Straight Completion

Brian Alspach

17 April 2000

Abstract:

We compute the probabilities for a variety of beginning hands in hold'em finishing with a straight.

I wish to thank Barbara Yoon and Mike Caro for contributions to the results.

The rank set of a board is the set of ranks appearing in the board. For example, if the board has $2\clubsuit, 2\diamondsuit, 5\heartsuit,J\clubsuit,J\spadesuit$, then the rank set is $\{2,5,J\}$. We assume the player has two cards of distinct ranks x and y. For the first step, we classify boards according to the relationship their rank sets have to the rank set $\{x,y\}$. For each rank set of a given type, we count the number of boards with that rank set which do not give the player a flush. The latter numbers depend on whether the player's cards are suited or not suited. For the second step, we use this information to count the number of straights as x and y vary.

Type A.
A Type A rank set comes from a board which contains 5 cards of distinct ranks a,b,c,d,e none of which are either x or y. Clearly, there are 45 = 1,024 choices for the 5 cards. Now we must eliminate the flushes. If the 2 cards of ranks x and y are suited, say clubs, then we must eliminate all choices having 3 or more clubs, or all 5 in the same suit. There are 3 ways all 5 cards are in the same non-club suit. There are $3^2{{5}\choose{3}}+3{{5}\choose{4}}+{{5}\choose{5}}=106$ choices with 3 or more clubs. Thus, if the 2 cards are suited, there are 1,024 - 109 = 915 boards of Type A which do not produce a flush.

If the 2 cards of ranks x and y are offsuit, say one club and one spade, then we must eliminate those choices having all five cards in the same red suit, or at least four of them in the same black suit. There are 2 ways of having the 5 cards in the same red suit. There are ${{5}\choose{5}}+3{{5}\choose{4}}=16$ ways of having 4 or more of them in clubs. Thus, there are 1,024 - 34 = 990 choices producing no flushes when the 2 cards are not suited.

Type B.
A Type B rank set arises from a board containing 5 cards of 4 distinct ranks a,b,c,d none of which are x or y. This implies there is a pair amongst the 5 cards. There are 4 choices for the rank of the pair, 6 choices for the pair of the given rank, and 43 = 64 choices for the remaining 3 cards. This yields 1,536 boards and again we must eliminate those which include flushes. If the 2 cards in the hand are suited, say clubs, any flush must be in clubs since there are only 4 ranks represented amongst the 5 board cards. Once the rank of the pair is determined, for which there are 4 choices, if the 3 remaining cards in the board are clubs, there is a flush no matter which pair is chosen. This gives 6 flushes. If only 2 of the remaining cards are clubs, then there must be a club in the pair. This gives 27 more flushes because there are 3 choices for the pair, the 2 ranks which are clubs, and the suit of the remaining card. This produces 4(27+6) =132 flushes leaving us with 1,404 boards when the 2 cards are suited.

When the 2 cards are not suited, say one spade and one club, we can get a flush in either spades or clubs. There are 3 ways to get a flush in either suit after the rank of the pair is known. Since there are 4 choices for the rank of the pair, there are 24 flushes present. This gives us 1,512 boards when the 2 cards are offsuit.

Type C.
A Type C rank set comes from a board having 5 cards of 3 distinct ranks a,b,c none of which are x or y, and with two pairs in the board. There are 3 choices for the 2 ranks having pairs, 6 choices for each pair, and 4 choices for the remaining card. This yields $3\cdot 6^2\cdot 4=432$Type C boards. When the 2 cards in the player's hand are suited, a flush is possible if each of the pairs has a card of that suit, and the remaining card is also of that suit. This gives us 3 possible pairs for each rank for which there is a pair, and there are 3 choices for the ranks of the pairs. Hence, there are 27 of the boards which produce a flush. We then have 405 boards of Type C not producing a flush when the 2 cards are suited.

No flushes are possible when the 2 cards are offsuit implying there are 432 boards in this case.

