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Author Topic: Is this maths correct?  (Read 13401 times)

Offline Colin2B

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Re: Is this maths correct?
« Reply #25 on: 24/06/2015 11:59:26 »
You are confusing yourself and everyone else with you misunderstanding of probability. As I have said before, probability is a minefield for the unwary and clear definition of the scenario is essential.

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.


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.
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.

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?
No, because the phrase "By random timing this could be intercepted of the distribution" is incomprehensible.
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.
« Last Edit: 24/06/2015 12:02:45 by Colin2B »
 

Offline Thebox

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Re: Is this maths correct?
« Reply #26 on: 24/06/2015 19:10:34 »
Ok, I understand now you are not understanding me.  Starting from fresh is always a good place to start, so I will explain a scenario.

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, 5 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.







« Last Edit: 24/06/2015 19:51:03 by Thebox »
 

Offline alancalverd

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Re: Is this maths correct?
« Reply #27 on: 25/06/2015 00:39:56 »
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.
 

Offline Thebox

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Re: Is this maths correct?
« Reply #28 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.








« Last Edit: 25/06/2015 19:00:02 by Thebox »
 

Offline Colin2B

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Re: Is this maths correct?
« Reply #29 on: 25/06/2015 21:02:34 »
Can you confer my scenario is the correct logic and maths?
........
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
I don't know about the logic because you haven't fully described the scenario.
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?
 

Offline Thebox

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Re: Is this maths correct?
« Reply #30 on: 26/06/2015 05:28:38 »
Can you confer my scenario is the correct logic and maths?
........
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
I don't know about the logic because you haven't fully described the scenario.
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?

« Last Edit: 26/06/2015 05:53:33 by Thebox »
 

Offline alancalverd

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Re: Is this maths correct?
« Reply #31 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.
 

Offline Colin2B

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Re: Is this maths correct?
« Reply #32 on: 26/06/2015 15:37:47 »

akj
akj
jka


play the x axis has 3 decks, or play the Y axis as 3 decks, see the difference?
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.
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!
 

Offline Thebox

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Re: Is this maths correct?
« Reply #33 on: 26/06/2015 16:22:36 »
hi Colin, you do understand.  Scenario

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
« Last Edit: 26/06/2015 16:47:16 by Thebox »
 

Offline Colin2B

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Re: Is this maths correct?
« Reply #34 on: 26/06/2015 17:32:47 »
the first card, the odds of an ace are 4/52.

 the second card.  Again 4/51odds of receiving a ace.

 the 3 rd card my odds are also 4/50
These probabilities are only valid if the first 2 players do not receive an ace. Ok with that?
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 shuffled
So 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 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
What 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.
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.
 

Offline Thebox

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Re: Is this maths correct?
« Reply #35 on: 26/06/2015 18:10:08 »
Hi Colin you are getting colder and away from the thinking, the new deck your table receives is distributed the same way except the small blind moves position clockwise, so card one moves a seat, it is diagonal in card order meaning player one receives card one then card two, the card three, as their first card from the deck,
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.
« Last Edit: 26/06/2015 18:41:07 by Thebox »
 

Offline Colin2B

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Re: Is this maths correct?
« Reply #36 on: 26/06/2015 22:39:00 »
Hi Colin you are getting colder and away from the thinking
Oh 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.

 

Offline Thebox

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Re: Is this maths correct?
« Reply #37 on: 27/06/2015 06:46:33 »
Hi Colin you are getting colder and away from the thinking
Oh 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.
That would not be the same game as I am talking about.

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,
« Last Edit: 27/06/2015 07:58:19 by Thebox »
 

Offline Colin2B

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Re: Is this maths correct?
« Reply #38 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 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,
There you have it, we both have the same chance of winning for any particular game.

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?

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
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.

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 , ...................
.............., x is short term and y is long term I do not have two lifetimes,
I don't know what you think science is, but this is not it.
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.
 

Offline Thebox

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Re: Is this maths correct?
« Reply #39 on: 27/06/2015 10:06:22 »
I am not in any pitfall colin, and cross odds is a new concept.

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}~
t

« Last Edit: 27/06/2015 10:15:37 by Thebox »
 

Offline Colin2B

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Re: Is this maths correct?
« Reply #40 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. 

it would be  would it not?
a={x+y}=
{4/52}~
t
No it wouldn't. Mathematical gibberish.

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 understand
if I rolled a number 1 with  a dice, the next go my odds are 1/6^2 to roll another number 1.
This is wrong, if you have already thrown a 1, the odds of your second throw being a 1 is 1/6.
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.


 

Offline alancalverd

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Re: Is this maths correct?
« Reply #41 on: 27/06/2015 16:40:11 »
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.
 

Offline Colin2B

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Re: Is this maths correct?
« Reply #42 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.
 

Offline alancalverd

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Re: Is this maths correct?
« Reply #43 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.
 

Offline Thebox

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Re: Is this maths correct?
« Reply #44 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.

« Last Edit: 28/06/2015 07:02:06 by Thebox »
 

Offline jccc

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Re: Is this maths correct?
« Reply #45 on: 28/06/2015 07:14:19 »
the wind took the note.

gone with the wind.

don't play online poker.
 

Offline Thebox

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Re: Is this maths correct?
« Reply #46 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.

« Last Edit: 28/06/2015 07:29:15 by Thebox »
 

Offline alancalverd

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Re: Is this maths correct?
« Reply #47 on: 28/06/2015 09:16:55 »
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.   
 

Offline Thebox

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Re: Is this maths correct?
« Reply #48 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.
















 

Offline Colin2B

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Re: Is this maths correct?
« Reply #49 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!
 

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Re: Is this maths correct?
« Reply #49 on: 28/06/2015 10:58:58 »

 

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