**Poker Digest Vol. 4, No. 9, April 20 - May 3, 2001**

I have seen some recent messages on the internet dealing with questions about boards for hold'em. Consequently, I have decided to take a detailed look at a variety of probability questions revolving around boards for both hold'em and Omaha. Much of this information has appeared elsewhere, but I intend to consider some questions which have not been answered before. I want to make this series a definitive guide to boards.

It is likely everyone reading this knows what is meant by a board, but let's define it anyway. In both hold'em and Omaha there are five cards dealt face-up which are community cards for all players to use in forming their hands. A player may use three, four or five of the board cards in forming her hand in hold'em, but in Omaha there is the additional stipulation that a player must use three of the community cards plus two of her four hole cards in forming her hand. The five community cards are referred to as the board.

How many possible boards are there? This question has an easy answer.
One simply counts the total number of ways of choosing five cards from 52.
I have been using the notation *C*(52,5). This number is
2,598,960 and probably is one of the most widely known numbers in poker
as it is just the number of five-card poker hands for a standard 52-card
deck.

The first step we take is partitioning the possible boards into a variety of subsets. These subsets, which we call parts, form the cast of characters under discussion, and will be used to answer certain questions later. Altogether there are 40 straight flushes. Part SF1 consists of the four royal flushes. Part SF2 consists of the 12 straight flushes whose smallest card is either a seven, eight or nine. Finally, part SF3 consists of the remaining 24 straight flushes.

There are 624 four-of-a-kind hands. Those of rank distinct from ace with an ace kicker, and four aces with a king kicker are type Q1. The remaining four-of-a-kind boards are of type Q2. There are 52 of type Q1 and there are 572 boards of type Q2.

There are 3,744 full houses which we leave as one part called FH.

Altogether there are 5,108 flushes which are going to be broken into many
parts according to the ranks of the five cards comprising the flush. A
flush has five cards of distinct ranks. Note that the number of ways of
choosing five distinct ranks from the thirteen available ranks is
*C*(13,5) = 1,287. Ten of these rank sets consist of consecutive ranks,
that is, they have the form
.
They give rise to
the forty straight flushes leaving 1,277 rank sets not forming a straight.
Multiplying 1,277 by four yields 5,108 and this explains how the number of
flushes is determined. There are 79 rank sets containing four consecutive
ranks. For example,
is such a rank set. Of these 79 rank
sets, 48 allow a low and 31 do not. Let F1 be the 192 flushes with four
consecutive ranks allowing a low, and F2 be the 124 flushes with four
consecutive ranks not allowing a low.

There are 85 rank sets having three consecutive ranks together with another two consecutive ranks but not five consecutive ranks. For example, the rank set is such a rank set. There are 57 allowing a low and 28 not allowing a low. Let F3 be the part consisting of the 228 flushes with this kind of rank set allowing a low and let F4 be the 112 flushes not allowing a low.

There are 266 rank sets with three consecutive ranks and two isolated ranks. An example of such a rank set is . There are 178 allowing a low and 88 not allowing a low. Let F5 be the part consisting of the 712 flushes with the rank set under discussion which allow a low and F6 be the part corresponding to the 352 flushes not allowing a low.

There are 280 rank sets with two disjoint sets of two consecutive ranks and one isolated rank. An example of such a rank set is . For this type of rank set, 215 allow a low and 65 do not. Let F7 be the 860 flushes allowing a low, and F8 be the 260 flushes not allowing a low for this kind of rank set.

There are 476 rank sets with exactly two consecutive ranks and three isolated ranks such as . From this collection of rank sets, 380 allow a low and 96 do not. So we let F9 be the 1,520 flushes with this kind of rank set which allow a low, and F10 be the 384 flushes not allowing a low.

There are 91 rank sets with no consecutive ranks. At this point let me
show you a nice little trick for obtaining the 91 without using a lot of
cases. Modifications of this trick were used elsewhere in this article's
calculations. For each of the 35 subsets
of
(there are
*C*(7,4) = 35 of them), where
*b* < *c* < *d* < *e*, take the
corresponding rank set
,
and for each of the 56
subsets
of
,
where
*a* < *b* < *c* < *d* <
*e*, take the corresponding rank set
.
Since we
are adding increasing numbers to elements which are in increasing order
already, the resulting rank sets must have gaps between successive ranks.
It is not hard to see the preceding correspondence is one-to-one so that
there are
35 + 56 = 91 rank sets with no successive ranks. This is
the kind of trick which makes a mathematician smile. Of these 91 rank
sets, 81 allow a low and 10 do not. Let F11 be the 324 flushes of this
type allowing a low and F12 be the 40 flushes not allowing a low.

The preceding takes care of the 5,108 flushes. There are 10 rank sets corresponding to straights. There are four possible cards of each rank yielding 1,024 straights (four raised to the fifth power) for a fixed rank set. Multiplying by 10 and removing the 40 straight flushes yields 10,200 straights. If the smallest rank in the straight is six or less, then the straight allows a low. That is, 6,120 of the straights allow a low. This gives us one partition of the straights into two parts. We also want to partition the straights according to suit distribution because this is what determines whether or not a player can make a flush.

For a given rank set corresponding to a straight, that is, , four choices have all five cards in the same suit, but these already have been removed because they correspond to straight flushes. There are five choices of four ranks to be in the same suit, four choices for that suit, and three choices for the remaining suit. Multiplying gives us 60 straights with four suited cards. Similarly, there are 10 choices of three ranks to be in the same suit and four choices for the suit. Thus, there are 120 straights with three cards in one suit and the other two cards in a different suit but the same, and 240 straights with three cards in the same suit and the remaining two cards in distinct suits. Thus, for a given rank set, there are 420 straights allowing a player to have a flush. The remaining 600 straights with that rank set do not allow a flush. We now are ready to describe the partition of the 10,200 straights.

Let S1 be the 600 ace high straights not allowing a flush. Let S2 be the 1,800 straights not allowing a low and not allowing a flush. Let S3 be the 1,680 straights not allowing a low and allowing a flush. Let S4 be the 2,520 straights allowing a low and allowing a flush. Let S5 be the 3,600 straights allowing a low but not allowing a flush.

There are 54,912 boards with three-of-a-kind. Such a board has only three distinct ranks and allows a low only if the three ranks are chosen from A, 2,...,8. This implies that 10,752 of the boards allow a low. For each rank set there are 36 of the three-of-a-kind boards with that rank set which allow a flush. So let T1 be the 2,016 boards allowing a low and a flush, let T2 be the 8,736 boards allowing a low and no flush, let T3 be the 8,280 boards allowing no low but allowing a flush, and let T4 be the 35,880 boards allowing neither a low nor a flush.

Since a board of two pair also has three ranks, we partition them as we did in the preceding case. Let D1 be the 6,048 boards allowing both a low and a flush, let D2 be the 18,144 boards allowing a low but not a flush, let D3 be the 24,840 boards allowing no low but allowing a flush, and let D4 be the 74,520 boards allowing neither a low nor a flush.

Boards with a single pair and high card boards need to be partitioned into many parts. We shall discuss this in the next article.

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