I'm probably over thinking this but, if a single player in a full ring game is dealt a pair roughly 6% of the time and all players at the table won't get their pair at the same time, obviously. What's the % someone at the table is holding a pocket pair after each dealing of the cards?

If each person got their cards from a separate deck, the chances someone would have a pair would be 1-[(16/17)to the ninth power]. With one deck that is slightly incorrect because the probabilities are not quite independent. I will let others elaborate.

This has been a topic of conversation several times over the years in this forum.

Here is a post I made with tons of results from a couple of years ago.

Furthermore, it's always slightly incorrect in the same direction: it's too small, because in reality, when on player lacks a pair, that increases the chance of another player having one.

There is another very simple estimate which is always wrong in the opposite direction: simply multiply one player's chance by 9, resulting in 9/17.

Thus, the answer lies somewhere between 1-(16/17)⁹ (42%) and 9/17 (53%).

The second estimate is too high because it overcounts the chance of multiple pairs occurring. What I'll show is a way to refine it until the exact answer is reached, using the principle of inclusion-exclusion, which happens to be the most elegant solution to this problem that I know of.

I'll illustrate why multiplying by 9 over-counts. For now, focus only on AA. When we multiplied 1/17 by 9, we added, among other things, P(sb gets AA) + P(bb gets AA) + ... + P(btn AA). Each of those probabilities ignores what the other players have, so we added P(AA, xy, xy, ...) + P(xy, AA, xy, ...) and so on. Can you see that by doing that, we've double-counted the possibility of (AA, AA, xy, ...)? Since the x's and y's are allowed to be A's, P(AA, xy, ...) and P(xy, AA, ...) each counted P(AA, AA, xy, ...) once. The same is true of every other pair.

Now let's look at why (AA, KK, ...) is over-counted. By virtue of multiplying by 9, we added P(AA, xx, ...) + ... + P(xx, KK, ...). Notice how they both count the (AA, KK, xx, ...) possibility? So overall, (AA, KK) was counted twice. Same is true of all pair-vs-pair matchups.

So now let's come up with another lower-bound estimate by taking our upper bound (9/17) and subtracting the repeated pair and two-pair probabilities (so that, instead of being counted twice, they're counted once).

9/17 - C(9,2)[13 + C(13,2)*C(4,2)[sup]2[sup]/3]/C(52,4) =~ 40.3%

It turns out, this lower bound is lower than our previous one and therefore doesn't replace it. We went from too high to too low because we've over-subtracted. Our next step will be to add, and then subtract again, and so on, until we've either converged enough, or until the process runs its full course (having accounted for cases like 9 pocket pairs) and arrives at the exact answer.

I'm falling asleep and will finish this tomorrow.

All I can say is wow! Thank you for taking the time (a couple years back) and providing such in an in depth thoughtful answer. Much appreciated!!!

Thank you! Looking forward to hearing the rest!

I left off at
9/17 -
C(9,2)*[13 + C(13,2)*C(4,2)²/3]/C(52,4)

Before I go further, I'll explain that 2nd line.

There are 13 possible repeating pp's.

There are C(13,2) possible two-pairs and C(4,2)² possible suits for them, but then we divide this all by 3 because only one grouping of AA|KK will result in pairs, out of 3 possible ways to divide the four cards into groups of 2 (that is, 2/3 of the groupings result in each player getting AK instead of pairs). Later we'll be dealing with 3-player groupings and will be dividing by 15 (5*3); for n-player groupings we'll divide by (2n-1)!! where !! is "double-factorial", e.g. 7!! = 7*5*3. Bonus points if you can see why that formula counts the number of ways to group the cards (when player identity/order doesn't matter).

We only need to divide by C(52,4) and only account for 2 players in the numerator because the other 7 players would cancel out anyway. We're allowing them to have any cards, so we'd just be dividing something by itself.

Moving on: let's take inventory of how many times things have been counted after the first two steps.

