because you have been much clearer in post #117, I will make one last attempt to show why the probability in x & y are the same, but the sequences are not.
nnnWe have previously pointed out that the probability of an ace in the y direction is the same as in the x see posts.#54, 55 and 105. What differs, as we have explained, are the sequences.
nnn
nnn
we know left to right of each row is still 1/3, but we do not know the columns values.
However, these are not unknown because probability helps us to understand the unknown.
OK, let's take your example as it is fairly simple.
nnn
nnn
nnn
You have 3 decks y1, y2, y3. Rather than use numbers for the cards I am going to use letters because otherwise we get confused when talking about 1st cards, 2nd card etc. So the 3 cards are A, B, C = Ace, Bravo, Charlie
So in the x direction P(x1)=P(x2)=P(x3)=1/3 and we agree on that.
So if you shuffle a deck of these cards and pick the first there is a 1/3 chance it is A, but if you had picked the bottom card that would also be probability of 1/3, same with the middle card.
So in the x direction we are agreed 1/3 1/3 1/3.
So what about y.
If we shuffle your 3 decks and take the top 3 cards y1 y2 y3 then there are 3 possibilities for each y and so the number of possible ways these cards can fall = 3*3*3=27 possibilities.
Let's look at the probability that in these 3 cards the 1st card only is A.
We can show this with maths using an X to show “not an A” = a 2 or a 3
So, AXX
here the probability of A is 1/3, but there are 2 possible cards in y2, a B or a C so the probability is 2/3, same with y3. So total probability = 1/3*2/3*2/3= 4/27.
We can also do this by writing out the possibilities for this deal:
A in y1
C B B C
B C B C
A A A A
a total of 4 ways out of 27 =4/27
We can also look at the probability that one of the cards out of the 3 is an A regardless of position. To do this we need to look at Ain y2 and A in y3
A in y2
We can also do this for y2 = A
C B B C
A A A A
B C B C
Another 4 ways
A in y3
and for y3 = A
A A A A
C B B C
B C B C
yet another 4 ways
Total 12
So if we want to know the probability of a single A anywhere in the 3 cards it is = 12/27=4/9
We would also want to look at the other ways for completeness:
A B C
A B C
A B C
3 more ways
and the ones we missed:
A C B C
A C C B
B B B B
A A B C
B C A A
A A A A
B A B C
B A C B
C C C C
Total 27
In order to do this many ways we have had to lay out 3x27 decks = 81
so in all those top cards how many were As?
Count them A = 27
So probability of A occurring in y = 27/81 = 1/3
the same as in x.
You will also note that AAA occurs once so probability = 1/27
but so does CBC, and BCC and BBB. All have equal probability.
Only humans decide that A is special. That an ace wins over a 6 or a 5. In reality their probabilities are all the same!
Now, the chances are that you would not get this exact match in exactly 81 decks, but as Alan said if you do enough games (trials) say >1000, then the actuality will get close to these probabilities.
What this also tells us is that for any 3 decks all arrangements are equally likely, so it doesn't matter if you skip decks or do them in sequence, the probability is the same. Out of a 1000 decks you could choose 3 from anywhere and the probability would be the same for each. If you think otherwise, you do not understand what probability is telling us.
You can obviously do the calculation shown above for any number of decks and cards, but the numbers become large, so I'll leave that to you.
Thank you Colin, interesting that probabilities may not be the answer I am looking for,I would agree. Probability tells us that there is no difference between shuffled decks, so skipping some of the decks has no effect on the outcome of games from a probability point of view.
Yes, there are other scenarios. Your's sounds closer to predestination, or "that one was intended for me" type of luck or fortune. I know something about probability, permutation and sequences, but nothing about predestination, fortune or luck, so I can't help you with that.
Post by: alancalverd on 12/07/2015 18:41:23
, I would bet you could not throw another heads,
Gotcha! You are ignoring the evidence in favour of your preconception. The perfect mug.
You have not got me, I do not ignore solid evidence, I know the coin still has 1/2 chance.
But the evidence to date suggests that it doesn't! The probability of scoring 5H in a row in inependent trials of a fair coin is (1/2)^5, around 3%, so the evidence suggests a 97% probability that the coin is not fair. On that basis it would be foolish to bet against it.
Post by: guest39538 on 13/07/2015 14:52:42
Colin calls it pre destined, I call it probabilities over a time period.
This is my whole point and the only part you are not understanding, the part about interception.
y1,y2,y3,y4,y5 where y is deck
t1,t2,t3,t4,t5 where t is table
randomly distributing y to each table is not the same.
You say you know sequences Colin, then you understand spacing between values, spacing being equivalent to time.
ace.........end.......ace..............end.................ace..end. end being end of sequence.
Above I show 3 shuffles and 3 decks, Player 1 table one receives on average 1/221 aces/t
divided by time being a key element when considering probabilities and certainly poker.
Post by: Colin2B on 13/07/2015 17:39:12
Thank you Alan and Colin for your persistence. Your maths is accurate and true if y1,y2 and y3 were all coming to the same table.No, that's incorrect. As I explained in my post, deck skipping (spacing, interception or timing as you call it) does not affect the outcome.
If you have an infinite number of decks and you wish to choose 3 for each table it does not matter which order you distribute them in or by what timing. You could take the 5th, the 10th and the 81st for one table; the 93rd, 4th and 100006th for the 2nd table etc. All are as equal as if you had taken them in order 1 2 3 4 5 6 etc.
If you had understood my post you would realise that what determines the randomness is not the distribution in the y direction, but the random shuffle of the decks. This makes all decks equal from a probability point of view.
Any other belief is not probability. And as I have said before you cannot divide a probability by time.
So let's leave it that I believe in probability and maths, and you believe in something else.
Post by: guest39538 on 14/07/2015 16:54:23
Thank you Alan and Colin for your persistence. Your maths is accurate and true if y1,y2 and y3 were all coming to the same table.No, that's incorrect. As I explained in my post, deck skipping (spacing, interception or timing as you call it) does not affect the outcome.
If you have an infinite number of decks and you wish to choose 3 for each table it does not matter which order you distribute them in or by what timing. You could take the 5th, the 10th and the 81st for one table; the 93rd, 4th and 100006th for the 2nd table etc. All are as equal as if you had taken them in order 1 2 3 4 5 6 etc.
If you had understood my post you would realise that what determines the randomness is not the distribution in the y direction, but the random shuffle of the decks. This makes all decks equal from a probability point of view.
Any other belief is not probability. And as I have said before you cannot divide a probability by time.
So let's leave it that I believe in probability and maths, and you believe in something else.
Again from your wording I see my very own idea,
''You could take the 5th, the 10th and the 81st for one table; the 93rd, 4th and 100006th for the 2nd table etc. All are as equal as if you had taken them in order 1 2 3 4 5 6 etc.''
This is not true,they are not equal.
y1
y2
y1
y2
y1
y2
That is a completely different distribution pattern if you deck skip .
lets change the order,
y1
y1
y1
y2
y2
y2
lets change the angle for you
y1y1y1y2y2y2
t...................t
can you not see the difference to a unified distribution and skipping time distribution?
You receive the first deck out of a stack of 1,000,000 decks, you receive an ace, the next turn in this example 100,000 of the decks have an ace as the second card, what is the chance you will receive a deck that the ace is the second card?
Post by: Colin2B on 14/07/2015 18:56:28
Not from a probability point of view. They are all the same.
This is not true,they are not equal.
y1
y2
y1
y2
y1
y2
That is a completely different distribution pattern if you deck skip .
lets change the order,
y1
y1
y1
y2
y2
y2
lets change the angle for you
y1y1y1y2y2y2
t...................t
can you not see the difference to a unified distribution and skipping time distribution?
I was beginning to suspect this is where the core of the problem lies, but let me answer your other question first
You receive the first deck out of a stack of 1,000,000 decks, you receive an ace, the next turn in this example 100,000 of the decks have an ace as the second card, what is the chance you will receive a deck that the ace is the second card?Probability deals with the likelihood of events happening where we do not know the outcome, but it is based on the knowledge we have at the time.
Ok, to be sure I understand your scenario.
I have just received the 1st deck from the stack and received an ace. You are asking the probability that the 2nd card is an ace. I don't know that that is the way the cards have fallen, so probability says the chance of a 2nd ace is 3/51.
If all the decks have been correctly shuffled the chance is still 3/51.
If you tell me before the game that these decks have not been correctly shuffled, and you have prior knowledge of a different proportion existing, then that changes the probability.
However, I don't think that is what you mean. I think you imagine the decks being set out, with their shuffled order and by chance, unknown to all the players, 10% of the decks have aces 2nd card. This is a possible scenario, but if the players do not know this then it does not change the probability of an ace 2nd card.
I will repeat "Probability deals with the likelihood of events happening where we do not know the outcome, but it is based on the knowledge we have at the time"
As I have described in previous posts, we can see the likely outcomes as probability distributions. What you are talking about are actuality distributions - in other words, what really happens.
This is akin to the QM situation where we do not know the position of an electron, say, but we know the probability of it being in a certain area. It is only when it is observed or captured that its location is known.
In the same way, probability tells us the likely values and locations of cards, but it is only when they are played that we know their true location and value.
