7 December 2001
We first determine the probability for a 10-handed hold'em game that
there at least two players holding pocket pairs of the same rank. Since
we do not care which players have the pairs, we are interested in
player-semideals. The total number of player-semideals is given by
(1) |
We use inclusion-exclusion because there may be more than one rank with
two pairs of that rank dealt. The maximum number of such ranks is 5.
We are then choosing 5 ranks from 13, and for each rank there are 3 ways
we may partition the 4 cards of that rank into two hands. Thus, the total
number player semi-deals with 5 ranks, where there are two pairs dealt for
each of the 5 ranks, is given by
(2) |
Now consider player-semideals with 4 ranks having two pairs of the ranks
dealt. There are 4 choices of ranks from 13, there are 3 ways of choosing
the two pairs for a given rank, there are 4 choices from 36 for the other
two hands, and there are 3 ways of choosing the two hands from the latter
4 cards. This gives us
10,232,885,520 | (3) |
We now move to 3 ranks. There are 3 choices from 13 ranks, 3 ways of
choosing the pairs of each of the ranks, 8 choices of cards from 40,
and 7!! ways to partition the 8 cards into 4 hands. Multiplying all
of this gives player-semideals with 3 or more sets of pairs of the same
ranks. Those counted in (2) are counted 10 times and those in (3) are
counted 4 times. Thus, the number of player-semideals with precisely 3
ranks having two players with pairs of these ranks is given by
(4) |
When we get to 2 ranks, there are 78 choices for the ranks, 3 ways of
choosing two pairs of each of the ranks, 12 choices from 44 cards for
the other hands, and 11!! ways of partitioning the 12 cards into 6 hands.
This gives us
153,717,823,465,652,160 | (5) |
This brings us to one rank for which two players are dealt pairs of that
rank. There are 13 choices for the rank, 3 ways of partitioning the 4
cards into two hands, 16 choices of cards from 48 for the remaining cards,
and 15!! ways of partitioning them into 8 hands. The product is
177,947,147,017,903,760,640 | (6) |
We take the sum of the player-semideals in (2) through (6) obtaining
178,100,927,165,755,586,421 | (7) |
Dividing the number in (7) by the number in (1) gives a probability of
Another question arising now and then is how often two players will be dealt pocket aces. In other words, how often will two players be dealt pairs of the same fixed rank? We may as well use aces to illustrate this, but the answer is the same for any fixed rank.
Each rank occurs equally often in the player-semideals counted in (6). So dividing by 13 gives the number of player-semideals for which two players are dealt pocket aces, and no other players have pocket pairs of the same rank as anybody else.
What about the player-semideals counted in (5)? Here there are two ranks
involved. There are 78 ways of choosing two ranks from 13 and 12 of those
choices involve aces. Thus, 2/13-ths of the player-semideals counted in
(5) involve aces. Similarly, it is 3/13 for (4), 4/13 for (3), and 5/13
for (2). Taking these proportions of the numbers, we obtain
Finally, let's turn our attention to conditional probabilities. This sometimes can produce surprising results. Suppose you are playing hold'em and find that you have been dealt a pair of rank x. What is the probability someone else also has been dealt a pair of rank x? We may as well use aces and again observe the result is the same for any fixed rank.
In this case, there have been 18 cards chosen from 50. Thus, there are
(8) |
Suppose a player has been dealt a pocket pair of rank x in a 10-handed hold'em game. In the last section we saw there is a probability of about 1/136 that another player has been dealt a pocket pair of the same rank. That is a small probability. What is of more interest is the probability the player is facing one or more pocket pairs of larger rank. This is the problem we deal with in this section.
Of course, if the player has pocket aces, no one can have a pair of larger rank. So the probability is 0 when the player has pocket aces.
We are going to work down the ranks. The number given in (8) above is the number of completions to player-semideals. We use it over and over.
Let's suppose the player has pocket kings. Two players can have pocket
aces. There are 3 ways the four aces can be split into two hands, and
the remaining 7 hands can happen in
ways. This
gives us
Now let's count the completions where there is at least one player with
pocket aces. There are 6 choices for the pair of aces and
choices for the remaining 8 players. Multiplying yields
Performing the indicated subtraction yields
27,326,563,159,384,514,025 | (9) |
For a player holding pocket queens, there are now two larger ranks. This
means the number of subcases begins to proliferate. Let f_{4} denote
the number of completions with 4 players holding pocket pairs of larger
ranks. There are 3 ways to split the aces into two pairs, and similarly
for the kings. The remaining 5 hands arise from 10 cards being chosen
from 42, and then splitting the 10 cards in 9!! ways into 5 hands. This
yields
Let f_{3} denote the number of completions for which a particular subset of 3 players receive pocket pairs of larger ranks. This means that each completion counted in f_{4} contributes 4 to f_{3}.
