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

....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 5deck one deck 2 deck 3 deck 4 deck 5........................deck 6......................table 3 finishes their hand first so get the next deck.

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.

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.

Quote from: Thebox on 29/06/2015 18:18:46So 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. Unfortunately you don't understand as much as you think you do.This is conditional probability. You don't play with 2 decks, so you choose one and the probability for that game is 4/52

of cause you can not see the values in reality, player 1 or player 2 has the better chance of picking an (n)?

Quote from: Thebox on 30/06/2015 16:42:17of cause you can not see the values in reality, player 1 or player 2 has the better chance of picking an (n)?Yes, the way you have 'rigged' it, player 2 has a better chance.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 etcHave fun with your new theory

What is the chance of x axis shuffling randomly its 52 variants of each row and aligning 1-52 to each players column?

Quote from: Thebox on 30/06/2015 17:59:26What 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!=10^{68} 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.

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?

player 1 . please pick from the x axis,player 2, please pick from the y axis,

Quote from: Thebox on 01/07/2015 05:55:29on 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.Quote from: Thebox on 30/06/2015 16:42:17player 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 da 132b 123c 132d 321Each player will receive 2 cardsPlayer 1 is going to use the x direction and is given deck a and draws 13Player 2 is going to use the y direction chooses 2 decks b&c and hence draws 11This 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 distributionEDIT:Just to make sure I understand what you mean by using the y axis.If we have 4 decks of cards:y4 d d xy3 d d xy2 d d xy1 d d x x1 x2 x3 x4 x5 x6 etc ............ x52If 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.

1. A bad workman blames his tools.timing is not a tool2. In poker, you are playing against a whole lot of other people.yes3. In online poker you don't know who they are.yes4. 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 online7. Therefore anyone who is less than very good is likely to lose money.not trueSee? No complicated maths required!untrue

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.

Now we will add some players for our game, 3 players on the Y axis and 3 players along the x axisyp3 {1. 2. 3}p2 {1 .2. 3}p1 {1. 2. 3}....p1 p2 p3p1y 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.

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.

Quote from: alancalverd on 01/07/2015 18:23:421. A bad workman blames his tools.timing is not a tool2. In poker, you are playing against a whole lot of other people.yes3. In online poker you don't know who they are.yes4. 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 online7. Therefore anyone who is less than very good is likely to lose money.not trueSee? No complicated maths required!untrue

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 explainQuote from: Thebox on 01/07/2015 16:19:32Now we will add some players for our game, 3 players on the Y axis and 3 players along the x axisyp3 {1. 2. 3}p2 {1 .2. 3}p1 {1. 2. 3}....p1 p2 p3p1y 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.In 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.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).Quote from: Thebox on 01/07/2015 16:19:32So 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.No, your scenario doesn't show this, it shows only a matrix which the players cannot access for a single deck.Quote from: Thebox on 01/07/2015 16:19:32My 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 Quote from: Thebox on 01/07/2015 16:19:32would 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.

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

Quote from: Thebox on 02/07/2015 05:47:53pocket 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/tAs 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?4/52Q2 shuffle the deck but do not look at the top card, what is probability of an ace?4/52Q3 still without looking, waste this pack, shuffle and turn over top card, probability of an ace?4/52Q3 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.

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.