If you wish to deal with actuality distributions, you need to record them, probability can tell you whether that was a likely event and in some cases will be able to tell you how likely it is that the deck was fairly shuffled. Beyond that probability in card games does not deal with actuality.
It may be that the 1st 10 decks all have an ace on top, yes if you get the 11th deck then you have missed out. Probability still says all the decks are equal and before the game was played you all had a probability of 4/52 for an ace. After the game has been played, probability will tell you that 10 aces in a row was unlikely to happen by chance.
As I have said before you are not looking at probability, you are looking at something more akin to predestination which cannot be described by probability.
Post by: alancalverd on 14/07/2015 23:29:58
You receive the first deck out of a stack of 1,000,000 decks, you receive an ace, the next turn in this example 100,000 of the decks have an ace as the second card,
If you know this, the decks have probably not been shuffled fairly. Out of 1,000,000 decks the expected number with an ace as second card is 76,923, not 100,000. A 24% discrepancy over a million trials is good evidence of an irregularity.
Post by: Colin2B on 15/07/2015 13:39:10
It is also interesting to note that if TB understood probability he would also understand that even assuming we know the higher incidence of aces, deck skipping has no effect on the probability of receiving an ace.You receive the first deck out of a stack of 1,000,000 decks, you receive an ace, the next turn in this example 100,000 of the decks have an ace as the second card,
If you know this, the decks have probably not been shuffled fairly. Out of 1,000,000 decks the expected number with an ace as second card is 76,923, not 100,000. A 24% discrepancy over a million trials is good evidence of an irregularity.
Post by: guest39538 on 15/07/2015 19:03:14
It is also interesting to note that if TB understood probability he would also understand that even assuming we know the higher incidence of aces, deck skipping has no effect on the probability of receiving an ace.
Thank you Colin and Alan for continuing with this conversation and at the same time giving me more knowledge.
Colin, you are truly starting to understand me, and you are working out what I mean in my posts.
''I think you imagine the decks being set out, with their shuffled order and by chance, unknown to all the players, 10% of the decks have aces 2nd card. This is a possible scenario,''
The reason I mentioned the second card is that for each hand, a player receives a different card position of the sequence relative to their seat position relative to the small blind, the small blind always gets the first card dealt, then the big blind gets the second card, and so up to 9 players. Every hand the small blind moves around a seat clockwise.
''If you know this, the decks have probably not been shuffled fairly. Out of 1,000,000 decks the expected number with an ace as second card is 76,923, not 100,000. A 24% discrepancy over a million trials is good evidence of an irregularity.''
My example was just an example, 100,000 was just made up for my example. but now you have gave me the correct values, thanks.
Colin mentions what actually happens, yes of cause that is what I am talking about like always with all my science, I always talk about what actually happens or what we actually observe and what we actually do know.
I was never insane, I just always see reality and the truths.
7
6
9
2
3
y axis
4
x axis
would you agree that in this scenario that x≠y?
Post by: alancalverd on 15/07/2015 23:57:24
Post by: guest39538 on 16/07/2015 20:47:12
It doesn't matter where you sit or when the cards are dealt. The probability of receiving an ace in any chair at any nominated point in any deal is 1/13.
I asked a question, and that does not answer my question Alan. Can you please answer my question, in the scenario I provided I asked if x was not equal to y,
y axis
7
6
9
2
3
x axis
4
x≠y would this equation be true for my provided scenario,
p.s y could have 13/13
Post by: alancalverd on 16/07/2015 23:41:26
If you write down any finite set of integers, you can always find an integer that is not a member of that set. And you can also find a number that is. So what?
Post by: guest39538 on 17/07/2015 08:06:20
None of the numbers in the set you call y is 4. So what?
If you write down any finite set of integers, you can always find an integer that is not a member of that set. And you can also find a number that is. So what?
''Out of 1,000,000 decks the expected number with an ace as second card is 76,923''
7
6
9
2
3
4
You still avoided the actual question, can you not see the significant difference between x and y?
X is obviously not equal to y, I am obviously correct about online poker, I do not know what we are talking about, it may be probability or it may be what actual happens, but either way playing just x is not the same as playing y. This is apparent.
We can not say that y is equal to x in any sense, y will have a certainty of repeat values, i.e ace of diamonds lets say 100 times where x can only have one.
Post by: alancalverd on 17/07/2015 08:50:59
There is no certainty in a fair shuffle, but because y is infinite and x is finite, the actual distribution in y is more likely to be close to the calculated probability than the actual distribution in any one x.
This kind of second-order statistics comes under the heading of confidence limits:the greater the number of trials, the greater the confidence we can have that the actual and expected distributions will converge.
However if all the trials are independent random shuffles, the expected distributions must be identical.
Post by: guest39538 on 17/07/2015 09:06:23
You can't "play y". In poker you play one hand at a time. Obviously x is not equal to y because you are comparing apples and chickens.
There is no certainty in a fair shuffle, but because y is infinite and x is finite, the actual distribution in y is more likely to be close to the calculated probability than the actual distribution in any one x.
This kind of second-order statistics comes under the heading of confidence limits:the greater the number of trials, the greater the confidence we can have that the actual and expected distributions will converge.
However if all the trials are independent random shuffles, the expected distributions must be identical.
One hand at a time of the x axis and not of the y axis,
if you are picking a deck, you are playing the y axis, which is not equal to x.
123
123
123
123
123
we can clearly see this in simple diagram form, we firstly play the Y axis then play the x axis, that is why I called it cross odds or interception , the chance of receiving a deck that the second card is an ace using 1,000,000 decks, is obviously (76,923/1,000,000)/t= 0.076923/t and then once the choice is made of deck (4/52)/t=0.07692307692.This is all good as long as no other tables are involved.
(0.07692307692/t)(/n)/t where n is table.
added - sorry that does not work,
P(x)=0.07692307692∩n
..................t
Post by: Colin2B on 17/07/2015 12:21:24
If you are going to convince people you are right you need to think like a scientist and follow the scientific method. There are 2 things you need to do.
1. Get the maths right, otherwise people will just criticise your maths and not understand your ideas
2. You need to devise an experiment either using actual data from your games or via a large scale test so that your theory can be compared to what probability predicts.
First the maths:
the chance of receiving a deck that the second card is an ace using 1,000,000 decks, is obviously (76,923/1,000,000)/t= 0.076923/t and then once the choice is made of deck (4/52)/t=0.07692307692.This is all good as long as no other tables are involved.
(0.07692307692/t)(/n)/t where n is table.
added - sorry that does not work,
P(x)=0.07692307692∩n
..................t
As I've said before you can't divide a probability by time, What do you divide by? 10 mins, 15s?
In your system time defines the move to the next deck. So you need to find the average game length, max & min times and devise a formula the will advance the deck selection to the next deck. So your formula would have to have a result something like =yi+1
Also people will pick holes in P(x)=0.07692307692∩n. If n is table number it equals a set of integers, this will not intersect with a probability, so you need a different way of expressing what you want to say.
Experimental Verification
Ideally you would take all the data provided by the gaming company and analyse it to compare with the predictions of probability theory and look at the differences.
If you don't have access to that you would need to either keep records as you play or devise an experiment to prove your theory.
You ought to be able to set up an experiment either using small packs of cards eg deck of 4 cards containing a heart, diamond, spade and club.
Or you could set up a spreadsheet to create random decks of cards and compare sequential vs deck skipping. You would need to have it go through a large number eg 1000.
What do you think? You could offer genuine scientific proof of your theory.
Post by: guest39538 on 18/07/2015 08:06:57
What do you think? You could offer genuine scientific proof of your theory.
Thank you Colin, I also understand this is not standard and is an advancement on probabilities . It is new and therefore the maths will be new. I understand I must be explicit in explanation and maths, I do agree.
I know from observation I can observe the anomaly and witness the receiving of, example - 3 clubs 5/7 hands. This happens regular, repeat values, I even have had the exact same hand from previous of both cards.
I go on holiday in the next hour so will be unable to reply for a week.
please see this,
http://www.badscience.net/forum/viewtopic.php?f=3&t=36878&p=1391841#p1391841
#14844 page 594
a wave function difference of probability.
and I am not trying to explain divided by time, but over a period of time.
cross odds= P(n) from (x)≠P(n) from (y) in a period of time.
..yyy
x123x
x231x
x123x
..yyy
What else can we call this?
How else can we describe it?
And is it ''none standard deviance'' I am looking for?
added - in my link , x is constant over time where y is random over time.
Post by: guest39538 on 24/07/2015 18:51:32
Post by: alancalverd on 26/07/2015 12:49:22
It's pretty clear that you want to be dealt a winning hand.
It also seems that you have an idea that, having entered the game and played a hand, you can maximise your probability of being dealt a winning hand by not playing the next hand, but choosing another one.
So please complete the sentence "in order to maximise the probability of being dealt a winning hand on my second play, I should...."
Post by: guest39538 on 27/07/2015 20:30:04
OK, let's try this.
It's pretty clear that you want to be dealt a winning hand.
It also seems that you have an idea that, having entered the game and played a hand, you can maximise your probability of being dealt a winning hand by not playing the next hand, but choosing another one.
So please complete the sentence "in order to maximise the probability of being dealt a winning hand on my second play, I should...."
Hi Alan, completely different to what I have said. I do not want to be dealt a winning hand, I want to be dealt winning hands over x amount of time and not winning hands over y amount of time.
