# All About Boards: Part VII

Brian Alspach

Poker Digest Vol. 4, No. 16, July 27 - August 9, 2001

I have received several queries concerning the method for computing the probability that at least one player holds a certain hand once the board is known. I am going to present three sample calculations ranging from fairly easy to fairly difficult.

Let's start with an easy example. Suppose the board has the plus some stray card, and we want the probability that at least one player has a straight flush in hold'em (remember we are discussingsemideals-- that is, 10 hands and the board are dealt at random with all 10 hands staying to the end). If either the deuce or seven of hearts has been dealt to a player, then someone has a straight flush. So this is one of those situations in which it is easier to compute the probability no one has a straight flush. Clearly, no one has a straight flush if neither the deuce nor the seven are included among the 20 cards dealt to the players. There are C(47,20) choices for 20 cards to be dealt to the players and there are C(45,20) choices of 20 cards that do not include either the deuce or seven of hearts. The ratio of these two numbers is 351/1,081 which is the probability no one has a straight flush. Thus, the probability of at least one player having a straight flush is 730/1,081 which is approximately .675.

Now let's look at a more complicated example for hold'em. Suppose the board is two pair--that is, the rank multiset has the form . We would like to know the probability of someone having a full house or quads. The total number of ways of choosing 10 two-card hands is C(47,20)19!! and let me remind you that . If any of the four remaining cards of rank x or rank y is dealt to a player, then that player has at least a full house. The easiest way to count these semideals is to count the number of semideals in which no one is dealt a card of rank x or y and subtract it from C(47,20)19!! We remove the four cards of ranks x and y from 47, leaving 43 cards. This implies the number of semideals with no cards of rank x or y is C(43,20)19!! However, some of the latter semideals still have a player with a full house. This is the case if there is a player holding a pair of cards of rank z.

To count the number of semideals with a player holding a pair of rank z, we observe there are three choices for the pair. We set aside such a pair for one hand. The remaining 18 cards are chosen from 41 cards and split into nine hands which may be done in 17!! ways. This means the number of full house semideals with no one holding a card of rank x or rank y is 3C(41,18)17!! Thus, the number of semideals with at least one player holding a full house or quads is given by

C(47,20)19!! - C(43,20)19!! + 3C(41,18)17!!

Upon dividing by the total number of semideals we get a probability of .9049 that at least one player has a full house or quads given that the board is a two pair hand.

The reason the second example is more complicated than the first is because we have to break up the objects being counted into two pieces with one piece being counted directly and the other piece being counted indirectly.

Our final example is even more complicated because we have to keep careful track of what is being counted. Suppose we have a board with and a stray card, and we want to know the probability of at least one player having a straight flush in Omaha. Any of the three following combinations of cards produce a straight flush: , , or . There are several complicating factors: Two players may have straight flushes; the is special because if some player holds these two cards, it is impossible for anyone else to have a straight flush; and players have four cards in their hands so that a player can hold more than two of the cards .

Once the five board cards are determined in Omaha, the total number of ways of dealing out 10 Omaha hands (without regard to order of the hands--that is, player semideals) is determined as follows. There are C(47,40) choices of 40 cards to be dealt. There are then C(40,4) choices for the first hand, C(36,4) choices for the third hand, and so on. However, we divide by 10! because we don't care in which order the hands are dealt. Multiplying all these numbers leads to considerable cancellation and we end up with

player semideals. There is an obvious analogous number for nine hands, eight hands and so on. For example, when the number of hands is nine, 47 is replaced by 43, 10! is replaced by 9!, and the exponent on 4! is 9.

Let P(A,2,6,8) denote the probability of one player having as her hand. We set aside those four cards to go into one hand. We then count the number of ways of dealing nine Omaha hands from 43 cards and divide by the total number of Omaha player semideals. Doing this arithmetic produces a probability of 2/35,673.

Now let P(A,2,6) denote the probability of one player having in her hand but not the eight of hearts. We set aside those three cards to go into one hand and choose any of 43 cards to complete the hand (we cannot use the eight of hearts). We then count the number of ways of dealing nine Omaha hands from 43. Dividing by the total number of semideals gives a probability of 86/35,673. P(A,2,8), P(A,6,8), P(2,6,8) are defined analogously and also equal 86/35,673.

Let P(A-2;6-8) denote the probability of one player having in her hand and a second player having in her hand. We set the four cards aside and make C(43,2) choices for two cards to form a hand with , and C(41,2) choices for two cards to join the . We then form eight more Omaha hands from the 39 remaining cards. Multiplying and dividing by the total number of player semideals yields P(A-2;6-8) = 36/11,891.

Defining P(A,2) as the probability of a player having but neither six nor eight of hearts, we first set aside the . We then choose two cards from 43 in C(43,2) ways. We choose another nine Omaha hands in the usual way from 43 cards. This produces the number 602/11,891 but also includes a contribution from P(A-2;6-8), so we subtract that to obtain P(A,2) = 566/11,891. The value for P(6,8) is the same, but the value of P(2,6) = 602/11,891because it does not include any contribution from P(A-2;6-8).

Adding the various probabilities gives us a probability of 5,656/35,673, which is approximately .1586, that one or more players have a straight flush in Omaha when the board is as described above.

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