There are two ways to split the ranks for contributing to f_{3}. Either
two players have kings or two players have aces. There are then 6 choices
for the isolated pair, 3 ways of splitting the four cards of the rank for
which there are two pairs, and 6 additional hands coming from 44 cards.
This yields
Now let f_{2} denote completions for one pair of pocket kings and one
pair of pocket aces. Since there are 6 choices for each pair, we see
that
Let g_{2} denote completions for two pairs of the same rank. Since there
are two choices for the rank and 3 choices for splitting the 4 cards into
two pairs, the 36 in the calculation for f_{2} is replaced by 6. This
means g_{2} is one-sixth of f_{2}, that is,
Finally, f_{1} denotes the completions with a fixed pair of larger rank.
There are two choices for rank and 6 choices for a pair of that rank.
This produces
Now we want to use the values determined above to work out some probabilities for what the player holding Q-Q is facing.
By working through the various patterns of occurrences of pocket pairs
of bigger rank, it is easy to verify that the expression
f_{1}-g_{2}-2f_{2}+2f_{3}-2f_{4} = 52,334,945,062,562,602,440 | (10) |
Similarly, the expression
f_{2}-f_{3}+f_{4} = 1,159,090,628,103,212,805 | (11) |
Dividing these numbers by the number in (8) gives us a probability of
Since there are getting to be more and more possible ranks available for larger pocket pairs, we are going to introduce different notation. Let f(i,j) denote incidences of i larger ranks for which there is one pair of the rank dealt to someone, and j ranks for which there have been two pairs of the rank dealt. Thus, f(2,1) denotes the number of incidences of player semideal completions where there are two ranks for which there is a single pair, and one rank for which two pairs of that rank have been dealt. This means four players have their hands determined ahead of time and no restriction is placed on what the other five players are dealt. There could be more pairs dealt to some of the other five players. This is why we say we are counting incidences of player semideal completions rather than semideals, because some semideals are being counted more than once.
Consider a player holding J-J. Then f(0,3) is counting player-semideal completions, where 3 larger ranks -- all of them available in this case -- have two pairs of each rank dealt. There are 27 ways of choosing the partitions into six pairs, ways of choosing 6 cards for the remaining 3 players, and 15 ways of partitioning the 6 cards into 3 hands. Thus, .
Similarly, , , , , , , , and .
The number of player-semideal completions with players holding pairs
of 3 bigger ranks is given by
f(0,3) + f(1,2) + f(2,1) + f(3,0) = 47,807,167,160,861,955. | (12) |
The number of player-semideal completions with players holding pairs
of 2 bigger ranks is given by
f(0,2)+f(1,1)+f(2,0)-3f(0,3)-3f(1,2) | |||
-3f(2,1)-3f(3,)=3,381,205,714,311,673,710. | (13) |
Finally, the number of player-semideal completions with a pair or two
pairs of a single bigger rank having been dealt is given by
f(1,0)+f(0,1)-2f(0,2)-2f(1,1)-2f(2,0)+3f(0,3)+3f(1,2) | |||
+3f(2,1)+3f(3,)= 75,657,341,882,411,690,940. | (14) |
Dividing the numbers in (12), (13) and (14) by the number from (8),
gives us probabilities of
We move to a player holding 10-10 and present the values for the various
values of f(i,j) as an array.
We use standard inclusion-exclusion formulations to obtain
Dividing the number just computed by the number from (8), we obtain probabilities of .0002844, .01044, and .15519 for pairs of three, two and one larger rank having been dealt. The corresponding odds against are 3,515-to-1, 94.8-to-1, and 5.44-to-1.
For the remaining ranks, the information is going to be presented in several tables. Also, note that the value of f(2,1), for example, changes as the rank of the pair under consideration changes. This follows because there are different numbers of possibilities for choosing the two ranks and the single rank as the rank under consideration varies.