(note this time not axis's)
Spacing is important to the fundamental structure of Texas holdem poker. Spacing of hands over a period of time/game is what makes losers or winners.
Players get huge stacks in minutes, then wait and wait and wait and nothing else comes. Of cause the vice versus, a player can receive no hands early on then a flourish of good hands in the later stages.
This does sometimes happen live as well, but there is differences online that destroy probabilities over a period of time/game.
I could get some hand history and show you, only the other night I received the same card several times in only several hands, an often occurrence that happens.
(I can not repeat this with a real deck of cards)
How can I explain it simply, the order of distribution probabilities online is not the same as in a live game using one deck. The spacing is different over time , sometimes there is no spacing,
Post by: alancalverd on 27/07/2015 21:59:50
Post by: Colin2B on 27/07/2015 23:23:31
The reason is simple. Live games do not employ truly random shuffles.This has been studied by games researchers who discovered all sorts of no random patterns. To truly shuffle a pack requires a careful technique to ensure no bias.
Look back at my post where I talk about experimental verification. You need to have detailed records of your games and compare them to the expected results from probability theory.
Post by: guest39538 on 28/07/2015 06:23:26
The reason is simple. Live games do not employ truly random shuffles. Cards stick together to some extent, the dealer's fingers are not symmetrical, the cards are collected in groups....in fact in some games the cards are never shuffled but just cut so the hands continue to improve, whereas your beneficiaries have gone to great lengths to show how the online game is completely random.
and truly random has no probabilities because it is infinite unlike a selective group of variants which is not infinite and a block of variants with a start and an end. 0 on a roulette wheel as got to come in time the same as an ace from a deck as to come in time.
I think you still are missing the point, lets say in an average live game of 100 hands I receive a total of 5 aces in that period of time, the aces are spaced out by other hands. Online it is possible to receive 100 aces in 100 hands from using a Y axis.
Post by: Colin2B on 28/07/2015 10:09:38
truly random has no probabilitiesNo, there are still probabilities that can be calculated. Anyway, random what?
Online it is possible to receive 100 aces in 100 hands from using a Y axis.Possible is not the same as probable.
Post by: guest39538 on 28/07/2015 18:00:15
truly random has no probabilitiesNo, there are still probabilities that can be calculated. Anyway, random what?Online it is possible to receive 100 aces in 100 hands from using a Y axis.Possible is not the same as probable.
Likely and unlikely spring to mind, it is likely after receiving an ace in a live game, that the next hand you will be unlikely to get an ace although not impossible.
It is just as likely from an internet game that the next card you receive is an ace from a multitude of decks and more than a possible 1 of 4 aces, been in the first card position of multiple decks. i.e ace of diamonds 2000 times as the first card.
if there were 100000 decks and 2000 top cards were the ace of diamonds, I have a 2000/100000 chance of receiving a top card of the ace of diamonds, I do not believe that is 1/52
I believe this is basic maths and undeniably obvious.
x=1/52
y=2000/100000
they are obviously not equal.
Post by: alancalverd on 28/07/2015 20:16:37
Likely and unlikely spring to mind, it is likely after receiving an ace in a live game, that the next hand you will be unlikely to get an ace although not impossible.
1 card in 13 is an ace. If you are dealt a hand of 5 cards the probability of getting 1 ace is therefore 5/13 in every hand you are dealt, if the pack is fully shuffled between hands. However if the pack is merely collected and cut, because nobody will discard the aces, so the collected hands will have more adjacent aces and
the probability of getting one ace in the next hand is less than 5/13 but
the probability of getting two aces in the next hand is greater than (5/13)^2.
But the fact remains that if the cards are fairly shuffled between deals, every hand has the same likelihood of any chosen distribution.
Post by: Colin2B on 30/07/2015 11:32:10
x=1/52They are not equal because 2000/100000 is a number you have made up, it is not the true probability in the y direction. We have told you before what it is.
y=2000/100000
they are obviously not equal.
Anyway, this is not the crux of your problem, it is the issue of deck skipping and it's effect on probability.
Come on old friend, we can't keep going around in circles , you need to provide actual game logs to show whether there is a fault in the RNG, or distribution. The problem does n't lie with probability.
Post by: alancalverd on 30/07/2015 16:53:57
Quote
if there were 100000 decks and 2000 top cards were the ace of diamonds, I have a 2000/100000 chance of receiving a top card of the ace of diamonds, I do not believe that is 1/52
If there were 100,000 fairly shuffled decks, the expectation is that top card would be the A◊ in 100,000/52 = 1923 cases. 1800 to 2000 cases in an actual 100,000 is not outside the realms of reasonable probability but I would suspect foul play at 1400 or 2400.
Post by: guest39538 on 31/07/2015 22:11:49
x=1/52They are not equal because 2000/100000 is a number you have made up, it is not the true probability in the y direction. We have told you before what it is.
y=2000/100000
they are obviously not equal.
Anyway, this is not the crux of your problem, it is the issue of deck skipping and it's effect on probability.
Come on old friend, we can't keep going around in circles , you need to provide actual game logs to show whether there is a fault in the RNG, or distribution. The problem does n't lie with probability.
There is no probability to the Y axis, it is ''random'' , where x is not random, we know that x is always 1/52, we also know that shuffling of x makes y and Y can have an uncertainty measurement.
For example purposes we will use 1 , 2 and 3 in an x axis and use 3 sets of
123
123
123
x always equals 1/3 where y has more possible combinations than x, x can never have 111,
We can say the probabilities of Y have a variable range.
I will get some game logs, and show you the fault, and the rng is not at fault, the rng is fine.
Post by: Colin2B on 31/07/2015 22:57:25
There is no probability to the Y axis, it is ''random'' , where x is not random.There is a calculatable probability for the Y axis.
Both Y and X comprise discrete random variables, so both are random.
You are confusing random events and combinatorics.
You need some real logs
Post by: guest39538 on 31/07/2015 23:35:54
Dealt to holdemace486 [4s Jh]
holdemace486: folds
PedroDeucher: folds
Uncalled bet (10) returned to Evgeni104
Evgeni104 collected 20 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 20 | Rake 0
Seat 1: holdemace486 (button) folded before Flop (didn't bet)
Seat 2: PedroDeucher (small blind) folded before Flop
Seat 3: Evgeni104 (big blind) collected (20)
*** HOLE CARDS ***
Dealt to holdemace486 [9h 2d]
PedroDeucher: folds
Evgeni104: calls 10
holdemace486: checks
*** FLOP *** [As Qs Jc]
Evgeni104: bets 20
holdemace486: folds
Uncalled bet (20) returned to Evgeni104
Evgeni104 collected 40 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Board [As Qs Jc]
Seat 1: holdemace486 (big blind) folded on the Flop
Seat 2: PedroDeucher (button) folded before Flop (didn't bet)
Seat 3: Evgeni104 (small blind) collected (40)
*** HOLE CARDS ***
Dealt to holdemace486 [Qd 3d]
Evgeni104: folds
holdemace486: folds
Uncalled bet (10) returned to PedroDeucher
PedroDeucher collected 20 from pot
*** SUMMARY ***
Total pot 20 | Rake 0
Seat 1: holdemace486 (small blind) folded before Flop
Seat 2: PedroDeucher (big blind) collected (20)
Seat 3: Evgeni104 (button) folded before Flop (didn't bet)
*** HOLE CARDS ***
Dealt to holdemace486 [4c Js]
holdemace486: folds
PedroDeucher: folds
Uncalled bet (10) returned to Evgeni104
Evgeni104 collected 20 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 20 | Rake 0
