Low Board Blues: Reprise and Coda

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 43 is 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 product

We now count semi-deals according to how many aces and deuces they contain.

A

. The number of 36-subsets containing no ace or deuce is C(37,36) = 37.

B

. The number of 36-subsets containing a single ace or single deuce is given by 6C(37,35) = 3,996 because there are six choices for the ace or deuce and 35 choices from 37 for the rest of the cards.

C

. The number of 36-subsets containing two aces or two deuces is given by 6C(37,34) = 46,620 because there are six ways to choose a pair of aces or a pair of deuces, leaving 34 other cards to choose.

D

. The number of 36-subsets containing one ace and one deuce is 9C(37,34) = 69,930.

E

. The number of 36-subsets containing either two aces and one deuce, or one ace and two deuces is 18C(37,33) = 1,188,810 because of similar reasoning as in the previous cases.

F

. The number of 36-subsets containing either three aces or three deuces is 2C(37,33) = 132,090.

G

. The number of 36-subsets containing either three aces and a deuce, or one ace and three deuces is 6C(37,32) = 2,615,382.

H

. The number of 36-subsets containing two aces and two deuces is 9C(37,32) = 3,923,073.

I

. The number of 36-subsets containing three aces and two deuces, or two aces and three deuces is 6C(37,31) = 13,948,704.

J

. The number of 36-subsets containing three aces and three deuces is C(37,30) = 10,295,472.

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.