17 April 2000
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 , then the rank set is . 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 . 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.
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 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.
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.
No flushes are possible when the 2 cards are offsuit implying there are 432 boards in this case.
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.
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 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.
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.
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.
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.
The total number of boards for any player's hand is
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
ways to pick 2 other
distinct ranks from
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
allows 8 values for x and
is the ninth.
There is exactly
one rank set of both Type C and Type D giving a straight, namely,
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,
The number of boards producing a straight from A,2 is then
As a second example, consider a player holding 5,6 offsuit. There are 3 rank sets of Type A producing isolated straights. There are rank sets of Type A not containing a 4 but containing 7,8 and 9 because we are choosing the remaining 2 ranks from . Of the Type A rank sets which contain 4,7,8, there are which do not use 3 as in the previous line. Similarly, there are rank sets of Type A containing 3,4,7 but not using 2. Finally, there are 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.
|6,7; 7,8; 8,9||93||29||4||4||58||8||8||4|
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.
|6,7; 7,8; 8,9||192,430||.0908||180,735||.0853|