Seat 1: holdemace486 (button) folded before Flop (didn't bet)
Seat 2: PedroDeucher (small blind) folded before Flop
Seat 3: Evgeni104 (big blind) collected (20)
*** HOLE CARDS ***
Dealt to holdemace486 [3h 8h]
PedroDeucher: folds
Evgeni104: calls 10
holdemace486: checks
*** FLOP *** [td 2d Kc]
Evgeni104: bets 20
holdemace486: folds
Uncalled bet (20) returned to Evgeni104
Evgeni104 collected 40 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Board [td 2d Kc]
Seat 1: holdemace486 (big blind) folded on the Flop
Seat 2: PedroDeucher (button) folded before Flop (didn't bet)
Seat 3: Evgeni104 (small blind) collected (40)
*** HOLE CARDS ***
Dealt to holdemace486 [Qc 5h]
Evgeni104: folds
holdemace486: folds
Uncalled bet (10) returned to PedroDeucher
PedroDeucher collected 20 from pot
*** SUMMARY ***
Total pot 20 | Rake 0
Seat 1: holdemace486 (small blind) folded before Flop
Seat 2: PedroDeucher (big blind) collected (20)
Seat 3: Evgeni104 (button) folded before Flop (didn't bet)
*** HOLE CARDS ***
Dealt to holdemace486 [7d Qh]
holdemace486: folds
PedroDeucher: folds
Uncalled bet (10) returned to Evgeni104
Evgeni104 collected 20 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 20 | Rake 0
Seat 1: holdemace486 (button) folded before Flop (didn't bet)
Seat 2: PedroDeucher (small blind) folded before Flop
Seat 3: Evgeni104 (big blind) collected (20)
*** HOLE CARDS ***
Dealt to holdemace486 [Ah 2s]
PedroDeucher: folds
Evgeni104: calls 10
holdemace486: raises 420 to 440 and is all-in
Evgeni104: folds
Uncalled bet (420) returned to holdemace486
holdemace486 collected 40 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (big blind) collected (40)
Seat 2: PedroDeucher (button) folded before Flop (didn't bet)
Seat 3: Evgeni104 (small blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [4c 9c]
Evgeni104: folds
holdemace486: calls 10
PedroDeucher: raises 40 to 60
holdemace486: calls 40
*** FLOP *** [7s Kd 4h]
holdemace486: bets 80
PedroDeucher: calls 80
*** TURN *** [7s Kd 4h] [6h]
holdemace486: checks
PedroDeucher: bets 60
holdemace486: calls 60
*** RIVER *** [7s Kd 4h 6h] [9d]
holdemace486: bets 260 and is all-in
PedroDeucher: folds
Uncalled bet (260) returned to holdemace486
holdemace486 collected 400 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 400 | Rake 0
Board [7s Kd 4h 6h 9d]
Seat 1: holdemace486 (small blind) collected (400)
Seat 2: PedroDeucher (big blind) folded on the River
Seat 3: Evgeni104 (button) folded before Flop (didn't bet)
*** HOLE CARDS ***
Dealt to holdemace486 [Qc Ts]
holdemace486: calls 30
PedroDeucher: raises 260 to 290 and is all-in
Evgeni104: folds
holdemace486: folds
Uncalled bet (260) returned to PedroDeucher
PedroDeucher collected 90 from pot
*** SUMMARY ***
Total pot 90 | Rake 0
Seat 1: holdemace486 (button) folded before Flop
Seat 2: PedroDeucher (small blind) collected (90)
Seat 3: Evgeni104 (big blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [Ad 8s]
PedroDeucher: folds
Evgeni104: calls 15
holdemace486: checks
*** FLOP *** [6c Jc Th]
Evgeni104: bets 30
holdemace486: folds
Uncalled bet (30) returned to Evgeni104
Evgeni104 collected 60 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 60 | Rake 0
Board [6c Jc Th]
Seat 1: holdemace486 (big blind) folded on the Flop
Seat 2: PedroDeucher (button) folded before Flop (didn't bet)
Seat 3: Evgeni104 (small blind) collected (60)
*** HOLE CARDS ***
Dealt to holdemace486 [7s 3d]
Evgeni104: folds
holdemace486: folds
Uncalled bet (15) returned to PedroDeucher
PedroDeucher collected 30 from pot
*** SUMMARY ***
Total pot 30 | Rake 0
Seat 1: holdemace486 (small blind) folded before Flop
Seat 2: PedroDeucher (big blind) collected (30)
Seat 3: Evgeni104 (button) folded before Flop (didn't bet)
*** HOLE CARDS ***
Dealt to holdemace486 [6d 2c]
holdemace486: folds
PedroDeucher: raises 335 to 365 and is all-in
Evgeni104: folds
Uncalled bet (335) returned to PedroDeucher
PedroDeucher collected 60 from pot
*** SUMMARY ***
Total pot 60 | Rake 0
Seat 1: holdemace486 (button) folded before Flop (didn't bet)
Seat 2: PedroDeucher (small blind) collected (60)
Seat 3: Evgeni104 (big blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [2h Tc]
PedroDeucher: raises 41 to 71
Evgeni104: folds
holdemace486: folds
Uncalled bet (41) returned to PedroDeucher
PedroDeucher collected 75 from pot
*** SUMMARY ***
Total pot 75 | Rake 0
Seat 1: holdemace486 (big blind) folded before Flop
Seat 2: PedroDeucher (button) collected (75)
Seat 3: Evgeni104 (small blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [8d 5s]
Evgeni104: folds
holdemace486: folds
Uncalled bet (15) returned to PedroDeucher
PedroDeucher collected 30 from pot
*** SUMMARY ***
Total pot 30 | Rake 0
Seat 1: holdemace486 (small blind) folded before Flop
Seat 2: PedroDeucher (big blind) collected (30)
Seat 3: Evgeni104 (button) folded before Flop (didn't bet)
*** HOLE CARDS ***
Dealt to holdemace486 [6c 4h]
holdemace486: folds
PedroDeucher: folds
Uncalled bet (15) returned to Evgeni104
Evgeni104 collected 30 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 30 | Rake 0
Seat 1: holdemace486 (button) folded before Flop (didn't bet)
Seat 2: PedroDeucher (small blind) folded before Flop
Seat 3: Evgeni104 (big blind) collected (30)
*** HOLE CARDS ***
Dealt to holdemace486 [Kd 2s]
PedroDeucher: folds
Evgeni104: calls 15
holdemace486: checks
*** FLOP *** [5d 3s 7h]
Evgeni104: bets 30
holdemace486: folds
Uncalled bet (30) returned to Evgeni104
Evgeni104 collected 60 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 60 | Rake 0
Board [5d 3s 7h]
Seat 1: holdemace486 (big blind) folded on the Flop
Seat 2: PedroDeucher (button) folded before Flop (didn't bet)
Seat 3: Evgeni104 (small blind) collected (60)
*** HOLE CARDS ***
Dealt to holdemace486 [3h Ac]
Evgeni104: calls 30
holdemace486: calls 15
PedroDeucher: checks
*** FLOP *** [3d 2h 4s]
holdemace486: checks
PedroDeucher: checks
Evgeni104: checks
*** TURN *** [3d 2h 4s] [4d]
holdemace486: bets 480 and is all-in
PedroDeucher: folds
Evgeni104: folds
Uncalled bet (480) returned to holdemace486
holdemace486 collected 90 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 90 | Rake 0
Board [3d 2h 4s 4d]
Seat 1: holdemace486 (small blind) collected (90)
Seat 2: PedroDeucher (big blind) folded on the Turn
Seat 3: Evgeni104 (button) folded on the Turn
*** HOLE CARDS ***
Dealt to holdemace486 [2c Ad]
holdemace486: raises 60 to 90
PedroDeucher: folds
Evgeni104: folds
Uncalled bet (60) returned to holdemace486
holdemace486 collected 75 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 75 | Rake 0
Seat 1: holdemace486 (button) collected (75)
Seat 2: PedroDeucher (small blind) folded before Flop
Seat 3: Evgeni104 (big blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [4c Js]
PedroDeucher: raises 60 to 90
Evgeni104: raises 400 to 490 and is all-in
holdemace486: folds
PedroDeucher: calls 305 and is all-in
Uncalled bet (95) returned to Evgeni104
*** FLOP *** [Jd Tc 7h]
*** TURN *** [Jd Tc 7h] [Jc]
*** RIVER *** [Jd Tc 7h Jc] [Ad]
*** SHOW DOWN ***
Evgeni104: shows [Th Ts] (a full house, Tens full of Jacks)
PedroDeucher: shows [9s 9d] (two pair, Jacks and Nines)
Evgeni104 collected 820 from pot
PedroDeucher finished the tournament in 3rd place
*** SUMMARY ***
Total pot 820 | Rake 0
Board [Jd Tc 7h Jc Ad]
Seat 1: holdemace486 (big blind) folded before Flop
Seat 2: PedroDeucher (button) showed [9s 9d] and lost with two pair, Jacks and Nines
Seat 3: Evgeni104 (small blind) showed [Th Ts] and won (820) with a full house, Tens full of Jacks
*** HOLE CARDS ***
Dealt to holdemace486 [9c Jd]
Evgeni104: calls 20
holdemace486: raises 545 to 585 and is all-in
Evgeni104: folds
Uncalled bet (545) returned to holdemace486
holdemace486 collected 80 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 80 | Rake 0
Seat 1: holdemace486 (big blind) collected (80)
Seat 3: Evgeni104 (button) (small blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [5s Ks]
holdemace486 has timed out
holdemace486: folds
Uncalled bet (20) returned to Evgeni104
holdemace486 is sitting out
Evgeni104 collected 40 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (button) (small blind) folded before Flop
Seat 3: Evgeni104 (big blind) collected (40)
*** HOLE CARDS ***
Dealt to holdemace486 [Qh 5d]
