The reader is advised that this article and the succeeding article are nonsense and should be ignored. See Low Board Blues: Reprise and Coda for corrections.

Low Board Blues: Part IV

Brian Alspach

Poker Digest Vol. 1, No. 7, October 23 - November 5, 1998

I flopped the nut low, feelin' pretty good
I flopped the nut low, feelin' pretty good
To keep from getting quartered
Think I'll knock on wood

You look at your cards and find A,2 of hearts along with a pair of kings. The flop brings the 3,4 of hearts and another king. You like the flop a lot. The turn card is the miracle 5 of hearts. For the first time in your life you have a bicycle straight flush! There is a bet, you raise, everyone drops except one other player who calls. The original bettor re-raises! You cap it and the limper calls the two additional bets. Avarice takes over as you glance at the pot trying to estimate how many chips you are likely to stack. The river card is an innocent jack so nothing has changed. The betting proceeds exactly as in the previous round.

Poof! Your greed turns to ashes in your mouth as the lead bettor shows 6,7 of hearts and the limper shows A,2. From a possible scooping hand to one-quarter of the pot in a flash.

The scene disappears as you awaken realizing it was only a nightmare, and nightmare is the perfect word to describe the preceding scene. How realistic is it? Given all the Omaha high-low hands being dealt every day across North America, something similar is bound to happen. The following did happen in a home game in which I play. One player made a bicycle straight flush and declared both ways. Unfortunately for him, we have a rule that a two-way declarer must win both ways or else he wins nothing. He was tied for low and forfeited any claim to any portion of the pot. Unfortunately for the game, he got up and went home because he was upset.

This leads me to the topic of this article. Given that you have a nut low, what is the probability someone else does too? Like most questions of this type the answer is dependent on what is in your hand and the composition of the board. I shall show you how to do the computation in one or two cases and hope you can then do similar computations for situations of interest to you. There are many variations on this theme which could fill a chapter of a book.

Let's assume the board allows only one way of getting a nut low. For example, the board might have 5,7,8 and two other big cards. If you hold A,2, in order for someone else to tie she must have an A,2 as well. So given that you hold A,2,H,H, where H denotes a big card, let's determine the probability someone else also has A,2.

We are going to assume the game is 10-handed. What we are going to discover in this situation is that the numbers involved become extremely large. The remaining 9 players receive 36 cards from the 48 unseen cards. The number of ways of choosing 36 cards from 48 is C(48,36) = 69,668,534,468. We now introduce a new notion not discussed in earlier articles, namely, a partition of a set. We don't care which players get which hands because we are NOT asking the probability of a certain player getting an A,2; instead, we are asking the probability that someone has an A,2. Thus, once 36 cards have been chosen, we are interested in how many ways we can chop the 36 cards into 9 hands of 4 cards each. This is called a partition of a 36-set into 9 subsets each of size 4. This also is a good time to remind you that n! means $n\times (n-1)\times
(n-2)\times\cdots \times 2\times 1$.

Even though we don't care which player receives which hand, it is easier to count by counting how many ways the first player can get 4 cards from 36, the second player 4 cards from 32, and so on, and then divide by 9! because the same partition of 9 hands of 4 cards each can be distributed in 9! different orders to the players. The first hand can be chosen in C(36,4) ways, the second hand in C(32,4) ways (because 4 cards have been removed), the third hand in C(28,4) ways (because 8 cards have been removed), and so on. We multiply all these numbers because the connective is the word ``and'' obtaining

\begin{displaymath}C(36,4)\times C(32,4)\times\cdot\times C(8,4)\times C(4,4).\end{displaymath}

So the number of ways of getting 9 hands from 36 cards is 36! divided by $24^9\times 9!$. Evaluating this produces the astronomically large number


We shall use the term semi-deal for the number of different 4-card hands that can be held by the other 9 players, not counting the order in which the hands are held. So the number L of semi-deals is the latter large number multiplied by C(36,12). Calculation yields


Of the L semi-deals, we want to know how many have A,2 in the same hand. There are 3 aces and 3 deuces amongst the 48 cards not in your hand. You can be tied only if there is at least one ace and one deuce in the 36 cards comprising the semi-deal. So the patterns of aces and deuces contained in the semi-deal is important. For purposes of counting, the roles of ace and deuce in the same suit can be interchanged to simplify the counting. For example, any semi-deal containing A,A,2 corresponds to one containing A,2,2. So the distinct patterns to consider are A,2; A,2,2; A,2,2,2; A,A,2,2; A,A,2,2,2; and A,A,A,2,2,2. The pattern A,2 can occur in 9 ways as each can be chosen in 3 ways, the pattern A,2,2 can occur in 18 ways (remember we are including A,A,2 in this pattern because of the interchange of A and 2), the pattern A,2,2,2 can occur in 6 ways, the pattern A,A,2,2 can occur in 9 ways, the pattern A,A,2,2,2 can occur in 6 ways, and the pattern A,A,A,2,2,2 can occur in 1 way.

For each pattern, we count the number of semi-deals having A,2 in the same hand, and the number of semi-deals having two two hands with an A,2, and the number of semi-deals having three hands with an A,2.

In the next, and last, installment of the Low Board Blues series, I shall provide more details on obtaining the following probabilities, but here are the numbers to tantalize you. The probability of one other player also having a nut low is 0.298, the probability of two other players also having nut lows is 0.024, and the probability of three other players also having nut lows is 0.0006.

Home | Publications | Preprints | MITACS | Poker Digest | Poker Computations | Feedback
website by the Centre for Systems Science
last updated 12 January 2000