# All About Boards: Part VIII

Brian Alspach

Poker Digest Vol. 4, No. 17, August 10 - 23, 2001

Let's look for the probability of a semideal in hold'em when at least one player makes a full house, but no one makes four-of-a-kind. The reason we are excluding quads is because it is not difficult to make this exclusion. On the other hand, we are not going to exclude straight flushes because doing so requires fooling around with many subcases. Thus, the final probability will not be the probability of a semideal for which some kind of full house is the best hand, but it will not be far off since many of the boards allowing full houses do not allow a straight flush, and even when both a straight flush and full house is allowed by a board, the straight flush and full house occur simultaneously with very small probability.

A board allows a full house only if the board shows a full house, three-of-a-kind, two pair, or a single pair. For each of these four situations we determine the probability some player has the appropriate cards to complete the board to a full house without quads occurring, multiply by the probability of the appropriate board occurring, and take the sum to get the probability.

The probability the board will show a full house is 6/4,165. Assume the rank multiset of the board is . In order to avoid quads in hold'em, no player may be dealt the case card of rank x or a pair of rank y. We must exclude the card of rank x so that we are choosing the cards to be dealt to the players from 46 cards. The number of player semideals in hold'em is then C(46,20)19!!. Of these, C(44,18)17!! have the pair of rank yin the same player's hand. Subtracting the two gives us the number of player semideals for the board with no one having quads. Dividing by the total number of player semideals gives us a probability of 615/1,081 that no one has quads with a full house on board in hold'em. Multiplying by 6/4,165 yields a contribution of 738/900,473 towards the probability of a full house without quads in hold'em.

If there are trips on board, then the rank multiset has the form . As long as the card of rank x is not dealt to anyone, nobody has quads. The number of player semideals with no card of rank x being dealt is C(46,20)19!! If any card of rank y or z is dealt, then someone has a full house. The number of player semideals with no card of rank x, y or z is C(40,20)19!! Subtraction gives us the number of player semideals having a full house because some player has at least one card of rank y or z. However, we also have a full house if some player has a pocket pair of rank different from x or y. We use inclusion-exclusion to estimate the number of player semideals with at least one pocket pair. Adding to those counted above and dividing by the total number of player semideals, followed by multiplying by the probability of a three-of-a-kind board, yields a probability of .012 as the contribution coming from a board with trips.

Boards with two pair or a single pair are easier to handle because there are more limitations on how a player can have a full house. If the rank multiset of the board is of the form , then a player has a full house only if her hand is xy, xz, yz or zz. It is not difficult to enumerate the player semideals including one or more hands of these types that do not include quads. Similarly, if the board is of the form , then a player has a full house only with the hands xy, xz, xw or a pocket pair matching one of the singleton ranks on board.

Doing the enumeration produces contributions of .0081 for a full house without quads when the board is two-pair, and .0967 for a full house without quads when the board is a single pair. Adding the four contributions gives a probability of .1176 for a semideal with a full house and no quads in hold'em.

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