Evgeni104: calls 20
holdemace486: folds
Evgeni104 collected 80 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 80 | Rake 0
Seat 1: holdemace486 (big blind) folded before Flop
Seat 3: Evgeni104 (button) (small blind) collected (80)
*** HOLE CARDS ***
Dealt to holdemace486 [8s 7h]
holdemace486: folds
Uncalled bet (20) returned to Evgeni104
Evgeni104 collected 40 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (button) (small blind) folded before Flop
Seat 3: Evgeni104 (big blind) collected (40)
*** HOLE CARDS ***
Dealt to holdemace486 [8d 5d]
holdemace486 has returned
Evgeni104: calls 20
holdemace486: raises 505 to 545 and is all-in
Evgeni104: folds
Uncalled bet (505) returned to holdemace486
holdemace486 collected 80 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 80 | Rake 0
Seat 1: holdemace486 (big blind) collected (80)
Seat 3: Evgeni104 (button) (small blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [3s Qc]
holdemace486: folds
Uncalled bet (20) returned to Evgeni104
Evgeni104 collected 40 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (button) (small blind) folded before Flop
Seat 3: Evgeni104 (big blind) collected (40)
*** HOLE CARDS ***
Dealt to holdemace486 [2h Tc]
Evgeni104: folds
Uncalled bet (20) returned to holdemace486
holdemace486 collected 40 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (big blind) collected (40)
Seat 3: Evgeni104 (button) (small blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [4s Ah]
holdemace486: raises 545 to 585 and is all-in
Evgeni104: folds
Uncalled bet (545) returned to holdemace486
holdemace486 collected 80 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 80 | Rake 0
Seat 1: holdemace486 (button) (small blind) collected (80)
Seat 3: Evgeni104 (big blind) folded before Flop
holdemace486: posts big blind 40
*** HOLE CARDS ***
Dealt to holdemace486 [Qc Js]
Evgeni104: folds
Uncalled bet (20) returned to holdemace486
holdemace486 collected 40 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (big blind) collected (40)
Seat 3: Evgeni104 (button) (small blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [6c Qh]
holdemace486: folds
Uncalled bet (20) returned to Evgeni104
Evgeni104 collected 40 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (button) (small blind) folded before Flop
Seat 3: Evgeni104 (big blind) collected (40)
*** HOLE CARDS ***
Dealt to holdemace486 [8h 2s]
Evgeni104: folds
Uncalled bet (20) returned to holdemace486
holdemace486 collected 40 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (big blind) collected (40)
Seat 3: Evgeni104 (button) (small blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [6h Qc]
holdemace486: folds
Uncalled bet (20) returned to Evgeni104
Evgeni104 collected 40 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (button) (small blind) folded before Flop
Seat 3: Evgeni104 (big blind) collected (40)
*** HOLE CARDS ***
Dealt to holdemace486 [9h 2s]
Evgeni104: folds
Uncalled bet (20) returned to holdemace486
holdemace486 collected 40 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (big blind) collected (40)
Seat 3: Evgeni104 (button) (small blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [Qs 9d]
holdemace486: calls 20
Evgeni104: raises 815 to 855 and is all-in
holdemace486: folds
Uncalled bet (815) returned to Evgeni104
Evgeni104 collected 80 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 80 | Rake 0
Seat 1: holdemace486 (button) (small blind) folded before Flop
Seat 3: Evgeni104 (big blind) collected (80)
*** HOLE CARDS ***
Dealt to holdemace486 [Ks 6s]
Evgeni104: folds
Uncalled bet (20) returned to holdemace486
holdemace486 collected 40 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (big blind) collected (40)
Seat 3: Evgeni104 (button) (small blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [6s 5h]
holdemace486: folds
Uncalled bet (20) returned to Evgeni104
Evgeni104 collected 40 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (button) (small blind) folded before Flop
Seat 3: Evgeni104 (big blind) collected (40)
*** HOLE CARDS ***
Dealt to holdemace486 [3h Th]
Evgeni104: calls 20
holdemace486: checks
*** FLOP *** [7c 5s 4d]
holdemace486: checks
Evgeni104: checks
*** TURN *** [7c 5s 4d] [Qd]
holdemace486: checks
Evgeni104: bets 40
holdemace486: folds
Uncalled bet (40) returned to Evgeni104
Evgeni104 collected 80 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 80 | Rake 0
Board [7c 5s 4d Qd]
Seat 1: holdemace486 (big blind) folded on the Turn
Seat 3: Evgeni104 (button) (small blind) collected (80)
*** HOLE CARDS ***
Dealt to holdemace486 [5c 4c]
holdemace486: calls 20
Evgeni104: checks
*** FLOP *** [9d 6h Ts]
Evgeni104: bets 40
holdemace486: folds
Uncalled bet (40) returned to Evgeni104
Evgeni104 collected 80 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 80 | Rake 0
Board [9d 6h Ts]
Seat 1: holdemace486 (button) (small blind) folded on the Flop
Seat 3: Evgeni104 (big blind) collected (80)
*** HOLE CARDS ***
Dealt to holdemace486 [6s Ks]
Evgeni104: calls 20
holdemace486: checks
*** FLOP *** [Tc 8h 8c]
holdemace486: checks
Evgeni104: bets 40
holdemace486: folds
Uncalled bet (40) returned to Evgeni104
Evgeni104 collected 80 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 80 | Rake 0
Board [Tc 8h 8c]
Seat 1: holdemace486 (big blind) folded on the Flop
Seat 3: Evgeni104 (button) (small blind) collected (80)
*** HOLE CARDS ***
Dealt to holdemace486 [8s Qc]
holdemace486: folds
Uncalled bet (20) returned to Evgeni104
Evgeni104 collected 40 from pot
Evgeni104: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (button) (small blind) folded before Flop
Seat 3: Evgeni104 (big blind) collected (40)
*** HOLE CARDS ***
Dealt to holdemace486 [9c 2h]
Evgeni104: folds
Uncalled bet (20) returned to holdemace486
holdemace486 collected 40 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (big blind) collected (40)
Seat 3: Evgeni104 (button) (small blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [Qc As]
holdemace486: calls 20
Evgeni104: raises 975 to 1015 and is all-in
holdemace486: calls 445 and is all-in
Uncalled bet (530) returned to Evgeni104
*** FLOP *** [3s 2c Kh]
*** TURN *** [3s 2c Kh] [Kc]
*** RIVER *** [3s 2c Kh Kc]
| *** SHOW DOWN *** Evgeni104: shows [4c 4s] (two pair, Kings and Fours) holdemace486: shows [Qc As] (a pair of Kings) Evgeni104 collected 970 from pot holdemace486 finished the tournament in 2nd place Evgeni104 wins the tournament and receives $6.00 - congratulations! *** SUMMARY *** Total pot 970 | Rake 0 Board [3s 2c Kh Kc Td] Seat 1: holdemace486 (button) (small blind) showed [Qc As] and lost with a pair of Kings Seat 3: Evgeni104 (big blind) showed [4c 4s] and won (970) with two pair, Kings and Fours |
notice the high volume of queens and the repeat king spades 6 spades
I have a lot more of these all showing this type thing, try it with a real deck this does not happen, I have repeat variant every other game. Qc the main repeat in this game and consider how so few hands there is.
Post by: alancalverd on 01/08/2015 10:07:11
Post by: Colin2B on 01/08/2015 13:15:28
Two things:
You say the RNG is not at fault. What evidence are you quoting for this?
You need to tabulate all the results you have for the first 2 cards dealt.
List all the cards in a pack and at the side the number of times they occur as 1st and 2nd card dealt eg
Number. Number
Times as 1st. Times as 2nd
Card dealt Card dealt
ad
ah
ac
as
2d
2h
2c
Etc
Then we have a starting point
Post by: guest39538 on 01/08/2015 18:06:23
Dealt to holdemace486 [Ts 2d]
holdemace486: folds
VictoryDay45: raises 20 to 40
greenman370: calls 20
*** FLOP *** [6c 9c 6s]
VictoryDay45: checks
greenman370: checks
*** TURN *** [6c 9c 6s] [Jc]
VictoryDay45: checks
greenman370: checks
*** RIVER *** [6c 9c 6s Jc] [4c]
VictoryDay45: checks
greenman370: bets 40
VictoryDay45: calls 40
*** SHOW DOWN ***
greenman370: shows [4s Tc] (a flush, Jack high)
VictoryDay45: mucks hand
greenman370 collected 160 from pot
*** SUMMARY ***
Total pot 160 | Rake 0
Board [6c 9c 6s Jc 4c]
Seat 1: holdemace486 (button) folded before Flop (didn't bet)
Seat 2: VictoryDay45 (small blind) mucked [Ah 2c]
Seat 3: greenman370 (big blind) showed [4s Tc] and won (160) with a flush, Jack high
*** HOLE CARDS ***
Dealt to holdemace486 [3c 9s]
