# Computations for Losing Flushes

Brian Alspach

20 January 2003

### Abstract:

We provide details of the computations for a player making a flush in hold'em with two suited cards in his hand, three cards of that suit on board, and at least one other player making a bigger flush.

This file provides the details for the numbers and probabilities given in the article Losing Flushes'' that appeared in Poker Digest, Vol. 3, No. 22, October 20 - November 2, 2000.

We assume a 10-handed hold'em game in which one player holds two suited cards -- for simplicity we shall say two hearts -- and the board finishes with three hearts and two cards of other suits. The player is interested in the probability of someone having a bigger heart flush given nine other random hands. We now give the details of calculating this.

The one assumption we make is that we ignore straight flushes. There are 45 unseen cards from which we choose 18 to be dealt to the other players. There are ways to do this. Once the 18 cards have been chosen, there are 17!! ways to break them into nine hands of two cards each. Since we are interested only in whether or not other players also receive two hearts and not in which players get them, we are interested in semi-deals. The number of semi-deal completions then is

 (1)

The first step is to partition the choices of 18 cards to be dealt to the players into sets according to how many hearts they contain. Let denote the number of 18-card sets with exactly hearts. It is easy to see that

because we are choosing hearts from 8, and choosing cards from the 37 non-hearts. This yields

Summing the values gives us so that we know the numbers above are correct. Multiplying by 17!! gives us the number of semi-completions with exactly hearts distributed among the nine hands. Doing this and summing the numbers gives us (1) above.

The numbers in the table above tell us how many sets of 18-cards contain precisely hearts. For a given fixed set of 18 cards containing hearts, we know there are 17!! = 34,459,425 ways to partition the cards into nine two-card hands. Some of the partitions give one or more hands with two hearts in them. So our next step is to break the 34,459,425 partitions according to how hearts are distributed across the hands.

We illustrate this with one example and then describe the general calculation. As an example, suppose we have an 18-set with precisely 7 hearts in the 18 cards. How many of the partitions produce exactly two hands containing two hearts? First we choose four hearts from the seven to form the two hands with two hearts. This can be done in ways. The four hearts can form two hands in 3!! = 3 ways. There are three hearts left over and 11 non-hearts. We must make certain the three remaining hearts form hands with non-hearts. For the first heart there are 11 choices of a non-heart, for the second there are 10 choices of a non-heart, and for the last heart there are 9 choices for the non-heart. This leaves 8 non-hearts and there are 7!! = 105 ways to make 4 hands with these non-hearts. Altogether we have partitions producing exactly two hands with two hearts.

Now we describe the general procedure for an 18-set with hearts. If we want the number of partitions with exactly hands having two hearts, we must have so that there are enough hearts available. There are ways to choose the hearts. There are ways to form the hands with two hearts. There are hearts left over that must not be paired with hearts, and there are non-hearts. So there are choices of non-heart to be paired with the first heart, choices of non-heart to be paired with the second heart, and so on. We continue this times. Of course, when , the latter steps do not occur. After dealing with the hearts, there are non-hearts left. They can be formed into 2-card hands in ways. Multiplying the numbers gives the number of partitions with exactly hands having two hearts.

The next table gives the information for each in the range . The rows correspond to 18-sets with hearts, and the columns correspond to the number of partitions for which there are exactly hands with two hearts in the hand. In order to verify the arithmetic, note that the sum of the entries across any given row should sum to 17!! = 34,459,425.

 0 hands 1 hand 2 hands 3 hands 4 hands 34,459,425 0 0 0 0 34,459,425 0 0 0 0 32,432,400 2,027,025 0 0 0 28,378,350 6,081,075 0 0 0 22,702,680 11,351,340 405,405 0 0 16,216,200 16,216,200 2,027,025 0 0 9,979,200 18,711,000 5,613,300 155,925 0 4,989,600 17,463,600 10,914,750 1,091,475 0 1,814,400 12,700,800 15,876,000 3,969,000 99,225

We are now ready to determine the numbers of semi-deal completions with various numbers of other players having a heart flush as well as our given player. We simply use the tables above. First, the number of semi-deal completions with exactly four other players having flushes is obtained by multiplying 99,225 by because is the number of 18-sets containing 8 hearts, and 99,225 is the number of partitions for each of those sets having exactly four players also having flushes. We obtain

 (2)

semi-deal completions with four other players having flushes.

We look at the table above and find three values of for which there is an entry in the column for three hands having flushes. We multiply by the corresponding values of to obtain

 (3)

semi-deal completions with exactly three other players having flushes.