Type D.
A Type D rank set comes from 5 cards of 3 distinct ranks a,b,c none of which are x or y, and there is 3-of-a-kind (trips) in the board. There are 3 choices for the rank of the trips, 4 choices for the trips, and 4 choices for the cards of each of the other ranks. This gives us $3\cdot 4^3=192$ boards for a Type D rank set. If the 2 cards are suited, there are 9 of the choices which also are flushes because of 3 choices for the rank of the trips, and 3 choices for trips which include a card of the flush suit. This gives us 183 boards for 2 suited cards.

As in the preceding case, no flush is possible when the 2 cards are offsuit implying there are 192 boards when the 2 cards are not suited.

Type E.
A Type E rank set arises from 5 cards of distinct ranks a,b, c,d,e one of which coincides with either x or y. There are 3 choices for the card of rank x or y (we are assuming the duplicated rank is fixed) and 4 choices for each of the remaining 4 ranks. This produces $3\cdot 4^4=768$ boards from which we must remove the flushes. Let's suppose the 2 cards in the hand are clubs. As noted above, there are 3 choices for the card of rank x or y as it cannot be a club. We obtain a flush if 3 or more of the remaining cards are clubs. There is 1 choice for all 4 to be clubs and there are $3{{4}\choose{3}}=12$ choices for the cards if precisely 3 of them are clubs. Thus, there are 3(12+1)=39 ways of getting a club flush. In addition, if all 5 cards on board are in the same non-club suit, we obtain a flush giving us 42 flushes altogether. Hence, there are 768 - 42 = 726 boards when the 2 cards are suited.

If one card is a spade and the other is a club, a flush may arise if the pairing card is red, and either all 4 of the other cards are in the same red suit or in the same black suit. This gives 6 flushes when the pairing card is red. The other possibility is for the pairing card to be black (meaning only 1 choice since the duplicating rank is fixed) and at least 3 of the remaining cards are in the same black suit. We then get a flush in another $1+1+3{{4}\choose{3}})=14$ ways because the remaining 4 cards can be all clubs, all spades or 3 in the same black suit as the pairing card. This yields 20 flushes leaving 748 boards when the 2 cards are not suited.

Type F.
A Type F rank set comes from 5 cards of 4 distinct ranks a,b,c,d including one card of rank x or y. This means there is a pair in the board of rank different from x and y. There are 3 choices for the card of rank x or y, there are 3 choices for the rank of the pair, there are 6 choices for a pair of that rank, and there are 4 choices for each of the other 2 cards. Altogether there are $3^2\cdot 6\cdot 4^2 = 864$ boards. Again we do the usual job of removing flushes.

If the 2 cards are both clubs, then the 3 remaining ranks must all have clubs in order for a flush to arise. There are 3 choices for the card of rank x or y, there are 3 choices for the rank of the pair on board, and there are 3 pairs involving a club in the pair. This gives us 27 flushes leaving 837 boards when the 2 cards are suited.

If one of the cards is a spade and the other is a club, then the other card of rank x or y must be the remaining black card, say club, of that rank. In addition, the pair must include a club and the other 2 cards must be clubs. There are 3 choices for the rank of the pair and there are 3 choices for the pair. Hence, there are 9 flushes leaving us 855 boards when the 2 cards are offsuit.

Type G.
A Type G rank set comes from 5 cards of 4 distinct ranks but now the pair on board is of rank x or y. We assume the duplicating rank is fixed. There are 3 choices for the pair and there are 4 choices for each of the remaining cards. This gives $3\cdot 4^3=192$ boards. Let's remove the flushes. If both cards are clubs, there are 3 choices for the pair and no choice for the remaining 3 cards all of which must be clubs. This gives us 3 flushes leaving 189 boards when the 2 cards are suited.

When one card is a club and the other is a spade, then there are 2 pairs which include the right black card and the remaining 3 cards are forced to be in the same suit. This yields 190 boards when the 2 cards are not suited.

Type H.
A Type H rank set comes from 5 cards of 5 distinct ranks one of which is x and another which is y. Thus, the player has two pairs. There are 3 choices for each of the pairing cards, and there are 4 choices for each of the remaining cards. This gives $3^2 \cdot 4^3=576$ boards of this type. If the 2 cards both are clubs, there are 9 club flushes because there is a choice of 3 for each of the pairing cards, but all remaining cards must be clubs. There are 3 flushes in suits other than clubs. This gives 12 flushes leaving 564 boards when the 2 cards are suited.

