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I watch with some mild interest on TV the poker games when I have little better to do at night. And I notice these guys playing Texas Holdem Poker.

My question is, the TV has beside each players name a set of cards, beside this they have percentages, these percentages change as the game evolves. How are these percentages worked out.

If I choose to play online. Given I cant see other peoples cards, and without hacking or cheating in other ways. How close can I get to these percentages if I had software to read only my cards and the up cards on the table as the game progresses. Remember, no cheating or hacking.

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Huh, you could've just undeleted your last question... in any event, I was merely suggesting that it might be a better fit for here; moderators would know the best option. –  J. M. Aug 13 '11 at 18:46
    
Apologies J.M. I meant no offese to you personally. A while back I had a question on SO which was clearly a programming question and for no reason, or without warning it was moved to some database forum. I had no account with them, and couldnt even edit it. When I looked there are over 30 different SO forums, I contacted customer service, no response. I couldnt delete it, cause I was not a member, and couldnt even add comments to thank people for their responses. I was furious. This is a math question, not a card game question, so why should I move it to card games. Again apologies to you. –  Francis Rodgers Aug 13 '11 at 20:35
    
It's worth mentioning that there exists software to do calculations that are in the same spirit as the ones you want (StoxEV, for example), but typically the problems are complicated enough that the software is not used in real time. You have to do things like set the range of possible hands of your opponent, weighted according to how likely they are; to analyze a hand in full, you might have to do this multiple times. Accurately determining the range of your opponent is, of course, incredibly difficult to begin with. –  guy Aug 13 '11 at 21:19
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2 Answers

up vote 4 down vote accepted

The percent showing up in the corner of the screen and the percent in a player's head are actually answers to subtly different questions.

The television percent is answering the question "Given perfect information about all players' cards, what is the likelihood that a particular player will win?" This can be computed by considering all possible board (the face-up cards) outcomes. If 25% of these outcomes give you a winning hand, then you have a 25% chance of winning. There are too many possibilities for a human to consider, but a computer will handle it easily.

The player is answering the question, "Given that I don't know my opponent's hand, what is the likelihood that mine will prevail?" Here, we might try to proceed by considering all possible hands for the opponent and all possible board cards - still reasonable for a computer.

To appreciate the difference between these two questions, consider the following example. You have Kh Qh, your opponent has 4d 5d, and the board is 2h 6h 9h Jh As (so it is $K$-high flush versus high card). The television reports 100% in your favor and 0% for your opponent; all the cards are in play and you hold the winning hand. Since you don't know your opponent's hand, however, you still fear he may hold an Ah - the only hand that will beat you. Purely mathematically, you could estimate the likelihood of him holding the Ah as follows:

There are $\binom{45}{2} = 990$ two-card hands that can be formed from the 45 unknown cards. Of these, 44 contain the Ah (select the Ah, then pair it with any of the other 44 cards). Thus, the chance of his having the Ah is $$ \frac{44}{990} \approx .04. $$

So, while everyone watching the game on television knows you are the winner, you are roughly 96% sure that you are the winner. As you might suspect, the discrepancy between these two answers can be much larger when more cards remain to be placed on the board.

As a more extreme example of how greatly these two calculations can differ, suppose the entire board is in-play with you and your opponent both holding extremely poor hands (but yours is slightly better). The television will again say 100% in your favor, but you will figure your chances of winning at next to zero!

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The calculation that a player does at the table will be more complicated than that, though. At every stage, the betting either indicates something about the cards or is a bluff meant to indicate something inaccurate about the cards. Of course, this is inexact information, and it has to be filtered through how the opponent tends to bet. Still, when considering possible card combinations, you don't care what their absolute probabilities are unless they are consistent with the betting. –  Aaron Aug 13 '11 at 20:35
    
This is a nice answer. Thank you. I was woundering however that from the time the initial cards a dealt to each player. The initial percentages appear. Then as the game progresses (players folding, turn, river, flop ect) at each stage the percentages change. Is it possible for you to run through a simple example, like player x has 2 cards, which give him x% of winning, then the turn happens, this gives him a new X%, then the river which gives a new x% then the next card and so on. How are each one calculated. Perhaps its a bit much. But thanks for your answer so far. –  Francis Rodgers Aug 13 '11 at 20:46
    
I don't know if this is how it's done in practice, but the method I alluded to in the second paragraph works. At the flop, consider all $\binom{48}{5}$ possible end-game scenarios. This comes from the fact that, if there are two players, there are 48 cards in the deck and some 5 of them will make it onto the board. Some percent of these will result in a win for player A, some for player B, and some are ties. After the flop comes down, we similarly consider all $\binom{45}{2}$ possible end-game scenarios. After the turn, there are $\binom{44}{1}$ possible end-game scenarios. –  Austin Mohr Aug 13 '11 at 20:57
    
After the river, there is $\binom{43}{0} = 1$ possible scenario, agreeing with our intuition. –  Austin Mohr Aug 13 '11 at 20:57
    
@Austin Mohr - Thank you for explaning this so well. You made it look very simple! :-> thanks. –  Francis Rodgers Aug 13 '11 at 22:54
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It depends on your and your opponent's cards. For instance, let's say you have AK while your opponent holds KQ and the board is AQ7 with no chances of flush. Then your opponent wins if a Q appears on either the turn or river. Since there are two queens left on the deck, and 7 cards are already known, you'll be beat about $2 \times \frac{2}{52-7} = 8.888\dots \%$ of the time.

That's not the exact number though, since both a queen and an ace could appear, giving you the pot, or JT would give both players a straight. These cases, however, are really unlikely, so the percentage is not awfully off for a rough estimate.

You should read about combinatorics if you're interested in these types of problems.

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I did combinatons in college, we only covered the basics though because its a software course and there were lots of other topics on the math side to focus on in the short time we had. Thanks for your answer, its not as detailed as the first but is closer to what I was looking for. Thank you. However, I will look it up and get better at it. Thanks for pointing me in the right direction. –  Francis Rodgers Aug 13 '11 at 20:51
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