Multiple Pocket Pairs

Brian Alspach

7 December 2001

Abstract:

We examine various probabilities involving more than one player holding a pocket pair.

SAME RANK

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

$\displaystyle {{52}\choose{20}}19!! = 82,492,346,176,096,206,475,125$     (1)

because we are choosing 20 cards from 52 and there are 19!! ways to partition 20 cards into 10 hands of 2 cards each.

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

$\displaystyle 3^5{{13}\choose{5}} = 312,741.$     (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

\begin{displaymath}{{13}\choose{4}}{{36}\choose{4}}3^5\end{displaymath}

player-semideals with at least 4 sets of pocket pairs of the same rank. However, each of the player-semideals counted in (2), is counted 5 times in the number just calculated. So we must subtract 5 times the number from (2). Doing so gives us
10,232,885,520     (3)

player-semideals with precisely four ranks for which there are players holding pocket pairs of the four ranks.

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

$\displaystyle 3^3\cdot 7!!{{13}\choose{3}}{{40}\choose{8}}-4(3)-10(2)= 62,314,152,975,360.$     (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

\begin{displaymath}9\cdot 78{{44}\choose{12}}11!! = 153,904,827,325,018,770\end{displaymath}

player-semideals. Those counted in (4) are counted 3 times, those in (3) are counted 6 times, and those counted in (2) are counted 10 times. Performing the appropriate subtractions yields
153,717,823,465,652,160     (5)

player-semideals with exactly 2 two ranks for which there are players dealt two pairs of these ranks.

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

\begin{displaymath}39\cdot 15!!{{48}\choose{16}} = 178,254,769,648,227,096,825.\end{displaymath}

Because of overcounting for the player-semideals counted in (2) - (5), we must subtract twice the number in (5), 3 times the number in (4), 4 times the number in (3), and 5 times the number in (2). Doing so yields
177,947,147,017,903,760,640     (6)

player-semideals in which there is exactly one rank for which two people are dealt pairs of this rank.

We take the sum of the player-semideals in (2) through (6) obtaining

178,100,927,165,755,586,421     (7)

player-semideals in which there is at least one rank for which two players have pairs of that rank.


Dividing the number in (7) by the number in (1) gives a probability of

.002158999

that two or more players are dealt pairs of the same rank. This is approximately 1/463 so that about once in every 463 deals there will at least two players with pairs of the same rank.


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

13,711,905,357,555,930,525

player-semideals for which two players have been dealt pocket aces. Dividing this number by the number in (1) gives us a probability of

.000166

that two players have been dealt pocket aces. This is about 1/6,024 so that you should see two players dealt pocket aces about once every 6,024 deals.


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

$\displaystyle {{50}\choose{18}}17!! = 622,114,224,555,778,329,375$     (8)

player-semideals completing the deal in which you hold aces. The number of them in which the two remaining aces are in the same hand is simply

\begin{displaymath}{{48}\choose{16}}15!!\end{displaymath}

because you set aside the two remaining aces to go in one hand and choose the remaining 8 hands arbitrarily. Dividing the latter number by the number given in (8) yields a probability of

\begin{displaymath}\frac{9}{1,225}\end{displaymath}

that another player also has pocket aces. This is approximately 1/136. Thus, upon being dealt a pair of aces, you expect to run into another player with pocket aces only about once in every 136 times.


DIFFERENT RANKS

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 ${{46}\choose{14}}13!!$ ways. This gives us

\begin{displaymath}3\cdot 13!!{{46}\choose{14}}=97,247,555,727,347,025\end{displaymath}

completions to player-semideals with two players holding pocket aces.

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 ${{48}\choose {16}}15!!$ choices for the remaining 8 players. Multiplying yields

\begin{displaymath}6\cdot 15!!{{48}\choose{16}}=27,423,810,715,111,861,050,\end{displaymath}

but any completion for which two players have pocket aces is counted twice. Thus, we must subtract the number of completions for which two players have pocket aces.

Performing the indicated subtraction yields

27,326,563,159,384,514,025     (9)

completions in which there is at least one player with pocket aces. Dividing by the number in (8) gives a probability of

\begin{displaymath}\frac{2,529}{57,575}\end{displaymath}

that the player holding pocket kings is facing a larger pocket pair. Expressed as a decimal this is approximately .0439. It is easy to calculate this means the odds against someone having a larger pocket pair are about 21.8-to-1.

For a player holding pocket queens, there are now two larger ranks. This means the number of subcases begins to proliferate. Let f4 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

\begin{displaymath}f_4 = 9\cdot 9!!{{42}\choose{10}} = 12,514,622,485,365.\end{displaymath}

Let f3 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 f4 contributes 4 to f3.

