Low Board Probability

Brian Alspach

9 October 1998

Abstract:

We determine the probability of a low board occurring for a board with 5 cards from a standard deck of 52 cards.

In a game such as Omaha or hold 'em where 5 cards are displayed in the middle for everyone to use in her or his hand, we refer to the 5 cards in the middle as the board. When playing Omaha high-low, in order to qualify for low a hand must achieve an 8-low or better. Thus, a low may occur only if the board has at least 3 distinct ranks chosen from $\{A,2,3,4,5,6,7,8\}$ . We refer to such a board as a low board. Here we determine the probability of a low board occurring when the board is dealt from a standard 52-card deck. We do so by counting the total number of possible low boards and dividing by the total number of possible boards. The total number of possible boards is

$\begin{displaymath}{{52}\choose{5}} = \frac{52!}{5!47!}= 2,598,960.\end{displaymath}$

We first break the possible boards into six types depending on the number of big cards, denoted H, and the number of small cards, denoted L.

Type A: Boards of the form H,H,H,H,H.

Since there are 20 big cards altogether, there are

$\begin{displaymath}{{20}\choose{5}}= \frac{20!}{5!15!}= 15,504\end{displaymath}$

boards of type A.

Type B: Boards of the form H,H,H,H,L.

We are choosing 4 cards from the 20 big cards and 1 card from the 32 low cards. Thus, there are

$\begin{displaymath}32{{20}\choose {4}}=\frac{32\cdot 20!}{4!16!}= 155,040\end{displaymath}$

boards of type B.

Type C: Boards of the form H,H,H,L,L.

We are choosing 3 cards from the 20 big cards and 2 cards from the 32 small cards. Then there are

$\begin{displaymath}{{20}\choose{3}}{{32}\choose{2}}= 565,440\end{displaymath}$

boards of type C.

Type D: Boards of the form H,H,L,L,L.

Now we are choosing 2 cards from the 20 big cards and 3 cards from the 32 low cards. This can be done in

$\begin{displaymath}{{20}\choose{2}}{{32}\choose{3}}= 942,400\end{displaymath}$

ways. This is the number of boards of type D.

Type E: Boards of the form H,L,L,L,L.

We are choosing one card from the 20 big cards and 4 cards from the 32 low cards. This gives us

$\begin{displaymath}20{{32}\choose {4}}=\frac{20\cdot 32!}{4!28!}= 719,200\end{displaymath}$

boards of type E.

Type F: Boards of the form L,L,L,L,L.

There are

$\begin{displaymath}{{32}\choose{5}}=\frac {32!}{5!27!}= 201,376\end{displaymath}$

boards of type F.

Adding the numbers of boards of the six types yields 2,598,960 boards as it should.

Boards of types A, B, or C are not low boards because they have at most two low cards. We now break the boards of the three remaining types into subtypes, some of which are low and some of which are not.

D1.

The three low cards are of distinct ranks. There are 3 choices for the ranks from 8 ranks, there are 4 ways of choosing each of the low cards of a given rank, and we are choosing 2 big cards from 20. This yields

$\begin{displaymath}4^3{{8}\choose{3}} {{20}\choose{2}} = \frac{64\cdot 8!20!}{5!3!18!21} = 680,960\end{displaymath}$

boards of type D1.

D2.

The three low cards contain a pair. There are 8 choices for the rank to be paired, 6 ways of choosing a pair of that rank, 7 ways to choose the rank which is not paired, 4 choices for a card of that rank, and a choice of 2 cards from the 20 big cards. This yields

$\begin{displaymath}8\cdot 6\cdot 7\cdot 4{{20}\choose{2}}= 255,360\end{displaymath}$

boards of type D2.

D3.

The three low cards form 3-of-a-kind (trips). There are 8 choices for the rank of the trips, 4 choices for the trips of that rank, and 2 choices for the cards from the 20 big cards. This yields

$\begin{displaymath}8\cdot 4{{20}\choose{2}}= 6,080\end{displaymath}$

boards of type D3.

Adding the numbers for boards of types D1, D2 and D3, we obtain 942,400 as required.

Next we consider boards of type E.

E1.

