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Author Topic: Is this maths correct?  (Read 13356 times)

Colin2B

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Re: Is this maths correct?
« Reply #125 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.

Thebox

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Re: Is this maths correct?
« Reply #126 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?

« Last Edit: 14/07/2015 17:00:10 by Thebox »

Colin2B

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Re: Is this maths correct?
« Reply #127 on: 14/07/2015 18:56:28 »

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?
Not from a probability point of view. They are all the same.
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.

alancalverd

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Re: Is this maths correct?
« Reply #128 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.

Colin2B

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Re: Is this maths correct?
« Reply #129 on: 15/07/2015 13:39:10 »
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.
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.

Thebox

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Re: Is this maths correct?
« Reply #130 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?

alancalverd

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Re: Is this maths correct?
« Reply #131 on: 15/07/2015 23:57:24 »
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.

Thebox

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Re: Is this maths correct?
« Reply #132 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

« Last Edit: 16/07/2015 21:28:59 by Thebox »

alancalverd

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Re: Is this maths correct?
« Reply #133 on: 16/07/2015 23:41:26 »
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?

Thebox

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Re: Is this maths correct?
« Reply #134 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.

alancalverd

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Re: Is this maths correct?
« Reply #135 on: 17/07/2015 08:50:59 »
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.

Thebox

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Re: Is this maths correct?
« Reply #136 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

« Last Edit: 17/07/2015 09:19:50 by Thebox »

Colin2B

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Re: Is this maths correct?
« Reply #137 on: 17/07/2015 12:21:24 »
OK, Alan & I understand what you are trying to say, but as you say this is not standard probability but what you call cross-odds or interception.

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.

Thebox

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Re: Is this maths correct?
« Reply #138 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.

#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.

« Last Edit: 18/07/2015 08:14:27 by Thebox »

Thebox

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Re: Is this maths correct?
« Reply #139 on: 24/07/2015 18:51:32 »
Hello , I am back, no comments for the last post?

alancalverd

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Re: Is this maths correct?
« Reply #140 on: 26/07/2015 12:49:22 »
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...."

Thebox

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Re: Is this maths correct?
« Reply #141 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,

alancalverd

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Re: Is this maths correct?
« Reply #142 on: 27/07/2015 21:59:50 »
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.

Colin2B

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Re: Is this maths correct?
« Reply #143 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.

Thebox

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Re: Is this maths correct?
« Reply #144 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.

Colin2B

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Re: Is this maths correct?
« Reply #145 on: 28/07/2015 10:09:38 »
truly random has no probabilities
No, 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.

Thebox

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Re: Is this maths correct?
« Reply #146 on: 28/07/2015 18:00:15 »
truly random has no probabilities
No, 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.

alancalverd

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Re: Is this maths correct?
« Reply #147 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.

Colin2B

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Re: Is this maths correct?
« Reply #148 on: 30/07/2015 11:32:10 »
x=1/52
y=2000/100000

they are obviously not equal.
They 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.
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.

alancalverd

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Re: Is this maths correct?
« Reply #149 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.

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Re: Is this maths correct?
« Reply #149 on: 30/07/2015 16:53:57 »