12-22-2014 , 07:38 PM
I have derived the exact probabilities of two or more players being dealt a pocket pair among N players for N=2 to 9. Of course, for small values of N these probabilities can be derived fairly easily.

For large values of N these probabilities are difficult to derive and heretofore have been estimated using an assumption of independence across the players so that the binomial formula can be utilized.

Below I will give the exact number of deals for each N that have P players being dealt a pocket pair for P=0 to N. I will give the exact probabilities of each pair occurring and compare it to the probability approximated from using the independent binomial assumption (and give their ratio too).

I will report probabilities as percentages, giving four decimal places in most cases. Since some of these probabilities are quite small, I will always give at least two significant digits. As usual, displayed probabilities may on occasion not appear to add to one due to rounding.

For N players, the total number of deals consisting of each player receiving two cards is easy to calculate. I will use the C(X,Y) notation for the number of combinations of choosing Y items from X items, where the order is irrelevant. There are C(52, 2N) ways of choosing the 2N cards to be dealt to the N players. Then the number of ways these 2N cards can be dealt to the N players to form N 2-card hands is the indexed product of C(2J, 2) with index J going from N to 1. [For example, for N=3 players and 2N=6 cards, the first player can get C(6,2) hands, the second player C(4,2) hands, and the third player C(2,2) hands.] We then need to divide by N! (N factorial) since we are not interested in which player gets which hand, only the overall combination of N 2-cards hands.

As an aside, this last term simplifies to the product of the odd numbers up to 2N-1, or equivalently the product of the first N odd numbers, which using the double factorial notation is (2N-1)!!. So for N=3, this last term is 1*3*5 = 15, meaning that there are 15 ways to deal a given 6 cards to form 3 2-card hands where the order of the cards in each player's hand is irrelevant [(As, Kh) is an equivalent hand to (Kh, As)] and the order of the players' hands is irrelevant [the deal of {(As, Kh), (Qd, Jc), (6c, 3h)} is equivalent to {(Qd, Jc), (6c, 3h), (As, Kh)}].

So the total number of deals of N players each receiving two cards is C(52, 2N) * (2N-1)!!.

Let's start easy with N=1. Of course there are C(52,2) = 1,326 ways of one player being dealt two cards. 1,248 of these will result in no pair and the other 78 ways will result in a pair. Clearly, after any one of the 52 cards is first dealt to the player, 48 of the remaining 51 cards will give no pair and 3 of the remaining 51 cards will give one pair (we then need to divide by two in both cases since the order of the cards in the hand is irrelevant). I will record this as follows:

N=1 Player
----------
No Pair : 1,248 ( 94.1176%)
One Pair: 78 ( 5.8824%)

TOTAL DEALS: 1,326 (100%)

where the exact percentages are given in parentheses. In this case, I do not display the approximate probabilities given by the independent binomial assumption since the two cases are identical (there is only one player).

For two players, there are C(52,4) * (1*3) = 812,175 ways of dealing two players two-card hands.

N=2 Players
----------
No Pair : 719,472 ( 88.5858%; 88.5813%; 1.000051)
One Pair: 89,856 ( 11.0636%; 11.0727%; 0.999184)
Two Pair: 2,847 ( 0.3505%; 0.3460%; 1.013061)

TOTAL DEALS: 812,175 (100%)

where now in parentheses are shown the exact probability, the approximate probability using the independent binomial assumption, and the ratio of the exact probability to the approximate probability.

You will see that the independent binomial approximation underestimates the probability of 0 pair and 2 pair, and overestimates the probability of 1 pair. This tendency holds for all N as in reality one player holding a pocket pair makes it slightly more likely that any other player also holds a pocket pair, and one player not holding a pocket pair makes it slightly more likely that any other player also does not hold a pocket pair.

For three players, there are C(52,6) * (1*3*5) = 305,377,800 ways of dealing three players two-card hands.

N=3 Players
----------
No Pair : 254,634,432 ( 83.3834%; 83.3706%; 1.000153)
One Pair: 47,661,120 ( 15.6073%; 15.6320%; 0.998418)
Two Pair: 3,017,664 ( 0.9882%; 0.9770%; 1.011437)
Three Pair: 64,584 ( 0.0211%; 0.0204%; 1.039045)

TOTAL DEALS: 305,377,800 (100%)

For four players, there are C(52,8) * (1*3*5*7) = 79,016,505,750 ways of dealing four players two-card hands.