AA/AA was counted twice and then subtracted once, so it's now counted once (good). Same is true of AA/KK.
On the other hand, AA/KK/QQ was counted...
3x by the 1st line (see why?),
then subtracted 3x by the 2nd line (see why?),
for a net result of being not counted at all.

Therefore we need to add P(AA/KK/QQ) back in.

AA/AA/KK was counted...
2x by AA, once by KK = 3 times by the 1st line,
then subtracted once by AA/AA and twice by AA/KK = 3 times by the 2nd line,
for a net reuslt of 0.

So we also need to add P(AA/AA/KK) back in.

It turns out, the pattern will continue: after each addition step, we'll have something double-counted, then after the subtraction we'll have something else uncounted. Therefore, we'll alternate between + and - until eventually (once we've accounted for the 9-pair cases), everything is counted once.

However, each step is less important than the previous, and we will have converged very close to the answer well before completing all 9 lines.

After 3 lines, our new upper bound is
9/17 -
C(9,2)*[13 + C(13,2)*C(4,2)²/3]/C(52,4) +
C(9,3)*[13*12*C(4,2)*(3!!) + C(13,3)*C(4,2)³] / C(52,6) / 5!!
=~ 42.098%

That's rather close to the lower bound of 1-(16/17)⁹ = 42.052%, so we needn't bother doing any more steps.

Let's split the difference and call it a day: P(at least one pp 9-handed) = 42.075% +/- .0232%

(But if anyone wishes, they can try the 4th line and I'll proofread it.)

Edit: The reason the lower bound was so much closer to the answer is because the effect of card removal (the source of its error) is much smaller than the amount of over-counting error present in the upper bound. The chance of multiple players having a pocket pair is not very slim. If we wanted the chance of a player getting a pp and making quads, then the upper bound would be accurate, since the chance of two players doing that is rather slim.

I think we may have to give this some more thought. In general, a joint probability can be higher or lower than the probability calculated assuming the events are independent. This is because events can be positively or negatively correlated.

In some poker situations, Event A occurring can decrease the probability of Event B occurring. In other poker situations Event A occurring can increase the probability of Event B occurring.

One player being dealt a pocket pair actually increases the probability of another player being dealt a pocket pair.

As shown in my long post of all the true probabilities of any number of players getting dealt a pocket pair in any table with 2-9 players, we can see that the true prob of 0 players getting dealt a pocket pair is actually higher than the prob calculated via the "binomial-independence" assumption.

Since the true actual prob of 0 players getting dealt a pocket pair is underestimated by the binomial-independence assumption, the true actual prob of one or more players getting dealt a pocket pair is overestimated by the binomial-independence assumption, meaning, of course, that the true prob of one or more players getting dealt a pocket pair is less than the prob calculated via the "binomial-independence" assumption.

From the above post of the true actual tallies, we know that the correct prob of getting one of more pocket pairs dealt at a 9-handed table is exactly:

= 616,784,160,491,584,253,454 / 1,470,451,803,495,476,051,250

= 0.4194521432 (to ten significant digits)

Very good point, whosnext.

P(player 2 pp | player 1 pp) = (1+12*6)/1225 = 5.96% > 1/17

So indeed, 1-(16/17)⁹ is the upper bound, the opposite of what I said.

This means 9/17 - C(9,2)*[13 + C(13,2)*C(4,2)²/3]/C(52,4) = 40.32% was in fact a useful lower bound. Already after that, we can say the answer is 41.187% +/- .865%

The 4th line will provide a tighter lower bound. Our subtotal after the 3rd line was 42.098%. For the 4th line we subtract:

C(9,4)*[C(13,2)*(3!!)² + 3*C(13,3)*(3!!)*C(4,2)² + C(13,4)*C(4,2)⁴] / C(52,8) / 7!!

For a new lower bound of 41.93558452 %

Now we can adjust the estimate to 42.0% +/- 0.1162687%.

