Poker Digest Vol. 2, No. 13, June 18 - July 1, 1999
As mentioned in the previous article, there are some problems with what I wrote in Parts IV and V of the Low Board Blues series. I want to fix the problems in this article. Those two articles purported to be dealing with a player who is holding A-2-H-H in Omaha, where H denotes any card of rank 9,10,J,Q or K, and sees a board for which A-2 is a nut low. The player is interested in the probabilities of one or more other players also having a nut low. When I started doing the calculations, for some reason I started working with 48 instead of 43, the latter number being the correct number of unseen cards. So let's start over.
We are assuming there are 10 players, the board allows a low, and no other
aces or deuces are on the board. This means A-2, and only A-2, gives a
player a nut low. There are 43 unseen cards of which 36 are distributed to
the remaining nine players. The number of ways of choosing 36 cards from
C(43,36) = 32,224,114. For each subset of 36 cards, we wish to
count the number of ways of partitioning them into nine hands of four
cards each, that is, the number of semi-deals. We determine this by
choosing four cards from 36 in C(36,4) ways, then four cards from the
remaining 32 cards in C(32,4) ways, and so on. After taking the
We now count semi-deals according to how many aces and deuces they contain.
Here is how we get the answer we are seeking. We consider each type of 36-subset above and determine which semi-deals give aces and deuces in the same hand. For example, type A and B subsets never have an ace and deuce in the same hand because there are not enough of them. Let's do one example to illustrate the technique. Consider a type H 36-subset.
Type H subsets actually allow two additional people to have hands containing A-2. When counting semi-deals here, remember the aces are distinguishable from each other (by suits) so this must be taken into account. There is one way for A-A-2-2 to be in the same hand. There are ways for A-A-2 to appear in the same hand without the remaining deuce because there are two ways to choose the deuce and 32 cards to choose as the remaining card. There are C(32,2)=496 ways for A-A to be completed to a hand with no deuce. There are ways for A-2-2 to appear in the same hand without another ace. We multiply each of these numbers by the number of ways of partitioning the remaining 32 cards into eight hands to obtain the number of semi-deals with the aces and deuces forming the patterns above.
The remaining completions involve the aces being in separate hands and partitioning 28 cards into seven hands instead of partitioning 32 cards into eights hands as above. The number of ways of completing aces in separate hands to a single A-2 hand is because there are two choices for the ace and the deuce comprising the hand with the A-2, two choices out of 32 cards to complete it to a hand, and three choices out of the remaining 30 cards to complete the other ace to a hand without a deuce. The number of ways of completing aces in separate hands to two hands each with an A-2 is 2C(32,2)C(30,2)=431,520. Finally, the number of ways to complete aces in separate hands so that A-2 does not occur is C(32,3)C(29,3)=18,123,840. These completions need to be multiplied by the number of ways of partitioning 28 cards into seven hands.
Upon doing all the arithmetic, we find the following probabilities: The probability of three other players also having A-2 is 0.0012, the probability of precisely two other players having A-2 is 0.053, the probability of precisely one other player having A-2 is 0.377, and the probability of no other player having A-2 is 0.57.
Let me remind you the preceding probabilities are valid after the player has seen the board and observed there are no aces or deuces in the board and the player has no further aces or deuces. What about before the player sees the board. That is, suppose the player looks in her hand and sees two cards of rank x and rank y and no other cards of those ranks. She wonders what the probabilities are that other players also hold cards of ranks x and y. We shall use A-2, but the probabilities are the same for any two ranks. Now the preceding computations must be done using 48, as I started in the earlier articles, instead of 43 and they lead to different results. The probability of three other players having A-2 is 0.00059, the probability of precisely two other players having A-2 is 0.035, and the probability of precisely one other player having A-2 is 0.32.