VictoryDay45: raises 20 to 40
greenman370: calls 30
holdemace486: folds
*** FLOP *** [Tc Ts 5d]
greenman370: checks
VictoryDay45: bets 40
greenman370: folds
Uncalled bet (40) returned to VictoryDay45
VictoryDay45 collected 100 from pot
VictoryDay45: doesn't show hand
*** SUMMARY ***
Total pot 100 | Rake 0
Board [Tc Ts 5d]
Seat 1: holdemace486 (big blind) folded before Flop
Seat 2: VictoryDay45 (button) collected (100)
Seat 3: greenman370 (small blind) folded on the Flop
*** HOLE CARDS ***
Dealt to holdemace486 [Kc Qh]
greenman370: folds
holdemace486: raises 40 to 60
VictoryDay45: folds
Uncalled bet (40) returned to holdemace486
holdemace486 collected 40 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (small blind) collected (40)
Seat 2: VictoryDay45 (big blind) folded before Flop
Seat 3: greenman370 (button) folded before Flop (didn't bet)
*** HOLE CARDS ***
Dealt to holdemace486 [Qs 6d]
holdemace486: folds
VictoryDay45: raises 40 to 60
greenman370: calls 40
*** FLOP *** [5d Jh 4c]
VictoryDay45: bets 60
greenman370: folds
Uncalled bet (60) returned to VictoryDay45
VictoryDay45 collected 120 from pot
VictoryDay45: doesn't show hand
*** SUMMARY ***
Total pot 120 | Rake 0
Board [5d Jh 4c]
Seat 1: holdemace486 (button) folded before Flop (didn't bet)
Seat 2: VictoryDay45 (small blind) collected (120)
Seat 3: greenman370 (big blind) folded on the Flop
*** HOLE CARDS ***
Dealt to holdemace486 [5s 7d]
VictoryDay45: folds
greenman370: raises 20 to 40
holdemace486: folds
Uncalled bet (20) returned to greenman370
greenman370 collected 40 from pot
greenman370: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (big blind) folded before Flop
Seat 2: VictoryDay45 (button) folded before Flop (didn't bet)
Seat 3: greenman370 (small blind) collected (40)
*** HOLE CARDS ***
Dealt to holdemace486 [3s 9d]
greenman370: folds
holdemace486: folds
Uncalled bet (10) returned to VictoryDay45
VictoryDay45 collected 20 from pot
VictoryDay45: doesn't show hand
*** SUMMARY ***
Total pot 20 | Rake 0
Seat 1: holdemace486 (small blind) folded before Flop
Seat 2: VictoryDay45 (big blind) collected (20)
Seat 3: greenman370 (button) folded before Flop (didn't bet)
*** HOLE CARDS ***
Dealt to holdemace486 [Ad 7h]
holdemace486: raises 40 to 60
VictoryDay45: folds
greenman370: folds
Uncalled bet (40) returned to holdemace486
holdemace486 collected 50 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 50 | Rake 0
Seat 1: holdemace486 (button) collected (50)
Seat 2: VictoryDay45 (small blind) folded before Flop
Seat 3: greenman370 (big blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [4c 9d]
VictoryDay45: folds
greenman370: calls 10
holdemace486: checks
*** FLOP *** [Tc 5h 8c]
greenman370: bets 20
holdemace486: folds
Uncalled bet (20) returned to greenman370
greenman370 collected 40 from pot
greenman370: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Board [Tc 5h 8c]
Seat 1: holdemace486 (big blind) folded on the Flop
Seat 2: VictoryDay45 (button) folded before Flop (didn't bet)
Seat 3: greenman370 (small blind) collected (40)
*** HOLE CARDS ***
Dealt to holdemace486 [Th 2d]
greenman370: folds
holdemace486: folds
Uncalled bet (10) returned to VictoryDay45
VictoryDay45 collected 20 from pot
VictoryDay45: doesn't show hand
*** SUMMARY ***
Total pot 20 | Rake 0
Seat 1: holdemace486 (small blind) folded before Flop
Seat 2: VictoryDay45 (big blind) collected (20)
Seat 3: greenman370 (button) folded before Flop (didn't bet)
*** HOLE CARDS ***
Dealt to holdemace486 [Jd 8c]
holdemace486: folds
VictoryDay45: raises 40 to 60
greenman370: calls 40
*** FLOP *** [7s 3d 7h]
VictoryDay45: bets 60
greenman370: calls 60
*** TURN *** [7s 3d 7h] [Ac]
VictoryDay45: checks
greenman370: bets 80
VictoryDay45: folds
Uncalled bet (80) returned to greenman370
greenman370 collected 240 from pot
greenman370: doesn't show hand
*** SUMMARY ***
Total pot 240 | Rake 0
Board [7s 3d 7h Ac]
Seat 1: holdemace486 (button) folded before Flop (didn't bet)
Seat 2: VictoryDay45 (small blind) folded on the Turn
Seat 3: greenman370 (big blind) collected (240)
*** HOLE CARDS ***
Dealt to holdemace486 [3c 9c]
VictoryDay45: calls 20
greenman370: calls 10
holdemace486: checks
*** FLOP *** [5h 2d 3s]
greenman370: bets 20
holdemace486: raises 40 to 60
VictoryDay45: raises 330 to 390 and is all-in
greenman370: folds
holdemace486: folds
Uncalled bet (330) returned to VictoryDay45
VictoryDay45 collected 200 from pot
VictoryDay45: doesn't show hand
*** SUMMARY ***
Total pot 200 | Rake 0
Board [5h 2d 3s]
Seat 1: holdemace486 (big blind) folded on the Flop
Seat 2: VictoryDay45 (button) collected (200)
Seat 3: greenman370 (small blind) folded on the Flop
*** HOLE CARDS ***
Dealt to holdemace486 [td Qc]
greenman370: raises 60 to 90
holdemace486: folds
VictoryDay45: folds
Uncalled bet (60) returned to greenman370
greenman370 collected 75 from pot
greenman370: doesn't show hand
*** SUMMARY ***
Total pot 75 | Rake 0
Seat 1: holdemace486 (small blind) folded before Flop
Seat 2: VictoryDay45 (big blind) folded before Flop
Seat 3: greenman370 (button) collected (75)
*** HOLE CARDS ***
Dealt to holdemace486 [Tc 9s]
holdemace486: folds
VictoryDay45: raises 60 to 90
greenman370: folds
Uncalled bet (60) returned to VictoryDay45
VictoryDay45 collected 60 from pot
VictoryDay45: doesn't show hand
*** SUMMARY ***
Total pot 60 | Rake 0
Seat 1: holdemace486 (button) folded before Flop (didn't bet)
Seat 2: VictoryDay45 (small blind) collected (60)
Seat 3: greenman370 (big blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [9s Jh]
VictoryDay45: calls 30
greenman370: calls 15
holdemace486: checks
*** FLOP *** [Qc 8h 2d]
greenman370: checks
holdemace486: checks
VictoryDay45: checks
*** TURN *** [Qc 8h 2d] [Ac]
greenman370: checks
holdemace486: checks
VictoryDay45: bets 45
greenman370: folds
holdemace486: folds
Uncalled bet (45) returned to VictoryDay45
VictoryDay45 collected 90 from pot
VictoryDay45: doesn't show hand
*** SUMMARY ***
Total pot 90 | Rake 0
Board [Qc 8h 2d Ac]
Seat 1: holdemace486 (big blind) folded on the Turn
Seat 2: VictoryDay45 (button) collected (90)
Seat 3: greenman370 (small blind) folded on the Turn
*** HOLE CARDS ***
Dealt to holdemace486 [8h Jd]
greenman370: raises 30 to 60
holdemace486: folds
VictoryDay45: folds
Uncalled bet (30) returned to greenman370
greenman370 collected 75 from pot
greenman370: doesn't show hand
*** SUMMARY ***
Total pot 75 | Rake 0
Seat 1: holdemace486 (small blind) folded before Flop
Seat 2: VictoryDay45 (big blind) folded before Flop
Seat 3: greenman370 (button) collected (75)
*** HOLE CARDS ***
Dealt to holdemace486 [8d Kc]
holdemace486: folds
VictoryDay45: raises 60 to 90
greenman370: folds
Uncalled bet (60) returned to VictoryDay45
VictoryDay45 collected 60 from pot
VictoryDay45: doesn't show hand
*** SUMMARY ***
Total pot 60 | Rake 0
Seat 1: holdemace486 (button) folded before Flop (didn't bet)
Seat 2: VictoryDay45 (small blind) collected (60)
Seat 3: greenman370 (big blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [2s 6d]
VictoryDay45: folds
greenman370: calls 15
holdemace486: checks
*** FLOP *** [5s 9s Jd]
greenman370: bets 30
holdemace486: folds
Uncalled bet (30) returned to greenman370
greenman370 collected 60 from pot
greenman370: doesn't show hand
*** SUMMARY ***
Total pot 60 | Rake 0
Board [5s 9s Jd]
Seat 1: holdemace486 (big blind) folded on the Flop
Seat 2: VictoryDay45 (button) folded before Flop (didn't bet)
Seat 3: greenman370 (small blind) collected (60)
*** HOLE CARDS ***
Dealt to holdemace486 [6d 9d]
greenman370: folds
holdemace486: folds
Uncalled bet (15) returned to VictoryDay45
VictoryDay45 collected 30 from pot
VictoryDay45: doesn't show hand
*** SUMMARY ***
Total pot 30 | Rake 0
Seat 1: holdemace486 (small blind) folded before Flop
Seat 2: VictoryDay45 (big blind) collected (30)
Seat 3: greenman370 (button) folded before Flop (didn't bet)
*** HOLE CARDS ***
Dealt to holdemace486 [5c 5h]
holdemace486: raises 60 to 90
VictoryDay45: folds
greenman370: folds
Uncalled bet (60) returned to holdemace486
holdemace486 collected 75 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 75 | Rake 0
Seat 1: holdemace486 (button) collected (75)
Seat 2: VictoryDay45 (small blind) folded before Flop
Seat 3: greenman370 (big blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [6h Js]