The further we go, the closer we'll get to whosnext's exact answer, and after 9 lines they'll match.

My guesstimate is ~42.1%.

This is what I calculated for probability of each player (Pn) not being dealt a pair, given that Pn-1 (and Pn-2, etc.) was/were not dealt pair(s):

P1 = .94118
P2 = .94122
P3 = .94138
P4 = .94158

If P5-P9 stayed at P4's rate of .94158, that would yield 41.9% probability of no players being dealt a pair. If P5-P9, kept increasing at the ratio of P4/P3, that would yield 41.7% probability of no pairs. Since the ratio of Pn/Pn-1 seemed to increase, that would suggest the possibility of a probability < 41.7% of no pairs.

You mean of a pair? If it were <41.7% of no pairs then that would mean >58.3% of a pair, which is definitely wrong.

If you meant "of a pair", then when I feel like it, I'll finish my solution to get the exact answer and test your theory. I'm inclined to believe it's what whosnext said, 41.94521432 %

Here are the results of five simulations of 10 Million trials each. The number of 9-handed deals with at least one pair was:

4193446
4196093
4195596
4192983
4191053

Avg = 41.93834%

Of course, simulations are not perfect. They can be mis-coded and/or their random number generator can be "biased". But I am pretty confident that the above results are accurate and representative.

Yes, sorry for the confusion.

Fwiw, I get a different prob for P3. I didn't calculate P4 but I am guessing that I would get a different answer for P4 as well.

If you are still interested, you can review your derivation of P3 and maybe find a mistake yourself. Or feel free to show your derivation of P3 and I would be able to show how we differ.

I calculated P3 from scratch. I got .941273. Is that what you got?

I was pretty sure I made an error for P4, because the calculations were so complex.

OK, third time's a charm?

P1 = .941176
P2 = .941224
P3 = .941273

If we create a progression, whether we use the arithmetic differences (.00004802, .00004805...) or the ratios (1.00005102, 1.00005105...), it seems to yield the same result to the rounded eighth decimal place:

Prob. of pair = 1 - (Prob. of no pair) = 1 - 0.58054805 = 0.41945195

Of course, this is a very close approximation, not an exact answer.

Yes, I get P3 to be 1,224,204 / 1,300,584 = .941272536...

As you undoubtedly realize, these derivations can be performed with the assistance of a computer quickly tallying various cases. Then the prob is simply a weighted average of the respective tallies. The smaller numbers can be done by hand but the computer is useful as the number of players increases.

Full PIE solution to OP:

9/17 -
C(9,2)*[13 + C(13,2)*C(4,2)²/3]/C(52,4) +
C(9,3)*[13*12*C(4,2)*(3!!) + C(13,3)*C(4,2)³] / C(52,6) / 5!! -
C(9,4)*[C(13,2)*(3!!)² + 3*C(13,3)*3!!*C(4,2)² + C(13,4)*C(4,2)⁴] / C(52,8) / 7!! +
C(9,4)*[3*C(13,3)*6*3² + 4*C(13,4)*3*6³ + C(13,5)*6⁵] / C(52,10) / 9!! -
C(9,3)*[C(13,3)*3³ + C(4,2)*C(13,4)*3²*6² + 5*C(13,5)*3*6⁴ + C(13,6)*6⁶] / C(52,12) / 11!! +
C(9,2)*[4*C(13,4)*6*3³ + C(5,2)*C(13,5)*3²*6³ + 6*C(13,6)*3*6⁵ + C(13,6)*6⁷] / C(52,14) / 13!! -
9*[C(13,4)*3⁴ + C(5,2)*C(13,5)*6²*3³ + C(6,2)*C(13,6)*6⁴3² + 7*C(13,6)*3*6⁶ + C(13,5)*6⁸] / C(52,16) / 15!! +
[5*C(13,5)*6*3⁴ + C(6,2)*C(13,6)*6³*3³ + C(7,2)*C(13,6)*6⁵*3² + 8*C(13,5)*3*6⁷ + C(13,4)*6⁹] / C(52,18) / 17!!