The four low cards have distinct ranks. There are 4 choices for the ranks from 8 ranks, there are 4 choices of card for each chosen low rank, there are 20 choices for the big card. This gives us

$\begin{displaymath}4^4{{8}\choose{4}}\cdot 20 = \frac {20\cdot 4^4\cdot 8!}{4!4!}= 358,400\end{displaymath}$

boards of type E1.

E2.

There is one pair amongst the four low cards. There are 8 choices for the rank of the pair, there are 6 choices of pairs of the given rank, there are 2 ranks to be chosen from the remaining 7 low ranks, there are 4 cards for each of the other 2 low ranks, and there are 20 choices for the big card. Altogether this gives

$\begin{displaymath}8\cdot 6\cdot 4^2\cdot 20{{7}\choose{2}}= 322,560\end{displaymath}$

boards of type E2.

E3.

The four low cards contain trips. There are 8 choices for the rank of the trips, there are 4 choices for the trips, there are 28 choices for the remaining low card, and there are 20 choices for the big card. This produces

$\begin{displaymath}8\cdot 4\cdot 28\cdot 20 = 17,920\end{displaymath}$

boards of type E3.

E4.

The four low cards form two pairs. There are 2 choices of ranks from 8 for the two pairs, each pair can be chosen in 6 ways, and there are 20 choices for the big card. Multiplying yields

$\begin{displaymath}6^2\cdot 20{{8}\choose{2}}= 20,160\end{displaymath}$

boards of type E4.

E5.

The four low cards form 4-of-a-kind (quads). There are precisely 8 choices for the quads, and 20 choices for the big card. This yields $8\cdot 20=160$ boards of type E5.

Adding the numbers of boards of types E1, E2, E3, E4, and E5 produces 719,200 which is the number of boards of type E as determined earlier.

Finally we consider boards of type F.

F1.

The five low cards have distinct ranks. There are 5 choices for the ranks from 8 possible ranks, and for each rank there are 4 choices for cards. This produces

$\begin{displaymath}4^5{{8}\choose{5}}= 57,344\end{displaymath}$

boards of type F1.

F2.

There is a single pair among the 5 low cards. There are 8 choices for the rank that is paired, there are 6 choices for the pair, there are 3 choices for ranks among the remaining 7 ranks, and there are 4 choices for each of the latter 3 ranks. We obtain

$\begin{displaymath}8\cdot 6\cdot 4^3{{7}\choose{3}} = 107,520\end{displaymath}$

boards of type F2.

F3.

There are trips occurring and two other low cards of different ranks. There are 8 ways of choosing the rank of the trips, there are 4 ways of choosing trips, the other 2 ranks are chosen from 7 ranks, and there are 4 choices for each of the other two ranks. This yields

$\begin{displaymath}8\cdot 4\cdot 4^2{{7}\choose{2}}= 10,752\end{displaymath}$

boards of type F3.

F4.

There are 2 pairs amongst the 5 low cards. There are 2 choices from 8 for the ranks of the 2 pairs, there are 6 ways of choosing each pair, and there are 24 choices for the remaining card. This yields

$\begin{displaymath}6^2\cdot 24{{8}\choose{2}} = 24,192\end{displaymath}$

boards of type F4.

F5.

The five low cards form a full house. There are 8 choices for the rank of the trips, 7 choices for the rank of the pair, 4 choices for the trips of the chosen rank, and 6 choices for the pair of the chosen rank. This produces

$\begin{displaymath}8\cdot 7\cdot 6\cdot 4= 1,344\end{displaymath}$

boards of type F5.

F6.

There are quads occurring amongst the 5 low cards. There are 8 ways of choosing quads and 28 ways of choosing the remaining card. This produces $8\cdot 28= 224$ boards of type F6.

We now add the F subtypes obtaining 201,376 which is the correct number of F boards. Therefore, all the boards have been accounted for.

The low boards are types D1, E1, E2, F1, F2, F3, and F4. Adding the numbers of boards of these types yields 1,561,728 low boards. Dividing this number by 2,598,960, the total number of boards, gives us 0.601 as the probability of a low board occurring for a standard 52-card deck. This implies the probability of 0.399 for no low being possible.

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