N=4 Players
----------
No Pair : 62,020,470,096 ( 78.4905 %; 78.4665 %; 1.000306)
One Pair: 15,464,756,736 ( 19.5716 %; 19.6166 %; 0.997702)
Two Pair: 1,467,494,496 ( 1.8572 %; 1.8391 %; 1.009865)
Three Pair: 62,764,416 ( 0.0794 %; 0.0766 %; 1.036600)
Four Pair: 1,020,006 ( 0.00129%; 0.00120%; 1.078153)

TOTAL DEALS: 79,016,505,750 (100%)

For five players, there are C(52,10) * (1*3*5*7*9) = 14,949,922,887,900 ways of dealing five players two-card hands.

N=5 Players
----------
No Pair : 11,046,277,262,592 ( 73.8885 %; 73.8508 %; 1.000511)
One Pair: 3,439,978,397,856 ( 23.0100 %; 23.0784 %; 0.997037)
Two Pair: 434,873,140,416 ( 2.9089 %; 2.8848 %; 1.008343)
Three Pair: 27,876,790,656 ( 0.1865 %; 0.1803 %; 1.034209)
Four Pair: 905,389,056 ( 0.0061 %; 0.0056 %; 1.074858)
Five Pair: 11,907,324 ( 0.000080%; 0.000070%; 1.130889)

TOTAL DEALS: 14,949,922,887,900 (100%)

For six players, there are C(52,12) * (1*3*5*7*9*11) = 2,145,313,934,413,650 ways of dealing six players two-card hands.

N=6 Players
----------
No Pair : 1,492,278,187,226,112 ( 69.5599 %; 69.5067 %; 1.000766)
One Pair: 557,175,599,735,040 ( 25.9718 %; 26.0650 %; 0.996423)
Two Pair: 87,971,700,939,408 ( 4.1006 %; 4.0727 %; 1.006872)
Three Pair: 7,512,990,046,848 ( 0.3502 %; 0.3394 %; 1.031871)
Four Pair: 365,736,653,568 ( 0.0170 %; 0.0159 %; 1.071618)
Five Pair: 9,613,334,016 ( 0.00045 %; 0.00040 %; 1.126693)
Six Pair: 106,478,658 ( 0.0000050%; 0.0000041%; 1.198023)

TOTAL DEALS: 2,145,313,934,413,650 (100%)

For seven players, there are C(52,14) * (1*3*5*7*9*11*13) = 239,049,266,977,521,000 ways of dealing seven players two-card hands.

N=7 Players
----------
No Pair : 156,549,075,187,355,136 ( 65.4882 %; 65.4180 %; 1.001073)
One Pair: 68,133,459,724,881,408 ( 28.5018 %; 28.6204 %; 0.995858)
Two Pair: 12,898,104,722,021,376 ( 5.3956 %; 5.3663 %; 1.005453)
Three Pair: 1,375,801,040,877,120 ( 0.5755 %; 0.5590 %; 1.029586)
Four Pair: 89,232,018,258,240 ( 0.0373 %; 0.0349 %; 1.068434)
Five Pair: 3,515,707,001,664 ( 0.0015 %; 0.0013 %; 1.122557)
Six Pair: 77,831,088,192 ( 0.000033 %; 0.000027 %; 1.192862)
Seven Pair: 746,037,864 ( 0.00000031%; 0.00000024%; 1.280607)

TOTAL DEALS: 239,049,266,977,521,000 (100%)

For eight players, there are C(52,16) * (1*3*5*7*9*11*13*15) = 21,006,454,335,649,657,875 ways of dealing eight players two-card hands.

N=8 Players
----------
No Pair : 12,952,160,151,935,374,656 ( 61.6580 %; 61.5699 %; 1.001431)
One Pair: 6,436,718,641,227,663,360 ( 30.6416 %; 30.7850 %; 0.995344)
Two Pair: 1,420,395,848,998,993,152 ( 6.7617 %; 6.7342 %; 1.004084)
Three Pair: 181,664,175,195,689,472 ( 0.8648 %; 0.8418 %; 1.027354)
Four Pair: 14,716,813,939,542,000 ( 0.0701 %; 0.0658 %; 1.065306)
Five Pair: 772,570,615,474,944 ( 0.0037 %; 0.0033 %; 1.118482)
Six Pair: 25,638,362,624,160 ( 0.00012 %; 0.00010 %; 1.187766)
Seven Pair: 491,218,535,808 ( 0.0000023 %; 0.0000018 %; 1.274393)
Eight Pair: 4,155,760,323 ( 0.000000020%; 0.000000014%; 1.380032)

TOTAL DEALS: 21,006,454,335,649,657,875 (100%)

For nine players, there are C(52,18) * (1*3*5*7*9*11*13*15*17) = 1,470,451,803,495,476,051,250 ways of dealing nine players two-card hands.