= 41.94521432 % (confirming whosnext's answer)

That's the probability of at least one player getting a pair. If one wants the probability of an exact #players getting a pair, one can use the above and simply multiply each line (after 9/17) by the right coefficient.

Whoever is curious can try doing that to figure out P(exactly 3 players dealt a pocket pair).

Very well done! Excellent work!!

Of course, the reason I am saying that has nothing whatsoever to do with the fact that your result (via PIE) matches my result (via combinatorics and computer-assisted brute force).

Since this type of question does arise quite often, I will sticky the thread since it has a bunch of useful information in it.

I will resurrect this thread to add some new results. I was at a home game a little while ago and we somehow wound up with 10 players. Nobody was thrilled with the prospect of playing 10-handed NLHE. People complained how slow it was, how few hands you get to play, etc. Somebody even mentioned that with 10 players it was likely that someone would have a pocket pair on every deal (why that was such a bad thing I cannot say).

My ears pricked up at that last comment and I remembered this thread. I pulled up the thread to confirm/deny the 50% likelihood of one or more pairs dealt with 10 players. But I was disappointed to see that we had gone from 2-9 player tables, but no results were posted for 10-player tables.

Since that time I have re-programmed the routines to derive the tallies of how many pairs will occur at an N-player table (0-N pairs are possible, of course). This time around I streamlined the programs and removed any quasi-manual elements, including the handling of very large numbers that bedeviled my initial efforts several years ago.

Before I ran my new routines for 10-player tallies, I tested them by trying to reproduce the earlier results of the number of pairs dealt at 2-player to 9-player tables. All of the earlier results were reproduced exactly from 2-player to 8-player (yea) ... except my new programs gave slightly different results than the earlier results for 9-player tables (boo). The tallies for the number of deals at 9-player tables with No Pair and One Pair were different (all the higher number of pairs were identical).

Since the current programs are 100% computer-based tallies from computer-based combinatorics (no manual calculations) I have greater faith in the new results than the programs from a few years ago which relied on, to some degree, manual calculations and procedures.

For comparison, if anyone is interested, here are the "old" results from a few years ago that I now think are incorrect in a few spots:

Here are the new (presumably correct) results:

N=9 Players
----------
No Pair : 853,667,643,098,977,394,688 ( 58.0548 %)
One Pair: 476,852,107,828,489,481,088 ( 32.4290 %)
Two Pair: 120,157,940,672,756,034,048 ( 8.1715 %)
Three Pair: 17,914,600,053,357,815,808 ( 1.2183 %)
Four Pair: 1,740,207,513,284,368,128 ( 0.1183 %)
Five Pair: 114,111,043,097,923,872 ( 0.0078 %)
Six Pair: 5,045,885,892,919,872 ( 0.00034 %)
Seven Pair: 144,930,110,637,312 ( 0.0000099 %)
Eight Pair: 2,450,932,035,552 ( 0.00000017 %)
Nine Pair: 18,577,440,882 ( 0.00000000126%)

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

Based on these new tallies, here is the percentage of deals at a 9-player table with one or more pairs:

41.945214316% (compared to the old result of 41.945214323%) different in the 11th decimal place.

Of course, nobody should care about the 11th decimal place of some obscure probability. However, if tallies are presented as being exact they should indeed be correct exact tallies.

Here are the new results for N=10 players at the table.

For ten players, there are C(52,20) * (1*3*5*7*9*11*13*15*17*19) = 82,492,346,176,096,206,475,125 ways of dealing ten players two-card hands.