VictoryDay45: folds
greenman370: calls 15
holdemace486: checks
*** FLOP *** [Tc Ah 3d]
greenman370: bets 30
holdemace486: folds
Uncalled bet (30) returned to greenman370
greenman370 collected 60 from pot
greenman370: doesn't show hand
*** SUMMARY ***
Total pot 60 | Rake 0
Board [Tc Ah 3d]
Seat 1: holdemace486 (big blind) folded on the Flop
Seat 2: VictoryDay45 (button) folded before Flop (didn't bet)
Seat 3: greenman370 (small blind) collected (60)
*** HOLE CARDS ***
Dealt to holdemace486 [Tc 8h]
greenman370: raises 30 to 60
holdemace486: folds
VictoryDay45: calls 30
*** FLOP *** [3d 7d Ad]
VictoryDay45: checks
greenman370: bets 60
VictoryDay45: calls 60
*** TURN *** [3d 7d Ad] [Qh]
VictoryDay45: checks
greenman370: checks
*** RIVER *** [3d 7d Ad Qh] [Qc]
VictoryDay45: bets 120
greenman370: folds
Uncalled bet (120) returned to VictoryDay45
VictoryDay45 collected 255 from pot
VictoryDay45: doesn't show hand
*** SUMMARY ***
Total pot 255 | Rake 0
Board [3d 7d Ad Qh Qc]
Seat 1: holdemace486 (small blind) folded before Flop
Seat 2: VictoryDay45 (big blind) collected (255)
Seat 3: greenman370 (button) folded on the River
*** HOLE CARDS ***
Dealt to holdemace486 [Ts 4h]
holdemace486: folds
VictoryDay45: raises 60 to 90
greenman370: folds
Uncalled bet (60) returned to VictoryDay45
VictoryDay45 collected 60 from pot
VictoryDay45: doesn't show hand
*** SUMMARY ***
Total pot 60 | Rake 0
Seat 1: holdemace486 (button) folded before Flop (didn't bet)
Seat 2: VictoryDay45 (small blind) collected (60)
Seat 3: greenman370 (big blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [5c Qd]
VictoryDay45: calls 40
greenman370: calls 20
holdemace486: checks
*** FLOP *** [Jc 4c 9h]
greenman370: checks
holdemace486: checks
VictoryDay45: checks
*** TURN *** [Jc 4c 9h] [6s]
greenman370: bets 80
holdemace486: folds
VictoryDay45: calls 80
*** RIVER *** [Jc 4c 9h 6s] [8s]
greenman370: bets 120
VictoryDay45: folds
Uncalled bet (120) returned to greenman370
greenman370 collected 280 from pot
greenman370: doesn't show hand
*** SUMMARY ***
Total pot 280 | Rake 0
Board [Jc 4c 9h 6s 8s]
Seat 1: holdemace486 (big blind) folded on the Turn
Seat 2: VictoryDay45 (button) folded on the River
Seat 3: greenman370 (small blind) collected (280)
*** HOLE CARDS ***
Dealt to holdemace486 [7s 4d]
greenman370: raises 40 to 80
holdemace486: folds
VictoryDay45: calls 40
*** FLOP *** [6h Qc 2h]
VictoryDay45: checks
greenman370: bets 80
VictoryDay45: calls 80
*** TURN *** [6h Qc 2h] [As]
VictoryDay45: checks
greenman370: bets 80
VictoryDay45: raises 80 to 160
greenman370: folds
Uncalled bet (80) returned to VictoryDay45
VictoryDay45 collected 500 from pot
VictoryDay45: doesn't show hand
*** SUMMARY ***
Total pot 500 | Rake 0
Board [6h Qc 2h As]
Seat 1: holdemace486 (small blind) folded before Flop
Seat 2: VictoryDay45 (big blind) collected (500)
Seat 3: greenman370 (button) folded on the Turn
*** HOLE CARDS ***
Dealt to holdemace486 [Qc 7d]
holdemace486: folds
VictoryDay45: raises 855 to 895 and is all-in
greenman370: folds
Uncalled bet (855) returned to VictoryDay45
VictoryDay45 collected 80 from pot
VictoryDay45: doesn't show hand
*** SUMMARY ***
Total pot 80 | Rake 0
Seat 1: holdemace486 (button) folded before Flop (didn't bet)
Seat 2: VictoryDay45 (small blind) collected (80)
Seat 3: greenman370 (big blind) folded before Flop
*** HOLE CARDS ***
Dealt to holdemace486 [7h Kd]
VictoryDay45: folds
greenman370: calls 20
holdemace486: checks
*** FLOP *** [9c 8c Ts]
greenman370: bets 300 and is all-in
holdemace486: folds
Uncalled bet (300) returned to greenman370
greenman370 collected 80 from pot
greenman370: doesn't show hand
*** SUMMARY ***
Total pot 80 | Rake 0
Board [9c 8c Ts]
Seat 1: holdemace486 (big blind) folded on the Flop
Seat 2: VictoryDay45 (button) folded before Flop (didn't bet)
Seat 3: greenman370 (small blind) collected (80)
*** HOLE CARDS ***
Dealt to holdemace486 [2c 3s]
greenman370: folds
holdemace486: folds
Uncalled bet (20) returned to VictoryDay45
VictoryDay45 collected 40 from pot
VictoryDay45: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (small blind) folded before Flop
Seat 2: VictoryDay45 (big blind) collected (40)
Seat 3: greenman370 (button) folded before Flop (didn't bet)
*** HOLE CARDS ***
Dealt to holdemace486 [3c 9c]
holdemace486: folds
VictoryDay45: calls 20
greenman370: raises 80 to 120
VictoryDay45: folds
Uncalled bet (80) returned to greenman370
greenman370 collected 80 from pot
greenman370: doesn't show hand
*** SUMMARY ***
Total pot 80 | Rake 0
Seat 1: holdemace486 (button) folded before Flop (didn't bet)
Seat 2: VictoryDay45 (small blind) folded before Flop
Seat 3: greenman370 (big blind) collected (80)
*** HOLE CARDS ***
Dealt to holdemace486 [2s Kc]
VictoryDay45: calls 40
greenman370: calls 20
holdemace486: checks
*** FLOP *** [Ks 8c Js]
greenman370: checks
holdemace486: checks
VictoryDay45: checks
*** TURN *** [Ks 8c Js] [2c]
greenman370: checks
holdemace486: bets 40
VictoryDay45: folds
greenman370: folds
Uncalled bet (40) returned to holdemace486
holdemace486 collected 120 from pot
holdemace486: doesn't show hand
*** SUMMARY ***
Total pot 120 | Rake 0
Board [Ks 8c Js 2c]
Seat 1: holdemace486 (big blind) collected (120)
Seat 2: VictoryDay45 (button) folded on the Turn
Seat 3: greenman370 (small blind) folded on the Turn
*** HOLE CARDS ***
Dealt to holdemace486 [3s Kh]
greenman370: folds
holdemace486: folds
Uncalled bet (20) returned to VictoryDay45
VictoryDay45 collected 40 from pot
VictoryDay45: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (small blind) folded before Flop
Seat 2: VictoryDay45 (big blind) collected (40)
Seat 3: greenman370 (button) folded before Flop (didn't bet)
*** HOLE CARDS ***
Dealt to holdemace486 [Qh 7s]
holdemace486: folds
VictoryDay45: folds
Uncalled bet (20) returned to greenman370
greenman370 collected 40 from pot
greenman370: doesn't show hand
*** SUMMARY ***
Total pot 40 | Rake 0
Seat 1: holdemace486 (button) folded before Flop (didn't bet)
Seat 2: VictoryDay45 (small blind) folded before Flop
Seat 3: greenman370 (big blind) collected (40)
*** HOLE CARDS ***
Dealt to holdemace486 [Kc 9c]
VictoryDay45: calls 40
greenman370: raises 80 to 120
holdemace486: folds
VictoryDay45: calls 80
*** FLOP *** [9d 5s 3d]
greenman370: bets 280 and is all-in
VictoryDay45: folds
Uncalled bet (280) returned to greenman370
greenman370 collected 280 from pot
greenman370: doesn't show hand
*** SUMMARY ***
Total pot 280 | Rake 0
Board [9d 5s 3d]
Seat 1: holdemace486 (big blind) folded before Flop
Seat 2: VictoryDay45 (button) folded on the Flop
Seat 3: greenman370 (small blind) collected (280)
*** HOLE CARDS ***
Dealt to holdemace486 [7s Ac]
greenman370: folds
holdemace486: raises 125 to 185 and is all-in
VictoryDay45: calls 125
*** FLOP *** [4h 5h 2d]
*** TURN *** [4h 5h 2d] [2c]
*** RIVER *** [4h 5h 2d 2c] [Qs]
*** SHOW DOWN ***
holdemace486: shows [7s Ac] (a pair of Deuces)
VictoryDay45: shows [Th 9h] (a pair of Deuces - lower kicker)
holdemace486 collected 370 from pot
*** SUMMARY ***
Total pot 370 | Rake 0
Board [4h 5h 2d 2c Qs]
Seat 1: holdemace486 (small blind) showed [7s Ac] and won (370) with a pair of Deuces
Seat 2: VictoryDay45 (big blind) showed [Th 9h] and lost with a pair of Deuces
Seat 3: greenman370 (button) folded before Flop (didn't bet)
*** HOLE CARDS ***
Dealt to holdemace486 [Ad 6s]
holdemace486: raises 60 to 120
VictoryDay45: folds
greenman370: calls 60
*** FLOP *** [3s 6d 2h]
greenman370: bets 60
holdemace486: raises 190 to 250 and is all-in
greenman370: calls 190
*** TURN *** [3s 6d 2h] [As]
*** RIVER *** [3s 6d 2h As] [9h]
*** SHOW DOWN ***
greenman370: shows [4c 5d] (a straight, Deuce to Six)
holdemace486: shows [Ad 6s] (two pair, Aces and Sixes)
greenman370 collected 770 from pot
holdemace486 finished the tournament in 3rd place
*** SUMMARY ***
Total pot 770 | Rake 0
Board [3s 6d 2h As 9h]
Seat 1: holdemace486 (button) showed [Ad 6s] and lost with two pair, Aces and Sixes
Seat 2: VictoryDay45 (small blind) folded before Flop
Seat 3: greenman370 (big blind) showed [4c 5d] and won (770) with a straight, Deuce to Six
nines this time, and it is nothing to do with the random generator because that only does the shuffle, it is all about frequency and the spacing is just not right.
