# Straight Shooting: Part II

Brian Alspach

Poker Digest Vol. 3, No. 10, May 5 - 18, 2000

In Part I of this series (Vol. 3/No. 9 of Poker Digest), we presented the following table which we shall also use in this article.

The left column of the table designates the eight types of rank sets defined in Part I.

The right column gives the number of boards for the given rank set which lead to a player holding two offsuit cards of distinct ranks not ending up with a flush.

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

We now assume the player is holding offsuit connectors of ranks xand x+1. The player can end up with a straight on the river only if the board is one of the rank sets of Types A through H. To determine the number of boards for which the player finishes with a straight, all we need do is count the number of rank sets of each type giving the two cards of ranks x and x+1 a straight and multiply that number by the entry in the table, and then sum these numbers. Dividing by 2,118,760, the total number of boards, gives us the probability the player's hand finishes with a straight. What could be easier?

We illustrate with an example which should clarify the procedure.

Suppose the player has offsuit A-2. The Type A rank sets giving the player a straight are , where x and y are chosen from ; , where z is chosen from ; and , where u is chosen from . There are C(8,2) = 28 choices for x and y; there are seven choices for z; and there are five choices for u. This gives us 40 Type A rank sets for A-2.

The Type B rank sets giving A-2 a straight are and , where x is chosen from . This gives us nine Type B rank sets. The only rank set of Types C and D is so that there is one of each.

A Type E rank set contains either A or 2 with the four remaining ranks forming a Type B rank set. Thus, there are twice as many Type E rank sets as there are Type B rank sets -- 18 of them. Similarly, there are two rank sets of Type F and two of Type G. Finally, is the only Type H rank set.

The next table gives the number of rank sets of each type for all possible connectors.

 Hands A B C D E F G H A,2; A,K 40 9 1 1 18 2 2 1 2,3; K,Q 55 15 2 2 30 4 4 2 3,4; Q,J 75 22 3 3 44 6 6 3 4,5; J,10 95 29 4 4 58 8 8 4 5,6; 9,10 94 29 4 4 58 8 8 4 6,7; 7,8; 8,9 93 29 4 4 58 8 8 4

We use the preceding table and the first table, as described above, to compute the number of completions of the given hands to straights. The next table gives the total number of boards for which the given hand achieves a straight (and not a flush). Let me remind you again that these hands are not suited.

 Hands Number of Completions Probability A,2; A,K 69,954 .033 2,3; K,Q 106,134 .05 3,4; Q,J 150,272 .0709 4,5; J,10 194,410 .0918 5,6; 9,10 193,420 .0913 6,7; 7,8; 8,9 192,430 .0908

For the remainder of this article we combine the preceding two tables into one table. The column labelled N is the total number of completions yielding a straight, while the column labelled P is the probability of finishing with a straight. Again, these hands are not suited.

 Hands A B C D E F G H N P A,3; A,Q 45 10 1 1 20 2 2 1 77,912 .0368 2,4; K,J 60 16 2 2 32 4 4 2 114,092 .0538 3,5; Q,10 80 23 3 3 46 6 6 3 158,230 .0747 4,6; J,9 79 23 3 3 46 6 6 3 157,240 .0742 5,7; 8,10 84 24 3 3 48 6 6 3 165,198 .078 6,8; 7,9 83 24 3 3 48 6 6 3 164,208 .0775 A,4; A,J 50 11 1 1 22 2 2 1 85,870 .0405 2,5; K,10 65 17 2 2 34 4 4 2 122,050 .0576 3,6; Q,9 64 17 2 2 34 4 4 2 121,060 .057 4,7; J,8 69 18 2 2 36 4 4 2 129,018 .0609 5,8; 7,10 74 19 2 2 38 4 4 2 136,976 .0646 6,9 73 19 2 2 38 4 4 2 135,986 .0642 A,5; A,10 55 12 1 1 24 2 2 1 93,828 .0443 2,6; K,9 49 11 1 1 22 2 2 1 84,880 .0401 3,7; Q,8 54 12 1 1 24 2 2 1 92,838 .0438 4,8; J,7 59 13 1 1 26 2 2 1 100,796 .0476 5,9; 6,10 64 14 1 1 28 2 2 1 108,754 .0513

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