If one card is a spade and the other a club, one way of getting a flush is for all 5 cards on board to be in the same red suit. This gives us 2 flushes. Another possibility is for one pairing card to be black and all 3 cards on board not of ranks x or y to be in the same black suit. This gives us a choice of 3 pairing cards for the other pair. Altogether this leads to 8 flushes. Hence, there are 568 boards of Type H which do not contain a flush when the 2 cards are offsuit.


Now we are going to determine the probability that a player holding 2 cards of distinct ranks x and y finishes with a straight. Notice that one can end up with a straight only if the board is one of the types described above. The following table summarizes the number of boards of the types described above which do not produce a flush when a player has two cards of distinct ranks. The column headed by ``Offsuit'' gives the number when the two cards are not suited, and the column headed ``Suited'' gives the number when the two cards are suited. We shall use this information repeatedly.


Type Offsuit Suited
A 990 915
B 1,512 1,404
C 432 405
D 192 183
E 748 726
F 855 837
G 190 189
H 568 564


The total number of boards for any player's hand is

\begin{displaymath}{{50}\choose{5}}=2,118,760\end{displaymath}

which we also use repeatedly to arrive at the probabilities below.


The completion of the problem is straightforward. For each player's hand with cards of ranks x and y, we count the number of rank sets of each type above which given the player a straight. We multiply the respective numbers by the corresponding entries from the preceding table and sum them to get the total number of ways for the player to finish with a straight. If x and y are not suited, we use the entries from the offsuit column, and if x and y are suited, we use the entries from the suited column. This means we are now interested in determining the number of rank sets of each type giving the player a straight.

We illustrate the process with two examples. First, consider a player holding A,2. Because of the ace in the hand, we can finish with big straights and small straights using cards in the player's hand. In order for her to finish with a straight, either there is a straight on board not using the deuce or ace (we call such a straight an isolated straight, or the board has ranks 3, 4 and 5 present, or the board has ranks 10, J, Q, K present. There are ${{8}\choose{2}}=28$ ways to pick 2 other distinct ranks from $\{6,7,...,Q,K\}$ so that there are 28 rank sets of Type A for A,2 which include 3, 4, 5. If ranks 10, J, Q, K are present, there are 7 choices for the rank of the remaining card. There are an additional 5 rank sets of Type A not already included producing isolated straights (for example, 5,6,7,8,9 is one such rank set). Altogether this gives us 40 rank sets of Type A. There are 9 rank sets of Type B since $\{3,4,5,x\}$ allows 8 values for x and $\{10,J,Q,K\}$ is the ninth. There is exactly one rank set of both Type C and Type D giving a straight, namely, $\{3,4,5\}$. There are 18 rank sets of Type E because there are 2 choices for the rank overlapping A or 2, and there are 9 choices for the remaining 4 ranks as in type B. There are 2 rank sets of Type F because there are 2 choices for the overlapping rank and the remaining ranks must be 3,4,5. There are 2 rank sets of Type G for the same reason. It is easy to see there is only 1 rank set of Type H, namely, $\{A,2,3,4,5\}$. The number of boards producing a straight from A,2 is then

\begin{displaymath}(40\cdot 990) + (9\cdot 1512)+432+192+(18\cdot 748)+(2\cdot 855)+ (2\cdot 190)+568 = 69,954.\end{displaymath}

As a second example, consider a player holding 5,6 offsuit. There are 3 rank sets of Type A producing isolated straights. There are ${{7} \choose{2}}=21$ rank sets of Type A not containing a 4 but containing 7,8 and 9 because we are choosing the remaining 2 ranks from $\{A,2,3, 10,J,Q,K\}$. Of the Type A rank sets which contain 4,7,8, there are ${{7}\choose{2}}$ which do not use 3 as in the previous line. Similarly, there are ${{7}\choose{2}}$ rank sets of Type A containing 3,4,7 but not using 2. Finally, there are ${{8}\choose{2}}$ rank sets of Type A containing 2,3 and 4. Altogether we obtain 94 rank sets of Type A for which the player ends up with a straight.