There are two ways to split the ranks for contributing to f3. 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

\begin{displaymath}f_3 = 36\cdot 11!!{{44}\choose{12}} = 7,892,555,247,436,860.\end{displaymath}

Now let f2 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

\begin{displaymath}f_2 = 36\cdot 13!!{{46}\choose{14}} = 1,166,970,668,728,164,300.\end{displaymath}

Let g2 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 f2 is replaced by 6. This means g2 is one-sixth of f2, that is,

g2 = 194,495,111,454,694,050.

Finally, f1 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

\begin{displaymath}f_1 = 12\cdot 15!!{{48}\choose{16}} = 54,847,621,430,223,722,100.\end{displaymath}

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

f1-g2-2f2+2f3-2f4 = 52,334,945,062,562,602,440     (10)

counts the number of completions for which there is exactly one rank with one or two players holding pocket pairs of that rank.

Similarly, the expression

f2-f3+f4 = 1,159,090,628,103,212,805     (11)

counts the number of completions in which the player with Q-Q is up against two bigger ranks with at least one pocket pair of that rank being present.

Dividing these numbers by the number in (8) gives us a probability of

.08412

that the player with Q-Q is facing one or two pocket pairs of exactly one bigger rank, and a probability of

.001863

that the player with Q-Q is facing pairs of two bigger ranks. The corresponding odds against this being the case are about 10.9-to-1 and 535.7-to-1, respectively.

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, ${{38}\choose{6}}$ ways of choosing 6 cards for the remaining 3 players, and 15 ways of partitioning the 6 cards into 3 hands. Thus, $f(0,3)=27\cdot 15{{38}\choose{6}}=$.

Similarly, $f(0,2)=27\cdot 945{{42}\choose{10}}$, $f(1,2)=162\cdot 105 {{40}\choose{8}}$, $f(2,1)=324\cdot 945{{42}\choose{10}}$, $f(3,0)= 216{{44}\choose{12}}11!!$, $f(1,1)=108{{44}\choose{12}}11!!$, $f(2,0)= 108{{46}\choose{14}}13!!$, $f(1,0)=18{{48}\choose{16}}15!!$, and $f(0,1)=9{{46}\choose{14}}13!!$.

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

.0000768

that a player holding J-J is facing at least one pocket pair of each of the ranks of queens, kings and aces;

.005435

that the player with J-J is facing at least one pocket pair of exactly two larger ranks; and

.12161

that he is facing at least one pocket pair of exactly one larger rank than jack. The corresponding odds against are 13,020-to-1, 183-to-1, and 7.2-to-1.

We move to a player holding 10-10 and present the values for the various values of f(i,j) as an array.

\begin{eqnarray*}f(0,4)=81{{34}\choose{2}}, & f(1,3) = 24\cdot 81{{36}\choose{4}... ...) = 24{{48}\choose{16}}15!!, & f(0,1)=12{{46}\choose{14}}13!! & \end{eqnarray*}


We use standard inclusion-exclusion formulations to obtain

176,917,579,707,312,576

player-semideal completions for which there are precisely 3 largers ranks with one or more pairs dealt to players. Similarly, we obtain

6,495,683,479,833,959,136

completions with precisely 2 ranks having one or more pairs dealt. Finally, there are

96,546,901,242,811,894,776

completions with exactly one rank for which there has been dealt at least one pair of that rank.

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 - - - 67 $21\cdot 6^6$ $189\cdot 6^5$
6-6 - 69 $24\cdot 6^7$ $8\cdot 6^7$ $168\cdot 6^6$ $28\cdot 54\cdot 6^5$
5-5 69 $9\cdot 6^8$ $27\cdot 8\cdot 6^7$ $36\cdot 6^7$ $27\cdot 28\cdot 6^6$ $36\cdot 189\cdot 6^5$
4-4 $10\cdot 6^9$ $45\cdot 6^8$ $30\cdot 36\cdot 6^7$ ${{10}\choose {3}}6^7$ $30\cdot 6^6{{9}\choose{6}}$ $405\cdot 56\cdot 6^5$
3-3 $55\cdot 6^9$ $165\cdot 6^8$ $33\cdot 120\cdot 6^7$ ${{11} \choose{4}}6^7$ $33\cdot 6^6{{10}\choose{4}}$ $495\cdot 6^5{{9}\choose {5}}$
2-2 $220\cdot 6^9$ $495\cdot 6^8$ $36\cdot 6^7{{11}\choose{4}}$ ${{12}\choose{5}}6^7$ $36\cdot 6^6{{11}\choose{4}}$ $594\cdot 6^5{{10} \choose{5}}$
constant 1 ${{34}\choose{2}}$ 1 $3{{36}\choose{4}}$ ${{34}\choose{2}}$ 1


