Poker Digest Vol. 3, No. 22, October 20 - November 2, 2000
In one of his articles, writer Roy Cooke discusses holding a suited hand in a loose hold'em game. He essentially said (I am paraphrasing) that one is often up against a player with a better hand of the same suit. I would like to quantify this situation in a very precise manner. The results, in fact, may surprise some of you, and a beautiful mathematical principle arises in the determination.
Let us make the context absolutely clear at the beginning. We're going to assume the game is 10-handed, each player is dealt two cards; a given player is dealt two suited cards (we shall say hearts, for simplicity, but the results are the same no matter the suit), and five community cards (the board) are dealt in the center in the usual manner. The question we address is the following: Given that the board allows a heart flush, what is the probability some other player has a better heart flush than our given player?
In actual play another player holding hearts may fold during the course of the hand. This has a tendency to increase the probability of a player holding a heart flush winning the hand. Thus, you should use the numbers we are about to determine accordingly. Also, we are going to ignore straight flushes. They have essentially no effect on the numbers and taking them into account needlessly complicates the mathematics.
First, let's assume there are precisely three hearts on the board. Let me describe how to determine the exact probability one of the additional nine players also has a hand with two hearts. There are 45 unseen cards of which eight are hearts and 37 are non-hearts. There are C(45,18) choices for the 18 cards which are dealt to the other nine players. Once the 18 cards have been chosen, we don't care which particular player of the remaining nine receives two hearts but we do care if someone else has two hearts; that is, we are interested in semi-deals rather than deals. The number of semi-deals for 18 cards to nine players is 17!! = 34,459,425. Thus, the total number of semi-deals is C(45,18) multiplied by 17!! Let's denote this product by A.
There can be any number of hearts from zero through eight included in the chosen set of 18 cards. So we break the problem into nine pieces depending on how many hearts are included. Let's look at an example of one of the cases. Suppose the eighteen cards contain exactly three hearts. One of the other players will have a heart flush if two of the three hearts are dealt to the same player. There are C(3,2) = 3ways of choosing two of the hearts together. Once they have been chosen to go in the same hand, the remaining sixteen cards can be partitioned into hands in 15!! = 2,027,025 ways. Hence, of the 17!! semi-deals for this particular set of eighteen cards, of them produce another player with a flush. Subtracting from 17!! gives 28,378,350 of the semi-deals not giving anyone else a flush.
We carry out a similar analysis for each of the nine cases and find the following. (I am going to give you the actual numbers now, even though they are big, because we are going to use a beautiful mathematical principle to solve the more interesting problem of determining the probability that someone else has a bigger flush.) The total number of semi-deals is denoted A (see above). The total number of semi-deals for which exactly one other player has two hearts is 13,101,833,111,984,555,400. The total number of semi-deals for which exactly two other players have two hearts is 949,042,314,710,582,400. The total number with exactly three other players holding two hearts is 16,935,898,294,854,000 and with exactly four other players holding two hearts is 34,563,057,744,600.
As a matter of curiosity, dividing the preceding numbers by A gives the probabilities of those events as follows: One other player with two hearts .22158; two other players with two hearts .01605; three other players with two hearts .000286; four other players with two hearts .00000058; and no one else with two hearts .76208.
The preceding numbers are interesting, but much more interesting is the problem of determining the probability that another player holds bigger hearts. At first glance one might think this would involve even more cases than the preceding computation. If it did, it would be reasonable to tell anyone interested in the values to simply forget it. It turns out we can use a principle I'll call the proportionality principle to solve the problem easily using only the numbers above and some proportions we'll derive.
Here is a step-by-step procedure. The given player holding the hearts first counts the number of bigger hearts amongst the eight unseen hearts. Call this value L. For example, suppose the player is holding 4-7 and the board has 5-10-J for its hearts. The unseen bigger hearts are 8-9-Q-K-A and L = 5 in this case. Among all the semi-deals with exactly one other player holding hearts, the striking fact is that each pair of hearts formed from the eight unseen hearts will occur in precisely the same number of boards. That is a simple and beautiful fact. So all we have to do is determine the proportion of pairs which involve one of the bigger cards. There are 28 pairs of unseen hearts and just six of them do not involve one of the five bigger hearts. That is, we multiply the number 13,101,833,111,984,555,400 by 22/28 to obtain the number of the semi-deals with a flush beating the player's flush. If the semi-deal has two or more players holding two hearts, then it is clear at least one of them must have a bigger flush. In this case we take the entire number above. In the end we divide by A to get the probability.
It's quite fast to work out the proportions and arrive at the various probabilities. The following table contains the information in a compact fashion. The probability is the probability that someone has a bigger flush and L is the number of unseen cards in the same suit which are bigger than the given player's bigger card.
We could look at boards with four or five hearts but will not do so.
First of all, of the boards which allow flushes for a given player,
more than 90 percent of the time there are only three cards of the player's
suit on board. Secondly, the computations are fairly simple when
there are four or five cards of the player's suit on board. Finally,
straight flushes need to be considered as there are situations where
the probability of a straight flush is significant in relation to
the probability of a flush.