# Straight Shooting: Part I

Brian Alspach

Poker Digest Vol. 3, No. 9, April 21 - May 4, 2000

First, I would like to acknowledge valuable contributions from Barbara Yoon and Mike Caro with regard to this topic. As usual I was running about 10 days late in reading the RGP newsgroup (rec.gambling.poker) and Barbara was the first person to alert me to a discussion revolving around the probability of a player holding unsuited connectors finishing with a straight. That is the first topic this mini-series addresses.

I don't intend to go through all the gory details of working out the probabilities, but I want to provide a clear framework for carrying out the computations so interested readers can fill in the details themselves. You also can check out http://www.math.sfu.ca/ãlspach under poker computations for further explanation. There are two main steps involved in working out the correct numbers.

First, we must determine the types of boards which allow a player's hand to be completed to a straight. Second, for each type of board determined in the preceding step, we must eliminate the corresponding boards which give the player a flush. The boards left over are the number which complete the player's hand to a straight. For each player's hand we multiply the number of boards of each type by the number of boards not producing a flush and sum the resulting products. This gives us the total number of boards completing the player's hand to a straight. We divide this by the total number of boards to get the probability. The total number of boards is C(50,2) = 2,118,760.

We're going to define the types of boards in terms of rank sets. The rank set of a board is the set of ranks appearing on board -- we don't count multiplicities. Thus, a board with 3-3-8-J-J has rank set 3,8,J.

We're interested in the rank sets and their relationships with the ranks of the player's two cards. Throughout the rest of this article, we assume the player has two cards of ranks x and y, where x and y are distinct.

Type A.
A Type A rank set comes from a board which contains five cards of distinct ranks a-b-c-d-e none of which is either x or y.
Type B.
A Type B rank set comes from a board containing five cards of four distinct ranks a-b-c-d none of which is x or y. This implies there is a pair among the five cards.
Type C.
A Type C rank set has five cards of three distinct ranks a-b-c none of which is x or y, and with two pair on the board.
Type D.
A Type D rank set has five cards of three distinct ranks a-b-c none of which is x or y, and there is three-of-a-kind (trips) on the board.
Type E.
A Type E rank set has five cards of distinct ranks a-b-c-d-e one of which coincides with either x or y.
Type F.
A Type F rank set has five cards of four distinct ranks a-b-c-d including one card of rank x or y. This means there is a pair on the board of rank different from x and y.
Type G.
A Type G rank set has five cards of four distinct ranks but now the pair on board is of rank x or y.
Type H.
A Type H rank set has five cards of five distinct ranks one of which is x and another which is y. Thus, the player has two pair.

For a player holding connectors, that is, two cards of ranks x and x+1, the only way to reach the river and make a straight is to have one of the rank sets described above. Next we determine the number of boards for each rank set which do not produce a flush. These numbers don't depend on the player having connectors. They are valid for any player hand of two distinct ranks x and y with the two cards not suited.

There are four choices for each of the five board cards in a Type A rank set giving 1,024 boards for a given Type A rank set. Now we must eliminate the flushes among these 1,024 boards.

In order to simplify the discussion, let's assume one of the cards of ranks x and y is a club and the other a spade. We get flushes when all five cards are in the same red suit, or at least four of them in the same black suit. There are two ways of having the five cards in the same red suit. There are C(5,5) + 3C(5,4) = 16 ways of having four or more of them in clubs because there are three choices for the suit of the card not in clubs. Thus, there are 1,024 - 34 = 990 choices producing no flushes since we have to double the 16 to eliminate flushes in both black suits.

Let's do one more explicitly which should give you the idea. Consider rank sets of Type C. They have the form , where none of the elements coincides with x or y, and there are two pair on board. We have three choices for the rank which is not paired; we have four choices for this card; and we have six choices for each of the two pair. This gives us 432 boards for a given rank set of Type C. However, in this case no flushes can occur because the cards of ranks x and y are not suited.

So we have 432 boards not producing a flush for our player when the board has rank set of Type C.

We perform similar calculations for the other types of rank sets and obtain the information in the table below. The first column indicates the type of the rank set. The second column gives the number of boards for a rank set of that type not producing a flush for a player holding two unsuited cards of different ranks. We shall use the information in this table repeatedly in subsequent articles.

 Type Number A 990 B 1,512 C 432 D 192 E 748 F 855 G 190 H 568

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