Cracking Aces: Part III

Recently, the following e-mail message came to Poker Digest from James Grosjean. I thank him for the message and included my response in this article.

Dear editors,

It may be a bit late to bring up, but I must comment on the ``Cracking Aces: Part II'' article in the``Mathematics and Poker With Brian Alspach'' column of the April 7-20 (Vol. 3/No. 8) issue of Poker Digest.

First, he says, ``... there is a 31.1 percent chance of winning with pocket aces against nine random hands played to the end.'' Later, ``the value .311, the probability of winning with pocket aces with nine opponents who stay all the way to the end, provides a lower bound in the following sense. Suppose you are dealt a pair of pocket aces against nine opponents.

If no one folds before the hand is played out, then the outcome is predetermined .... So at this juncture the probability you will win at the end of the hand is .311, and you will not do worse than this in the long-run ...''

When dealt pocket aces against nine opponents, if no one folds, then you are not playing against nine random hands. In a live game, if faced with nine opponents, the fact that they have chosen to play allows you to infer that (assuming that their decision to play is ``normal'' in the sense that it correlates to better hands, i.e., they are not complete morons playing a perverse strategy) they hold good hands, such as wired pairs, connectors, and suited cards. Then, the 0.311 figure is overly optimistic.

Neglecting your ability to subsequently bet opponents out of the pot, the 0.311 would be an upper bound, not a lower bound. Considering the effect of your opponents' responses to your aggressive betting, the 0.311 would be no bound of any kind, but merely a benchmark.

The critical fact is that Dr. Alspach has, at least in the first page of the article, confused the live, non-random choice of nine opponents, with the computer-simulated, forced play (and hence equivalent to random hands) of nine opponents. At the very end, he starts to allude to this point, suggesting that if we could program nine ``realistic'' opponents, we could then estimate, through simulation, the probability of winning with pocket aces when the nine opponents have chosen to play to the end. The figure would be clearly below 0.311. Keep up the good work.

Mr. Grosjean has paid much closer attention to my wording than I did and he is perfectly justified in interpreting the sentence containing the phrase ``If no one folds before the hand is played out'' as he did. However, in my defense, let me explain that I intended this to still be the random case. That is, when I talked about no one folding before the the hand was played out, I was thinking of this in terms of the following. Our hero finds pocket aces and she, together with all nine opponents, turn their hands face up. The dealer now properly deals the board cards with no further player action. The player has a probability of .311 of winning the hand. (We are assuming .311 is very accurate based on simulations because exact enumeration is computationally infeasible.)

This means a player who plays all pocket aces all the way to the end will win approximately 31.1 perecnt of these hands. The figure of 31.1 percent is a lower bound in the sense that a player with pocket aces who takes strong betting action and stays to the end will win more than 31.1 percent of the hands because some of the weak hands which would have won the pot will have folded.

Now let's get realistic about a lower bound for the percentage of pots won with pocket aces. Suppose as a player you would like to win the highest possible percentage of pots when holding pocket aces. If you follow an algorithm of staying all the way to the end, and betting and raising aggressively, you will definitely win at least 31.1 percent of the pots.

However, it's impossible to know what the percentage might be as it clearly depends on the players in the game. The more players in the game who are willing to throw away risky hands, the more the percentage will increase.

It also is clear that staying all the way to the end is a foolish algorithm to follow. There are going to be times when you are almost certain you are beat and staying to the end just because you have aces is going to cost you more pots than you would gain. This makes the problem even more hopeless because as soon as you are willing to toss the aces under certain conditions, the 31.1% is no longer a lower bound because you might mistakenly throw away more winning hands than losing hands. I suspect most reasonable players will have a net gain in percentage of pots won by being willing to throw their aces, but measuring the gain is impossible.

If you do not know what a tree (mathematical version) is, you might not gain much beyond this point. There are several ways we can model the process of what happens as a tree, where the root of the tree is our hero finding pocket aces and the leaves of the tree are all possible outcomes. We know that 31.1 percent of the leaves have a W for the player and the remaining leaves have an L for the player. Returning to the interpretation Mr. Grosjean discussed in his message, if the player is in the situation where all the other nine players have chosen to bet, call or raise, then she is at some internal vertex of the tree. The possible outcomes available for the player are those leaves lying in the subtree below her current location in the tree. The percentage of available leaves with a W no longer need be 31.1 percent as he correctly writes. Given reasonable players, it is highly likely it is strictly less. It is very difficult to say what the percentage is because this tree is dynamic in that it changes according to the other nine players. These problems are extremely difficult although simulation can give some valuable information.