Number of Pairs	Tally	Percent
0	45,094,385,528,141,909,566,464	__54.66493271010%
1	27,963,692,497,107,222,755,328	__33.89852973440%
2	7,920,400,008,908,797,046,208	___9.60137561369%
3	1,348,468,215,381,919,577,088	___1.63465858093%
4	152,703,280,565,074,407,168	___0.18511205905%
5	12,007,346,312,416,815,360	___0.01455570955%
6	663,260,602,641,869,232	___0.00080402684%
7	25,385,653,622,938,752	___0.00003077334%
8	643,603,899,839,472	___0.00000078020%
9	9,751,674,696,192	___0.00000001182%
10	67,026,963,861	___0.00000000008%

TOTAL DEALS: 82,492,346,176,096,206,475,125 (100%).

Based on these tallies, the percentage of deals at a 10-player table with one or more pairs is 45.33506728990%.

Of course, there are ways to determine how many players would have to be at the table before the probability of one or more pairs exceeds 50% should that question be of interest. (My best guess is 12.)

Since you bumped this thread I was browsing back through it and noticed this:

Which is of course true, but I'm not sure the simple reason for this was explained. It's not just a quirk due to how the joint probabilities work out, there is a logical reason.

We start with 78 possible pairs in a deck, from the 6 combinations of suits times 13 ranks. If you remove one pair from the deck, there is one possible pair left in that rank, eliminating 5 possible pairs and leaving 73. BUT you also eliminate 96 possible non-pairs, as those two cards could have been dealt with any of the 48 cards in other ranks.

So the chance for paired hole cards goes from 78/1326 or 5.882% initially, to 73/1225 or 5.959% after removing one pair. This same logic holds true for more pairs if they are in different ranks, the probability keeps going up.

Having nothing better to do, I applied my programs to the case of N=11 players at the table (knowing that 11-player poker tables are non-existent).

The same approach can apply. Break down all possible deals into cases. Then utilize the computer to brute-force tally how many pairs are made (from 0 to N) in each case. Some care has to be taken to handle these very large numbers.

Here are the results for N=11 players at the table. For eleven players, there are C(52,22) * (1*3*5*7*9*11*13*15*17*19*21) = 3,719,654,882,122,156,219,242,000 ways of dealing eleven players two-card hands when order of the hands dealt to the 11 players does not matter. (The pcts may appear not to add to 100% due to rounding.)

Number of Pairs	Tally	Percent	Approx Once in How Many Deals
0	1,914,716,483,621,475,679,641,600	__51.47564879807%	2
1	1,304,933,902,122,154,668,908,544	__35.08212303228%	3
2	410,326,228,671,817,898,778,624	__11.03129837781%	9
3	78,527,448,790,800,748,637,184	___2.11114878341%	47
4	10,155,161,125,756,530,284,544	___0.27301353076%	366
5	930,939,855,996,238,792,704	___0.02502758685%	3,996
6	61,667,439,163,551,277,056	___0.00165788067%	60,318
7	2,948,580,441,801,536,256	___0.00007927027%	1,261,507
8	99,620,303,721,434,496	___0.00000267821%	37,338,321
9	2,262,958,128,452,736	___0.00000006084%	1,643,713,525
10	31,091,439,240,576	___0.00000000084%	119,635,982,540
11	195,812,257,680	___0.00000000001%	18,996,026,735,981
.
TOTAL	3,719,654,882,122,156,219,242,000	_100.00000000000%

Based on these tallies, the percentage of deals at an 11-player table with one or more pairs is 48.52435120193%.

I confirm. I redid my PIE calculation in the Julia REPL. Last time, I used my calculator, which rounds to the 10th decimal place.

With PIE the 10-max case would require an additional line of work compared to 9max, which I won't bother doing, but I can get very close by taking my 9max solution and changing the lead coefficients to C(10,x) combos instead of C(9,x).

Result: .45335067289951864

The next line would be a subtraction, so the exact answer is slightly lower.

Based on that, I consider your 10-max result confirmed as well.