# Straight Shooting: Conclusion

Brian Alspach

Poker Digest Vol. 3, No. 11, May 19 - June 1, 2000

In this last part of my Straight Shooting series, we shall determine the probability that a player holding a variety of suited cards in hold'em will end up with a straight. Let's take a quick review of the types of boards under discussion. We are assuming the player is holding two suited cards of ranks x and y. A Type A rank set comes from a board having five cards of distinct ranks none of which are x or y; a Type B rank set comes from a board having five cards of four distinct ranks none of which are x or y; a Type C rank set comes from a board having five cards of three distinct ranks none of which are x or y and two pair on board; a Type D rank set is the same as Type C except there is a three-of-a-kind on board; a Type E rank set is the same as Type A except one of the board cards has rank x or y; a Type F rank set is like Type B except there is one card of rank x or y; a Type G rank set is like Type B except the pair on board has rank x or y; and a Type H rank set corresponds to five cards of five distinct ranks including one card of rank x and another card of rank y.

Let's quickly review how we determined the probability a player holding two unsuited cards ends up with a straight. For example, suppose the player is holding 6-8 offsuit. We first count the number of Type A rank sets producing a straight for 6-8. There are 1,024 boards for a given Type A rank set because there are four choices for each rank, but 34 of them give the player a flush. Thus, for each Type A rank set giving 6-8 a straight, there are 990 boards producing a straight. Multiplying the number of Type A rank sets by 990 gives the number of straights arising from Type A rank sets.

We then perform a similar computation for all of the other possible types of rank sets and sum the numbers to get the total number of boards producing a straight for a player holding an offsuit 6-8. The process is the same for all other hands.

Since the number of rank sets of each type producing a straight for our player holding two cards of ranks x and y is independent of whether or not the cards are suited, we already have much of the information we need to compute the desired probabilities. We simply use the numbers from Part II. However, there is one difference. The number of boards for a given rank set which do not produce a flush for the player does depend on whether or not the player's cards are suited. So we must re-do the table which was given in Part I for suited cards. We illustrate with an example.

For simplicity let's assume the player's two cards are hearts. The number of boards for a given Type A rank set is 1,024 because there are four choices for each card. If all five are in the same non-heart suit, the player has a flush on board. There are three such choices. The player makes a heart flush if three or more of the board cards are hearts. There is one way for all of them to be hearts. To get four hearts on board, there are five choices for the rank of the non-heart and three choices for the card of that rank. This gives us fifteen heart flushes with four hearts on board. Finally, to get three hearts on board, there are C(5,2) = 10 choices for the two ranks which are not hearts, and there are three choices for each card of those two ranks. This gives us ninety flushes with three hearts on board. Altogether there are 1 + 15 + 90 + 3 = 109 flushes for the player.

We must remove these 109 flushes from the 1,024 boards leaving 915 boards for which the player does not have a flush. Contrast this with the value of 990 when the player's two cards are not suited.

The next table gives the number of boards not giving the player a flush for each type of rank set when the player's two cards are suited. The numbers can be found in a manner similar to that just used.

 Type Number A 915 B 1,404 C 405 D 183 E 726 F 837 G 189 H 564

As mentioned above, the table in part II contains the information of how many rank sets of each type there are for particular hands. We replicate the table below for completeness. The total number of boards giving the player a straight is then calculated as described above and displayed in the column labelled N. The last column gives the probability. We remind you the hands are suited.

 Hands A B C D E F G H N Prob. A,2; A,K 40 9 1 1 18 2 2 1 65,508 .0309 2,3; K,Q 55 15 2 2 30 4 4 2 99,573 .047 3,4; Q,J 75 22 3 3 44 6 6 3 141,069 .0666 4,5; J,10 95 29 4 4 58 8 8 4 182,565 .0862 5,6; 9,10 94 29 4 4 58 8 8 4 181,650 .0857 6,7; 7,8; 8,9 93 29 4 4 58 8 8 4 180,735 .0853 A,3; A,Q 45 10 1 1 20 2 2 1 72,939 .0344 2,4; K,J 60 16 2 2 32 4 4 2 107,004 .0505 3,5; Q,10 80 23 3 3 46 6 6 3 148,500 .0701 4,6; J,9 79 23 3 3 46 6 6 3 147,585 .0697 5,7; 8,10 84 24 3 3 48 6 6 3 155,016 .0732 6,8; 7,9 83 24 3 3 48 6 6 3 154,101 .0727 A,4; A,J 50 11 1 1 22 2 2 1 80,370 .0379 2,5; K,10 65 17 2 2 34 4 4 2 114,435 .054 3,6; Q,9 64 17 2 2 34 4 4 2 113,520 .0536 4,7; J,8 69 18 2 2 36 4 4 2 120,951 .0571 5,8; 7,10 74 19 2 2 38 4 4 2 128,382 .0606 6,9 73 19 2 2 38 4 4 2 127,467 .0602 A,5; A,10 55 12 1 1 24 2 2 1 87,801 .0414 2,6; K,9 49 11 1 1 22 2 2 1 79,455 .0375 3,7; Q,8 54 12 1 1 24 2 2 1 86,886 .041 4,8; J,7 59 13 1 1 26 2 2 1 94,317 .0445 5,9; 6,10 64 14 1 1 28 2 2 1 101,748 .048

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