N=9 Players
----------
No Pair : 853,667,643,003,891,797,796 ( 58.0548 %; 57.9481 %; 1.001840)
One Pair: 476,852,107,923,575,077,980 ( 32.4290 %; 32.5958 %; 0.994880)
Two Pair: 120,157,940,672,756,034,048 ( 8.1715 %; 8.1490 %; 1.002766)
Three Pair: 17,914,600,053,357,815,808 ( 1.2183 %; 1.1884 %; 1.025174)
Four Pair: 1,740,207,513,284,368,128 ( 0.1183 %; 0.1114 %; 1.062234)
Five Pair: 114,111,043,097,923,872 ( 0.0078 %; 0.0070 %; 1.114466)
Six Pair: 5,045,885,892,919,872 ( 0.00034 %; 0.00029 %; 1.182736)
Seven Pair: 144,930,110,637,312 ( 0.0000099 %; 0.0000078 %; 1.268252)
Eight Pair: 2,450,932,035,552 ( 0.00000017 %; 0.00000012 %; 1.372645)
Nine Pair: 18,577,440,882 ( 0.00000000126%; 0.00000000084%; 1.498219)

TOTAL DEALS: 1,470,451,803,495,476,051,250 (100%)

From these tables, it is straightforward to calculate any other probability of interest, for example the probability of two-or-more players being dealt a pocket pair among N players. The table below gives the exact and approximate probabilities and their ratio for N=2 to 9.

Exact Prob Approx Prob
N 2+ Pairs (%) 2+ Pairs (%) Ratio
--------------------------------------
2 0.3505 0.3460 1.013061
3 1.0093 0.9974 1.012001
4 1.9379 1.9169 1.010976
5 3.1015 3.0708 1.009987
6 4.4684 4.4284 1.009032
7 6.0099 5.9616 1.008111
8 7.7004 7.6451 1.007225
9 9.5163 9.4560 1.006371

As mentioned above, the independence binominal approximation underestimates the likelihood of multiple players being dealt pocket pairs.
12-22-2014 , 10:03 PM
You've done a lot of work already but can you give an example how you calculate the nominators for example for N=2?

Did you use inclusion/exclusion? For example for N=4, 4 pair can be AA BB CC DD, AA AA BB CC, AA AA BB BB etc.
12-22-2014 , 10:15 PM
This looks good. I assume you used inclusion-exclusion? If you want, I'll check your exact results a little bit at a time.
12-23-2014 , 02:18 AM
I did not use inclusion/exclusion, though I imagine someone clever enough could do so. My approach was much more brute force, though utilizing straightforward combinatorics wherever I could.

For each N, I compiled a list of all the possible cases of quads, trips, doubles, and singletons being dealt among the 2N cards. For example, for N=5 (10 cards) there are 23 possible cases, ranging from 2 sets of quads and one double to 10 singletons. It should be clear that when I say that "I compiled a list" I mean that I programmed my computer to compile the list.

Then for each case it is straightforward combinatorics to derive how many different card combinations this case represents. That is the easy part.

The challenging part, at least it was challenging for me especially when N gets large, is to derive how many different ways that collection of quads, trips, doubles, singletons can form P pairs (P ranging from 0 to N).

The way I tackled it was to consider each quad-trip-double-single case one by one, and each P one by one (the pairs, of course, use up 2P cards). It is straightforward to calculate how many card "splits" these represent, where "split" is a specific way to deal a given 2N cards (so for N=5, the splits add up to 1*3*5*7*9 = 945).

To try to make it clear what I did, consider the case for N=5 and P=2 for some quad-trip-double-single combination. You want 10 cards to be dealt and 2 people to get pairs. Further consider the case of two quads and one double forming the 10 cards. One way to form two pair is to have one pair come from one of the quads and the other pair to come from the double.

In this approach, the most difficult aspect was to tally how many ways the remaining cards can be dealt to form no additional pairs. In the N=5, P=2, two quads and one double case, with one pair coming from one of the quads and the other pair coming from the double, this would mean how many ways can the 6 remaining cards (one complete set of quads and the other two cards from the quads that formed the pair) be dealt to not form another pair.

Of course, in this extreme example, it is clear that there is no way to deal these six cards to make three two-card hands without making another pair (there are too many of the quad ranks to go around). In most cases, there are many ways to make no additional pairs. I ran extensive programs to derive these "no additional pairs" tallies for each possible quad-trip-double-singleton case for each N.

With all that information at hand, I was able to combine it all to derive the number of deals for each {N, quad-trip-double-single, P} scenario. Then sum over all the quad-trip-double-single cases for a given P (for a given N, of course) and you have one result. Do this for all P (from 0 to N) for a given N, and then do it for all N from 2 to 9.

All of this was done in a programming language (Basic). First step was to derive all the "No Additional Pairs" tallies. Then these No Additional Pairs tallies were plugged into a straightforward routine that, for a given N and a given P, looped over all the possible quad-trip-double-single cases. Of course, the program then looped over all the P and then all the N.