pair | f(9,0) | f(8,0) | f(7,1) | f(7,0) | f(6,1) | f(5,2) |
9-9 | - | - | - | - | - | - |
8-8 | - | - | - | - | - | - |
7-7 | - | - | - | 6^{7} | ||
6-6 | - | 6^{9} | ||||
5-5 | 6^{9} | |||||
4-4 | ||||||
3-3 | ||||||
2-2 | ||||||
constant | 1 | 1 | 1 |
pair | f(6,0) | f(5,1) | f(4,2) | f(3,3) | f(5,0) | f(4,1) |
9-9 | - | - | - | - | 6^{5} | |
8-8 | 6^{6} | 6^{6} | ||||
7-7 | ||||||
6-6 | ||||||
5-5 | ||||||
4-4 | ||||||
3-3 | ||||||
2-2 | ||||||
constant | 1 |
pair | f(3,2) | f(2,3) | f(1,4) | f(4,0) | f(3,1) | f(2,2) |
9-9 | ||||||
8-8 | 14,580 | |||||
7-7 | 51,030 | |||||
6-6 | 136,080 | |||||
5-5 | 306,180 | |||||
4-4 | 612,360 | |||||
3-3 | 1,122,660 | |||||
2-2 | 1,924,560 | |||||
constant | 1 |
pair | f(1,3) | f(0,4) | f(3,0) | f(2,1) | f(1,2) | f(0,3) |
9-9 | 3,240 | 405 | 2,160 | 3,240 | 1,620 | 270 |
8-8 | 9,720 | 1,215 | 4,320 | 6,480 | 3,240 | 540 |
7-7 | 22,680 | 2,835 | 7,560 | 11,340 | 5,670 | 945 |
6-6 | 45,360 | 5,670 | 12,096 | 18,144 | 9,072 | 1,512 |
5-5 | 81,648 | 10,206 | 18,144 | 27,216 | 13,608 | 2,268 |
4-4 | 136,080 | 17,010 | 25,920 | 38,880 | 19,440 | 3,240 |
3-3 | 213,840 | 26,730 | 35,640 | 53,460 | 26,730 | 4,455 |
2-2 | 320,760 | 40,095 | 47,520 | 71,280 | 35,640 | 5,940 |
constant |
pair | f(2,0) | f(1,1) | f(0,2) | f(1,0) | f(0,1) |
9-9 | 360 | 360 | 90 | 30 | 15 |
8-8 | 540 | 540 | 135 | 36 | 18 |
7-7 | 756 | 756 | 189 | 42 | 21 |
6-6 | 1,008 | 1,008 | 252 | 48 | 24 |
5-5 | 1,296 | 1,296 | 324 | 54 | 27 |
4-4 | 1,620 | 1,620 | 405 | 60 | 30 |
3-3 | 1,980 | 1,980 | 495 | 66 | 33 |
2-2 | 2,376 | 2,376 | 594 | 72 | 36 |
constant |
We now are prepared to compute the probabilities we want. We use the information in the tables together with inclusion-exclusion to extract the exact number of player-semideal completions with the appropriate pocket pairs dealt. Here are the details on how to use the tables.
The columns are headed by some f(i,j) telling us that i ranks appear with one pair of that rank dealt, and j ranks appear with two pairs of that rank dealt. The actual entries in the tables correspond to how many ways the i and j ranks may be chosen from ranks larger than the rank corresponding to the row multiplied by the number of ways of choosing the various pairs. The number at the bottom of the column is the number of ways the deals may be completed to 9-i-2j players. Thus, we multiply the entry in the table by the number at the bottom of the column to get the actual value of f(i,j) for the pair corresponding to the row.
Thus, assume we are working with a pair of rank x-x. We let F(n)denote the sum of all f(i,j) such that i+j=n for the row corresponding
to rank x-x. Then
F(3)-4F(4)+10F(5)-20F(6)+35F(7)-56F(8)+84F(9) | (15) |
The number with exactly 2 larger ranks with one or two pairs dealt is
given by
F(2)-3F(3)+6F(4)-10F(5)+15F(6)-21F(7)+28F(8)-36F(9). | (16) |
Finally, the number for exactly one larger rank with one or two pairs of
that rank dealt is given by
F(1)-2F(2)+3F(3)-4F(4)+5F(5)-6F(6)+7F(7)-8F(8)+9F(9). | (17) |
Let's go through one example using the three formulas just presented.
Assume the player is holding 9-9. Looking at the tables, we find entries
for 9-9 under f(i,j) when i+j=5. Using them to calculate F(5) as
described above yields
Continuing, we obtain
We combine them according to (15), (16) and (17) and divide by the number from (8) to obtain the probabilities in the table below.
We continue this for all other ranks. This gives the table below. The column headed ``pair'' tells us the rank of the pair held by the player. The column headed ``one rank'' tells us the probability there is exactly one larger rank for which either one or two pairs of that rank have been dealt to other players. The column headed ``two ranks'' tells us the probability there are exactly two larger ranks for which either one or two pairs of each rank have been dealt to other players. There is an analogous meaning to the column headed ``three ranks''.
We have not bothered to separate out those completions for which two pairs of some larger rank have been dealt. The reason for this is that it is rare for two pairs of the same rank to have been dealt to players, and it follows that most of the contribution to the probabilities in the table arise from just one pair of each of the appropriate ranks to have been dealt.
pair | one rank | two ranks | three ranks |
K-K | .0439 | - | - |
Q-Q | .08412 | .001863 | - |
J-J | .12161 | .005435 | .0000768 |
10-10 | .15519 | .01044 | .0002844 |
9-9 | .1857 | .01662 | .0007109 |
8-8 | .2132 | .02397 | .001367 |
7-7 | .2380 | .03218 | .0023 |
6-6 | .2603 | .04114 | .003538 |
5-5 | .2801 | .05071 | .0051 |
4-4 | .2977 | .06076 | .007 |
3-3 | .3133 | .07118 | .009251 |
2-2 | .3269 | .08186 | .01186 |