pair f(6,0) f(5,1) f(4,2) f(3,3) f(5,0) f(4,1)
9-9 - - - - 65 $15\cdot 6^4$
8-8 66 $18\cdot 6^5$ $135\cdot 6^4$ $20\cdot 18^3$ 66 $15\cdot 6^5$
7-7 $7\cdot 6^6$ $126\cdot 6^5$ $21\cdot 45\cdot 6^4$ $35\cdot 108 \cdot 6^3$ $21\cdot 6^5$ $315\cdot 6^4$
6-6 $28\cdot 6^6$ $504\cdot 6^5$ $3,780\cdot 6^4$ $15,120\cdot 6^3$ $56\cdot 6^5$ $840\cdot 6^4$
5-5 $84\cdot 6^6$ $1,512\cdot 6^5$ $11,340\cdot 6^4$ $45,360\cdot 6^3$ $126\cdot 6^5$ $1,890\cdot 6^4$
4-4 $210\cdot 6^6$ $3,780\cdot 6^5$ $28,350\cdot 6^4$ $113,400\cdot 6^3$ $252\cdot 6^5$ $3,780\cdot 6^4$
3-3 $462\cdot 6^6$ $8,316\cdot 6^5$ $62,370\cdot 6^4$ $249,480\cdot 6^3$ $462\cdot 6^5$ $6,930\cdot 6^4$
2-2 $924\cdot 6^6$ $16,632\cdot 6^5$ $74,844\cdot 6^4$ $498,960\cdot 6^3$ $792\cdot 6^5$ $11,880\cdot 6^4$
constant $15{{38}\choose{6}}$ $3{{36}\choose{4}}$ ${{34}\choose{2}}$ 1 $105{{40}\choose{8}}$ $15{{38}\choose{6}}$


pair f(3,2) f(2,3) f(1,4) f(4,0) f(3,1) f(2,2)
9-9 $90\cdot 6^3$ $270\cdot 6^2$ $30\cdot 3^4$ $5\cdot 6^4$ $60 \cdot 6^3$ $270\cdot 6^2$
8-8 $540\cdot 6^3$ $1,620\cdot 6^2$ 14,580 $15\cdot 6^4$ $180\cdot 6^3$ $810\cdot 6^2$
7-7 $1,890\cdot 6^3$ $5,670\cdot 6^2$ 51,030 $35\cdot 6^4$ $420 \cdot 6^3$ $1,890\cdot 6^2$
6-6 $5,040\cdot 6^3$ $15,120\cdot 6^2$ 136,080 $70\cdot 6^4$ $840\cdot 6^3$ $3,780\cdot 6^2$
5-5 $11,340\cdot 6^3$ $34,020\cdot 6^2$ 306,180 $126\cdot 6^4$ $1,512\cdot 6^3$ $6,804\cdot 6^2$
4-4 $22,680\cdot 6^3$ $68,040\cdot 6^2$ 612,360 $210\cdot 6^4$ $2,520\cdot 6^3$ $11,340\cdot 6^2$
3-3 $41,580\cdot 6^3$ $124,740\cdot 6^2$ 1,122,660 $330\cdot 6^4$ $3,960\cdot 6^3$ $17,820\cdot 6^2$
2-2 $71,280\cdot 6^3$ $213,840\cdot 6^2$ 1,924,560 $495\cdot 6^4$ $5,940\cdot 6^3$ $26,730\cdot 6^2$
constant $3{{36}\choose{4}}$ ${{34}\choose{2}}$ 1 $945{{42} \choose{10}}$ $105{{40}\choose{8}}$ $15{{38}\choose{6}}$


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 $3{{36}\choose{4}}$ ${{34}\choose{2}}$ $10,395{{44}\choose {12}}$ $945{{42} \choose{10}}$ $105{{40}\choose{8}}$ $15{{38}\choose{6}}$


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 $135,135{{46}\choose{14}}$ $10,395{{44}\choose {12}}$ $945{{42} \choose{10}}$ $2,027,025{{48}\choose{16}}$ $135,135{{46}\choose{14}}$

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)

is the number of player-semideal completions with exactly 3 larger ranks for which one or two pairs of these ranks have been dealt. Some of the terms are zero depending on x.

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

F(5)=63,599,592,583,350.

Continuing, we obtain

\begin{eqnarray*}F(4) & = & 9,115,583,164,884,405,\\ F(3) & = & 478,071,671,608,... ...,385,980,865,250, and\\ F(1) & = & 137,605,291,354,196,040,375. \end{eqnarray*}


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

Home | Publications | Preprints | MITACS | Poker Digest | Poker Computations | Feedback
website by the Centre for Systems Science
last updated 7 December 2001