Post by: chiralSPO on 01/08/2015 19:14:58
Ac 1
Ad 2
Ah 2
As 1
2c 2
2d 1
2h 2
2s 2
3c 0
3d 2
3h 3
3s 1
4c 4
4d 0
4h 1
4s 2
5c 1
5d 2
5h 2
5s 2
6c 2
6d 1
6h 0
6s 3
7c 0
7d 1
7h 1
7s 1
8c 0
8d 2
8h 2
8s 3
9c 3
9d 1
9h 1
9s 0
Tc 2
Td 0
Th 1
Ts 1
Jc 0
Jd 1
Jh 1
Js 3
Qc 6
Qd 1
Qh 3
Qs 1
Kc 0
Kd 1
Kh 0
Ks 3
So how well does this match up to what we would expect from a completely random selection?
Frequency 0 1 2 3 4 5 6 7
observed: 10 19 14 7 1 0 1 0
expected: 10.0 18.2 14.3 7.4 2.9 0.9 0.2 0.002 (calculations shown below)
52*(50/52)42 = 10.0
52*42*((50/52)41)*(2/52) = 18.2
52*(42*41/2)*((50/52)^39)*(2/52)^2 = 14.3
52*(42*41*40/(3*2))*((50/52)^39)*(2/52)^3 = 7.4
52*(42*41*40*39/(4*3*2))*((50/52)^38)*(2/52)^4 = 2.9
52*(42*41*40*39*38/(5*4*3*2))*((50/52)^37)*(2/52)^5 = 0.9
52*(42*41*40*39*38*37/(6*5*4*3*2))*((50/52)^36)*(2/52)^6 = 0.2
52*(42*41*40*39*38*37*36/(7*6*5*4*3*2))*((50/52)^35)*(2/52)^7 = 0.002
This does not appear to be very different from a random distribution (based on this one simple test of a fairly small data set).
If you wanted to prove something about time dependence you would need to include timestamps in the data--but there is no point trying to show a pattern with respect to one variable, when the results are very well matched to a random distribution...
Post by: alancalverd on 02/08/2015 00:52:22
If Qc=6 is considered exceptional, then 4c = 4 is also exceptional. The problem is one of perception: minor but entirely normal deviations of obs/exp have an unreasonable significance, both psychologically and to the outcome of the game, if they involve aces or picture cards.
Feynman wrote a couple of good essays on the subject of significant coincidence.
Post by: guest39538 on 02/08/2015 09:41:00
The problem is clusters.
I will post some parts to show what I mean.
4hAs/6h6c/jh2h/kh6d/2hjh/qc2c/3d7h/kh3s/ks5s/td7s/ac7d/td4h/
I have done you the first few hands, below my game when I am playing it shows my cards I have received, it is much easier to see when there is colour there, notice in these first few hands the repeat values in a short space of time.
continued -
7dts/jc4h/4d5h/8d4h/2dac/asks/qdah/2has/5s7d/thts/9c6c/3c6c/
8s6c/qdah/3h3s/2h6h/kd8c/thkc/asks/7sqs/qsjd/jd9/3h2s/kd6d/
I would accuse a live dealer of not shuffling properly and cards being stuck together.
continued-
3h4h/jdjs/tsks/js2s/4d8h/5s2s/js5s/jc2c/8s2d/9htd/4h5s/9d/5d/
final part
5ckc/ks7h/6s9c/8d2s/7hqh/2d6c/kdqs/8cjd/6d2d/8sjs/jhkc/8c8h/6h7c/3c2d/3h8c/
4d6c/ks3s/ts9d/qdjs/ks2s/3d2d/tstc/3c4h/2s8c/3s6d/4d3c/5d9s/9ctd/2s6d/8hth/3d5d/3htc/3h9d/4das/5skh/ts5h/9hts/td6d/8sjd/6cqd/ksjd/8c4s/5hkc/jd3s/6d6h/ts2c/jdts/kdkc/5c5h/9d6h/jh6d/4s3c/7hjh/7hqc/8ckh/tdth/2d5c/9has/
Post by: guest39538 on 02/08/2015 10:53:33
6c
kd
7h
6c
5h
as
9d
6c
By deck skipping , it allowed to me receive deck 1, 4, 8 giving me 3 of the same values in a row.
It causes clusters of values by skipping the designated path over time.
Following a standard course over time without skipping deviation I would of received 6c,kd,7h,6c,5h,as,9d,6c over time which spaces out 6c over time in this example.
Please feel free to experiment using a single deck the patterns and spaces are different compared to multiple decks and choice.
I have just got out my deck of cards I will deal to the rules and record my cards I get with a live deck here and now.
hand 1 - jc8d
hand 2 - ts2d
hand 3 - thqs
hand 4 - 4s7s
hand 5 - tc6s
hand 6- asjc
hand 7- adts
hand 8- qh4h
hand 9- kc3h
hand10- 4s6d
we can see within ten hands a difference of pattern and a better variation of hands.
The next ten hands,
h1- 3h7d
h2- js8d
h3-6d2d
h4-4s7d
h5-jhjs
h6-3hah
h7-jh9d
h8- 5c8s
h9-5h8d
ht-kh3d
again much more even distribution over time.
n) over a period of time of X equals (a)
(n) over a period of time of Y equals (b)
t=............................................
x=.........n...................................n
y=...nn........n....n...........n.......n......
relative to (n) the probability spacing over time Y≠X
x=(a)
Y=(b)
(a)/t≠(b)/t
distance
Post by: Colin2B on 02/08/2015 16:49:21
First, is the distribution random. ChiralSPO has kindly done the donkey work for us and it is a good match.
Secondly, if it is random then there is the same probability that the same cards will occur side by side as they will be fully separated or indeed in any position.
As Alan says, confirmation bias and the brain's pattern seeking mechanism means we notice certain items and events more than others.
Every time you claim to be disadvantaged by the deck skipping, someone else will think you have been advantaged compared to themselves.
If you feel the need to try posting all your data, how about attaching a file? PDF?
Post by: alancalverd on 03/08/2015 22:51:22
Quote
By deck skipping , it allowed to me receive deck 1, 4, 8 giving me 3 of the same values in a row.
A priori and a posteriori are very different conditions. If you were skipping a priori, what made you choose 1, 4 and 8?
Post by: guest39538 on 04/08/2015 10:25:20
QuoteBy deck skipping , it allowed to me receive deck 1, 4, 8 giving me 3 of the same values in a row.
A priori and a posteriori are very different conditions. If you were skipping a priori, what made you choose 1, 4 and 8?
It would actually unknown which deck you receive, the distribution of the decks is defined by when a table finishes its hand it gets new deck from the system queue, my example was to show what happens. My observation shows this, with a single deck the distribution spacing of frequency rate of repeat occurrence is different to that of online as seen in my few examples, all my examples show the same thing, the last 2 examples I posted were fresh examples , I played them to get the examples, every time it is noticeable by observation.
I do not know what to even call it, by I am certain you understand exactly what I am saying.
I have found a way to combat the problem sort of, by using less time games , but this does not fix the problem .
I am unsure how to upload a pdf file to show that every game is the same by observation you can see this to be true.
Post by: guest39538 on 08/08/2015 09:06:26
can anyone please explain this, I can not understand it.
Post by: Colin2B on 08/08/2015 14:40:48
Else where has directed me to this-http://bactra.org/notebooks/ergodic-theory.htmlYou would need to learn a lot more maths before you could understand this.
can anyone please explain this, I can not understand it.
Post by: guest39538 on 09/08/2015 13:24:20
You would need to learn a lot more maths before you could understand this.
I am not sure that maths can ever explain moving values forward in time.
Post by: guest39538 on 09/08/2015 13:31:14
or can it?
You would need to learn a lot more maths before you could understand this.
I am not sure that maths can ever explain moving values forward in time.
P(n)/y=(x)∩(n)/t2
t1
this
in this grid the rows are randomly shuffled,
nnnnn
nnnnn
nnnnn
nnnnn
nnnnn
i can tell you that each row has a 1/5 chance, of any value being in the first column.
n
n
n
n
n
however we know that all the first column, could have the same value,
1
1
1
1
1
so if several rows of values were set, and you were designated a column, you are aligned an unknown chance of the values aligned.
what happens if we add a random distribution of your alignment to other outputs?
a
b
c
d
e
if we receive in order, edcba, that is a continuation of time, where as if we received e and then b, we would be bringing b, forward in time, to a specific point of time.
so what is the chances from the columns of the values?
P(n)/y=(x)∩(n)
y=decks
x=players
n=variant
P=probability
∩=intercept