We perform a similar calculation for Type B rank sets. There are 7 rank sets of Type B containing 7,8,9 but no 4. There are 7 rank sets of Type B containing 4,7,8 but no 3. There are 7 containing 3,4,7 but no 2, and there are 8 containing 2,3,4. This produces 29 rank sets of Type B for which the player end up with a straight.

It is easy to see there are 4 rank sets of Type C and 4 rank sets of Type D for which the player finishes with a straight. There are 58 rank sets of Type E because it is like type B except for 2 choices for the overlapping rank. Similar reasoning gives 8 rank sets of Type F, 8 rank sets of Type G, and 4 rank sets of Type H.

Before including all the information in a table, let us point out some symmetry in the results. The number of boards giving a straight for cards of ranks x and y is the same as the number of boards giving a straight for cards of ranks 15-x and 15-y. This follows by simply interchanging the roles of cards of rank x and cards of rank 15-x. Thus, for example, the number of completions to a straight for 3,4 is the same as the number for J,Q.

In the following table we include the numbers of rank sets of each type for a variety of player hands. The lettered column headings indicate the types of boards.

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
A,3; A,Q 45 10 1 1 20 2 2 1
2,4; K,J 60 16 2 2 32 4 4 2
3,5; Q,10 80 23 3 3 46 6 6 3
4,6; J,9 79 23 3 3 46 6 6 3
5,7; 8,10 84 24 3 3 48 6 6 3
6,8; 7,9 83 24 3 3 48 6 6 3
A,4; A,J 50 11 1 1 22 2 2 1
2,5; K,10 65 17 2 2 34 4 4 2
3,6; Q,9 64 17 2 2 34 4 4 2
4,7; J,8 69 18 2 2 36 4 4 2
5,8; 7,10 74 19 2 2 38 4 4 2
6,9 73 19 2 2 38 4 4 2
A,5; A,10 55 12 1 1 24 2 2 1
2,6; K,9 49 11 1 1 22 2 2 1
3,7; Q,8 54 12 1 1 24 2 2 1
4,8; J,7 59 13 1 1 26 2 2 1
5,9; 6,10 64 14 1 1 28 2 2 1


The final table gives the number of boards for which the player finishes with a straight. The column headed by ``Offsuit'' is the number of boards when the two cards are not suited, and the next column is the corresponding probability. The column headed by ``Suited'' is the number of boards when the two cards are suited, and the final column is the corresponding probability.

Hands Offsuit Prob. Suited Prob.
A,2; A,K 69,954 .033 65,508 .0309
2,3; K,Q 106,134 .05 99,573 .047
3,4; Q,J 150,272 .0709 141,069 .0666
4,5; J,10 194,410 .0918 182,565 .0862
5,6; 9,10 193,420 .0913 181,650 .0857
6,7; 7,8; 8,9 192,430 .0908 180,735 .0853
A,3; A,Q 77,912 .0368 72,939 .0344
2,4; K,J 114,092 .0538 107,004 .0505
3,5; Q,10 158,230 .0747 148,500 .0701
4,6; J,9 157,240 .0742 147,585 .0697
5,7; 8,10 165,198 .078 155,016 .0732
6,8; 7,9 164,208 .0775 154,101 .0727
A,4; A,J 85,870 .0405 80,370 .0379
2,5; K,10 122,050 .0576 114,435 .054
3,6; Q,9 121,060 .057 113,520 .0536
4,7; J,8 129,018 .0609 120,951 .0571
5,8; 7,10 136,976 .0646 128,382 .0606
6,9 135,986 .0642 127,467 .0602
A,5; A,10 93,828 .0443 87,801 .0414
2,6; K,9 84,880 .0401 79,455 .0375
3,7; Q,8 92,838 .0438 86,886 .041
4,8; J,7 100,796 .0476 94,317 .0445
5,9; 6,10 108,754 .0513 101,748 .048

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