We continue in this way using the entries from the two tables above to obtain

 (4)

semi-deal completions with exactly two other players having flushes,

 (5)

semi-deal completions with exactly one other player having a flush, and

 (6)

semi-deal completion with no other players having a flush. The sum of (2), (3), (4), (5) and (6) is (1) so that we know the arithmetic is valid.

The final step in this computation is to determine how many of the semi-deal completions lead to flushes that beat our given player. This is going to depend on how many missing hearts are bigger than the player's bigger heart in his hand. We are able to use a simple proportionality argument for this last step.

Let denote the number of unseen hearts of greater rank than the bigger of the two ranks in his hand. For example, if the player has the J-9 of hearts and the board has the 5-7-Q of hearts, then there are two hearts of larger rank among the unseen hearts. So in this case.

The next table gives the proportions of various semi-deal completions that beat the player. Again, we illustrate this with an example. Suppose we are in the situation where and we are working with semi-deal completions having exactly two other players with flushes. The essential principle involved here is that each possible combination of two 2-card heart hands from the 8 unseen hearts occurs the same number of times. Thus, all we need to do is work out the proportion of those involving one or more of the three cards of larger rank and multiply by (4). In this case, there are possible combinations of two 2-card hands. There are combinations of two 2-card hands from the five cards of smaller rank. Thus, 195 of the 210 combinations involve at least one card of bigger rank, and . Thus, 13/14 of the semi-deal completions in (4) produce another player with a higher flush.

The following table gives all the proportions. Note that a 0 entry means none of the semi-deal completions give a bigger flush (that is, the player's flush is the nut flush ignoring straight flushes), and an entry of 1 means that all the other flushes beat the player's flush. The rows correspond to the values of and the columns correspond to the number of other hands containing flushes.

 one hand two hands three hands four hands 0 0 0 0 1/4 1/2 3/4 1 13/28 11/14 27/28 1 9/14 13/14 1 1 11/14 69/70 1 1 25/28 1 1 1 27/28 1 1 1 1 1 1 1

We now are ready to obtain what we want. For each value of , we multiply the proportions in the preceding table by the appropriate values from (2), (3), (4) and (5) to get the number of semi-deal completions for which the player is facing a larger flush. For example, when , we multiply 9/14 and (5), 13/14 and (4), 1 and (3), 1 and (2), and take the sum. We divide the sum by (1) to get the probability.

The following table gives the results of these computations.

 Losing completions Probability of Losing 0 0 3,762,715,922,130,315,150 .06364 6,845,035,657,107,497,850 .11577 9,320,831,039,859,639,300 .15764 11,246,752,473,840,894,780 .19021 12,664,078,054,620,819,750 .21418 13,599,923,276,905,430,850 .23001 14,067,845,888,047,736,400 .23792

9-Handed Hold'Em

We now go through the same exercise for 9-handed hold'em. The total number of semi-deal completions in this case, replacing formula (1), is the product of and 15!! which is

 (7)

The next step is to break down the possible 16-sets according to how many hearts they contain. We let denote the number of sets of 16 cards with exactly hearts. Similar to the 10-handed case, we have

Any given 16-set can be partitioned in 15!! = 2,027,025 ways into eight 2-card hands. For each such set, we want to determine how many of them have different numbers of hands with flushes according to the number of hearts in the 16-set. We do the calculations similarly to the 18-set case. We list the results in the next table and note that each row must sum to 15!!

 0 hands 1 hand 2 hands 3 hands 4 hands 2,027,025 0 0 0 0 2,027,025 0 0 0 0 1,891,890 135,135 0 0 0 1,621,620 405,405 0 0 0 1,247,400 748,440 31,185 0 0 831,600 1,039,500 155,925 0 0 453,600 1,134,000 425,250 14,175 0 181,440 952,560 793,800 99,225 0 40,320 564,480 1,058,400 352,800 11,025

We now are ready to obtain the number of semi-deal completions having the various number of flushes for the other player. For example, multiplying 11,025 and gives the number of semi-deal completions with four other players having flushes. Performing the multiplication produces

 (8)

such semi-deal completions. As before, we obtain
 (9)

semi-deal completions with exactly three other players having flushes,
 (10)

semi-deal completions with exactly two other players having flushes,
 (11)

semi-deal completions with exactly one other player having a flush, and
 (12)

semi-deal completions with no one else having a flush.

The proportion table remains the same as before. We then multiply the appropriate proportions by the numbers of semi-deal completions in (8), (9), (10) and (ll) to get the table below for nine players.

 Losing completions Probability of Losing 0 0 74,142,185,657,740,200 .05657 135,229,434,970,429,800 .10317 184,529,392,470,423,000 .14078 223,020,029,727,439,560 .17015 251,416,923,652,055,700 .19181 270,191,158,262,417,700 .20614 279,578,275,567,598,700 .2133

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