In addition to the Basic programming, I was able to program this all up in a spreadsheet environment too (OpenOffice Calc). For the cases of N=2-8 players, I was able to derive the same results in both Basic and in Calc. The case of N=9 players was too large for the Calc spreadsheet environment but I am 99% confident in the results since the Basic program for N=9 is identical to that for N=8 (which was confirmed by the spreadsheet).

To pull back the curtain all the way, actually intermediate results were ported over from Basic to the Calc spreadsheet environment so I could keep track of (retain) all the digits in these super large numbers. Both the programming environment and the spreadsheet I used utilize double precision (15 significant digits). Since some of these super large numbers have more than 15 digits, I was able to massage the calculations so I could "manually" (via the computer) retain all the digits.

Hope this gives a sense of how these results were obtained.

P.S. It goes without saying that I am confident that these results are correct. However, any amount of independent checking would be welcome.
12-23-2014 , 09:12 PM
What OP did is quite impressive. From a practical point of view, here is what I take away from the results.

The most useful result IMO is the probability that at least one opponent has a pair. While this probability is dependent on whether you have a pair or not, the difference is so small that it can be ignored. This was validated using a simulation program I wrote. For example, if you are in a 6-max game, the chance that at least one of your five opponents has a pair is 1-0.7389 = 26.11% according to OPs result. In 5m runs, my simulation program also gave the same result. It showed a result of 26.08% if you did not have a pair and 26.43% if you did. Also, for all practical purposes, the binomial results are good enough, with absolute errors always in the 3rd or 4th decimal place.

I could be wrong, but Im guessing this project was not a one-day effort.
01-02-2015 , 09:48 PM
i think that correct answer is...

chance to one player have a pair is 0.0588

but to one of 9 players have a pair is 1- 0.9412^9=0.42 or 43%

so if you have a pocket pair.. another player can have pair in 9 hand game in
1- 0.9412^8=0.3842 or 38.42%
01-03-2015 , 12:34 AM
Quote:
Originally Posted by rebarca91
i think that correct answer is...
OP has given the exact results for all cases with no pre-set conditions.

Your results for more than one player are based on assuming independence, which is not true, but generally gives very close results, which was shown by OP.
01-03-2015 , 12:58 AM
Quote:
Originally Posted by statmanhal
OP has given the exact results for all cases with no pre-set conditions.

Your results for more than one player are based on assuming independence, which is not true, but generally gives very close results, which was shown by OP.
His math was wrong for the approximation too.
01-03-2015 , 02:14 AM
Quote:
Originally Posted by statmanhal
What OP did is quite impressive. From a practical point of view, here is what I take away from the results.

The most useful result IMO is the probability that at least one opponent has a pair. While this probability is dependent on whether you have a pair or not, the difference is so small that it can be ignored. This was validated using a simulation program I wrote. For example, if you are in a 6-max game, the chance that at least one of your five opponents has a pair is 1-0.7389 = 26.11% according to OPs result. In 5m runs, my simulation program also gave the same result. It showed a result of 26.08% if you did not have a pair and 26.43% if you did. Also, for all practical purposes, the binomial results are good enough, with absolute errors always in the 3rd or 4th decimal place.

I could be wrong, but Im guessing this project was not a one-day effort.
Yeah, so does that mean about a quarter of time (or one out of every four hands) somebody was dealt a pair?

People like me aren't going to check the math and I don't think even mathematicians can't calculate this stuff real time (or even need to?) during a game so, let's put into layman's terms, yeh?
05-27-2020 , 04:19 AM
Hi,

I am writing an internal assessment of poker probabilities for my math class. I was wondering if you could show me the workings and math behind how you arrived at the number of combinations to get a pair?

The number I need help with are:
N=2: 2,847
N=3: 64,584
and so on...

Regards, Kvars 05-27-2020 , 06:24 AM
Quote:
Originally Posted by Kvars
The number I need help with are:
N=2: 2,847
C(13,2)*C(4,2)² + 13*3 = 2847

When they share a pair, there are 3 valid ways to distribute the suits. His denominator counts the ways to group the cards into two, so his numerator counts that too when there is more than one valid grouping.

Quote:
N=3: 64,584
C(13,3)*C(4,2)³ + 13*12*C(4,2)*3 = 64584

There are (13P2)=13*12 ways to pick two ranks such that one repeats and the other doesn't. That distinction is why order matters. When there's a repeating rank, there are again 3 valid ways to group those 4 cards among the 2 player hands they're in.

Last edited by heehaww; 05-27-2020 at 06:33 AM. Reason: They could also share a pair
06-03-2020 , 03:27 AM

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