Computations for Losing Flushes

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 ${{45}\choose{18}}$ 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

$\begin{displaymath}{{45}\choose{18}}17!!=59,128,393,062,047,809,500. \end{displaymath}$

(1)

The first step is to partition the ${{45}\choose{18}}=1,715,884,494,940$ 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

$\begin{displaymath}H(i)={{37}\choose{18-i}}{{8}\choose{i}}17\end{displaymath}$

$\begin{eqnarray*} H(0) & = & 17,672,631,900\\ H(1) & = & 127,242,949,680\\ H(2... ...69,523,888\\ H(7) & = & 6,839,937,216\\ H(8) & = & 348,330,136 \end{eqnarray*}$

Summing the

values gives us ${{45}\choose{18}}$ 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 ${{7}\choose{4}}=35$ 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 $35\cdot 3\cdot 11\cdot 10\cdot 9\cdot 105=10,914,750$ 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 $2j\leq i$ so that there are enough hearts available. There are ${{i}\choose{2j}}$ 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 $0\leq i \leq 8$ . 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

$\begin{displaymath}99,225H(8)=34,563,057,744,600 \end{displaymath}$

(2)

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

$\begin{displaymath}3,969,000H(8)+1,091,475H(7)+155,925H(6)= 16,935,898,294,854,000\end{displaymath}$

(3)

$\begin{displaymath}13,101,833,111,984,555,400 \end{displaymath}$

(5)

$\begin{displaymath}45,060,547,174,000,073,100 \end{displaymath}$

(6)

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 $3{{8}\choose{4}}=210$ possible combinations of two 2-card hands. There are $3{{5}\choose{4}} =15$ 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.

	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

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 ${{45}\choose{16}}$ and 15!! which is

$\begin{displaymath}1,310,727,925,020,764,250. \end{displaymath}$

(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

$\begin{eqnarray*} H'(0) & = & 12,875,774,670\\ H'(1) & = & 74,913,598,080\\ H'... ...753,243,808\\ H'(7) & = & 995,228,960\\ H'(8) & = & 38,608,020 \end{eqnarray*}$

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

$\begin{displaymath}1,031,149,649,453,165,550 \end{displaymath}$

(12)

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

Computations for Losing Flushes

Abstract: