Brian Alspach

Poker Digest Vol. 4, No. 11, May 18 - 31, 2001

In the last issue we gave a fairly detailed table containing the numbers of boards with various properties. Now we are going to start looking at some probability questions dealing with boards and possible hands. We are going to start right at the top with straight flushes. Though they may be rare, straight flushes generate adrenaline rushes and are involved in bad-beat jackpot considerations.

In our table, the board types T1, T2, T3, T4, D1, D2, D3 and D4 do not mention straight flushes. Indeed, some of them do allow straight flushes so we must refine that portion of the table. Henceforth, we will use the type designations as shown below:

 Type Number Allows Low Clumping Hands Sequential Hands T1 1,008 yes quads straight flush T2 1,008 yes quads flush T3 4,368 yes quads straight T4 4,368 yes quads board plays T5 1,296 no quads straight flush T6 6,984 no quads flush T7 5,616 no quads straight T8 30,264 no quads board plays D1 3,024 yes quads straight flush D2 3,024 yes quads flush D3 9,072 yes quads straight D4 9,072 yes quads board plays D5 3,888 no quads straight flush D6 20,952 no quads flush D7 11,664 no quads straight D8 62,856 no quads board plays

The refinement is for boards which are either three-of-a-kind or two pair. Let's discuss how we carried it out. The rank sets for both three-of-a-kind and two pair have three elements. Hence, there are C(13,3) = 286 rank sets. For each such rank set, there are three choices for the rank which is trips, four choices for the trips of that rank, and four choices for each of the single cards of the remaining two ranks. This gives us 192 ways of forming a three-of-a-kind hand for a given rank set with three elements. Thus, the product of 192 and 286 is the number of three-of-a-kind hands. Similarly, for a given rank set with three elements, there are three choices for the two ranks which will have pairs, six choices for each of the pairs, and four choices for the singleton. This gives us 432 ways of forming two-pair hands from a given rank set. So the product of 432 and 286 is the number of two-pairs hands.

For a given rank set with three elements, the only way a board allows a flush is by having three suited cards of each of the ranks on the board. There are four choices for the suit, and in the case of trips, there are three choices for the rank of the trips and there are three choices for the nonsuited pair completing the trips. Thus, for each rank set, there are 36 boards with trips that allow flushes. This means there are boards with trips that allow flushes. Note that the type T1 and T3 from the old table sum to 10,296. Similarly, there are 108 boards with two pair which allow flushes for each rank set with three elements. This gives us boards with two pair allowing flushes -- that is exactly the sum of the numbers of types D1 and D3 from the old table. We want to extract those boards that allow straight flushes.

In order to do the extraction, we have to look at the rank sets with three elements and distinguish between those allowing straights and those not allowing straights. If a rank set with three elements allows a straight and allows a flush, then it must simultaneously allow a straight flush. So whether or not a rank set with three elements allows a straight is crucial.

There are 12 rank sets of the form , all of which allow straights and six of which allow lows. There are 118 rank sets of the form , where there is a gap between y and the two successive ranks. When x = A, y can take on 10 values, whereas, for all other values of x, y can take on only 9 values. When x is an ace, deuce, queen or king, there are only two values of y that allow straights. When x is three or jack, there are three values of y allowing straights. For all other values of x, there are four values of y that allow straights. This gives 42 rank sets of this form that allow straights. Finally, there are 156 rank sets with gaps, but of these only 10 allow straights because they must have the form to allow a straight.

Now let's look at the table in the previous article. There are 56 rank sets allowing a low. Of these, exactly one-half of them also allow straights. Therefore, the old T1 breaks into 1,008 boards with trips allowing lows and straight flushes, while the remaining 1,008 boards allow lows and flushes. In the same way, the old T2 breaks into 4,368 boards allowing lows and straights, and 4,368 boards allowing lows and no sequential hand. The old boards D1 and D2 split in half in the same way.

There are 230 rank sets that do not allow a low, with 36 of them allowing straights. We then split the old boards T3, T4, D3 and D4 proportionately to obtain the expanded portion of the above table.

Suppose we see a board which allows a straight flush. What is the probability someone actually has a straight flush? First of all, we must recognize that not all boards allowing straight flushes are created equal. For example, if a board has the 3-5-7 of hearts in it, plus two stray cards, then a player must have 4-6 of hearts in order to have a straight flush. If, instead, a board has 3-5-6 of hearts plus two stray cards, then a player has a straight flush by having either 2-4 of hearts or 4-7 of hearts. It is pretty clear that the probability of someone having a straight flush in the second scenario is twice that of the first scenario.

Let's consider hold'em first. Suppose we have a board for which there are two specific cards producing a straight flush. If we now deal 10 random hands of two cards each from the remaining 47 cards, what is the probability that some hand contains the two cards that make the straight flush? This is easy to determine. There are C(47,20) ways of choosing the 20 cards to form the 10 hands. Once the 20 cards are chosen, there are ways to partition the 20 cards into 10 hands of two cards each. The product of the two numbers gives us the total number of ways of dealing 10 random hold'em hands from 47 cards, where we ignore which hand goes to which player.

Next we count the number of semi-deals for which the two specific cards are in the same hand. The two cards are set aside and placed in one hand. The product of C(45,18) and 17!! is the number of ways of completing the semi-deal to nine other hands. We divide the latter product by the preceding product. Most of the terms cancel and we are left with a probability of 10/1081 = .009251 that some player holds the two required cards.

If the board allows a straight flush in two different ways, where each of the ways requires two cards (the second scenario above), then the probability of someone having a straight flush is 20/1081 = .018502. If the board allows straight flushes in three different ways, where each of the ways requires two cards, then it is possible for two players to have straight flushes simultaneously. In this case, the probability of at least one straight flush being present is 329/11891 = .02767.

Hold'em has one feature not found in Omaha, namely, that players may have straight flushes by playing the board or using only one card from their hands. The probability of a straight flush by playing the board is the probability of the board itself being a straight flush. This probability is 40/2598960 = .00001539. Of more interest is the situation in which only one card is required for a player to make a straight flush. For example, if the board has the 4-5-6-8 of hearts and a stray, a player has a straight flush if she has either the 2-3 or 7 of hearts. If there is a single card and no other giving a player a straight flush, the probability of a straight flush being present is 20/47 = .4255 because the probability of it not being present is simply C(46,20) divided by C(47,20). If there are two single cards which result in straight flushes, then the probability is 750/1081 = .6938 that someone has a straight flush. Finally, if there is a single card producing a straight flush and a combination of two cards producing a straight flush (for example, the 4-5-6-8 of hearts and a stray on board), then the probability of a straight flush being present is 466/1081 = .4311.

In the next article we shall complete our discussion of straight flushes.

Home | Publications | Preprints | MITACS | Poker Digest | Poker Computations | Feedback
website by the Centre for Systems Science
last updated 16 September 2001