Chances of Winning

Brian Alspach

Poker Digest Vol. 4, No. 3, January 26 - February 8, 2001

Bib Ladder blew into town from Vancouver for the recent Harvest Classic at Casino Regina. It was good to see the old guy again and watch him go up against some of Western Canada's better tournament players.

Shortly after he arrived, we had a chance to sit and talk for a while. He was telling me about his adventures earlier in the Summer at The Orleans Open and got around to asking a question which prompted me to write this article.

One off the button in one of the no-limit hold'em tournaments at The Orleans Open, he found himself staring at pocket 10s and contemplating what action to take in response to the under-the-gun player who had gone all in.

No one else had called, and Bib had twice the chips of the all in player. He decided to call as he thought the chances of being heads up were pretty good.

The button and both blinds did indeed fold, so he was heads up.

The under-the-gun player did not turn over his hand, leading Bib to follow the example and keep his cards face down as well. The flop came 10-J-Q, all in different suits, prompting the other player to dramatically expose his Q-J for what he thought was an initial lead. Bib slowly turned over his pocket 10s watching his opponent deflate when he saw he was up against trip 10s.

Bib then asked the simple question: How do you determine how much of a favorite he was after the flop?

The answer to this question is not hard to determine, although in the heat of battle, we would not expect someone to provide an immediate answer. Besides, with one player all in, the exact answer is not of much interest because que sera, sera. Nevertheless, let's work out the details of solving the problem.

Let's assume we have two players going against each other. Following the flop, there are two cards to come. If one of the players is all in, the order in which the two cards come is irrelevant because no further betting is allowed. However, if both players still have chips, then the order of the two cards is important because the betting action following the turn card may affect the players' decisions on whether or not to continue.

We are considering a case in which one of the players is all in. There are 45 possible turn cards and 44 possible river cards, giving 1,980 possibilities for the sequences of two cards, but since the order is unimportant, we divide by two and get 990 possible sets of two cards. In other words, there are C(45,2) = 990 sets of two cards with which to complete the board.

To complete the analysis of Bib's chances of winning, we simply examine the outcomes of all 990 sets of two cards. There are 44 sets of two cards that contain a 10. Any of these give Bib unbeatable quads. So we now remove the 10 and consider subsets of the 44 remaining cards. If any queen or jack comes, then Bib's opponent will make at least a full house and Bib cannot catch up (remember we have discarded the ten).

There are four unseen jacks and queens. This means there are C(4,2) = 6 ways for his opponent to get two of those four special cards, and there are $4\cdot 40 = 160$ ways for him to get one of the four special cards and some other card. This gives us 166 ways Bib's opponent wins.

Now we remove the four jacks and queens and consider subsets of the 40 cards left over. From these 40 cards, Bib's opponent has only one way of getting a hand better than two-pair; namely, the two cards might produce a straight. (In this particular confrontation, no flush was possible because of the suits the two players held and the suits of the cards in the flop. Frequently, flushes have to be considered in the mix.) However, if Bib's opponent makes a straight, then Bib also makes the same straight and they end up with a tie. Thus, if the last two cards are any of A-K; K-9; or 8-9, the players tie. There are sixteen choices for each of the rank pairs. This gives 48 sets which produce a tie. All of the remaining sets of two cards leave Bib ahead.

Summing the numbers above, we find Bib wins 776 times, his opponent wins 166 times, and they tie 48 times. Expressed in percentages, Bib wins 78 percent of the time, his opponent wins 17 percent of the time and they tie 5 percent of the time.

Bib originally asked how much of a favorite he was after the flop. Ties make this question somewhat confusing. For example, consider the following two extreme situations.

There are two contestants and, of the 300 possible outcomes, contestant A wins 200 times and contestant B wins 100 times. Alternately, there are two contestants and of the 300 possible outcomes, contestant A wins 2 times, contestant B wins 1 time and they tie 297 times. How should we think about these contestants in terms of favoring A over B?

One could argue that in both situations, contestant A wins twice for every time contestant B wins. It is clear that in the first situation, people would agree that A is a 2-to-1 favorite over B. However, most people are going to be uncomfortable saying that A is a 2-to-1 favorite over B in the second situation. There could be scenarios where the 2-to-1 measure makes sense, but somehow, we feel they are close to evenly matched and our measure should reflect this.

My point is that in most situations ties have to be given some kind of weight, but the weight may vary depending on the context. In Bib's situation discussed above, including ties as separate information is the way to go, and the viewer can make up her own mind as to its weight.

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