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 01-31-2017, 08:25 PM #1 SpewUrStack enthusiast     Join Date: Nov 2011 Location: Massachusetts Posts: 60 How often is a pair being dealt in a 9 handed game? 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?
 01-31-2017, 08:47 PM #2 David Sklansky Administrator     Join Date: Aug 2002 Posts: 13,891 Re: How often is a pair being dealt in a 9 handed game? 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.
01-31-2017, 10:46 PM   #3
whosnext
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Join Date: Mar 2009
Location: California
Posts: 3,789
Re: How often is a pair being dealt in a 9 handed game?

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.

Quote:
 Originally Posted by whosnext 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.

02-01-2017, 12:39 AM   #4
heehaww
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Join Date: Aug 2011
Location: It was an attractive goat in AR
Posts: 4,182
Re: How often is a pair being dealt in a 9 handed game?

Quote:
 Originally Posted by David Sklansky 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.
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.

02-01-2017, 07:04 AM   #5
SpewUrStack
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Join Date: Nov 2011
Location: Massachusetts
Posts: 60
Re: How often is a pair being dealt in a 9 handed game?

Quote:
 Originally Posted by whosnext 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.
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!!!

02-01-2017, 07:05 AM   #6
SpewUrStack
enthusiast

Join Date: Nov 2011
Location: Massachusetts
Posts: 60
Re: How often is a pair being dealt in a 9 handed game?

Quote:
 Originally Posted by heehaww 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.
Thank you! Looking forward to hearing the rest!

 02-01-2017, 03:16 PM #7 heehaww Pooh-Bah     Join Date: Aug 2011 Location: It was an attractive goat in AR Posts: 4,182 Re: How often is a pair being dealt in a 9 handed game? 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. Last edited by heehaww; 02-01-2017 at 03:34 PM.
 02-01-2017, 04:38 PM #8 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 3,789 Re: How often is a pair being dealt in a 9 handed game? 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) Last edited by whosnext; 02-01-2017 at 04:54 PM.
 02-01-2017, 06:59 PM #9 heehaww Pooh-Bah     Join Date: Aug 2011 Location: It was an attractive goat in AR Posts: 4,182 Re: How often is a pair being dealt in a 9 handed game? 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.
 02-01-2017, 09:01 PM #10 David Sklansky Administrator     Join Date: Aug 2002 Posts: 13,891 Re: How often is a pair being dealt in a 9 handed game? What about if you use my deck that has one each of deuces through kings and 40 aces?
02-02-2017, 01:36 AM   #11
heehaww
Pooh-Bah

Join Date: Aug 2011
Location: It was an attractive goat in AR
Posts: 4,182
Re: How often is a pair being dealt in a 9 handed game?

Quote:
 Originally Posted by David Sklansky What about if you use my deck that has one each of deuces through kings and 40 aces?
In this case, the binomial approximation (99.966%) is a lower bound because clearly, if one player has a non-ace, that improves another player's chance of getting two aces. Thus, we can estimate 99.983% knowing that the maximum error is .017%

But here's the exact answer below:

9*C(40,2) / C(52,2) -
C(9,2)*C(40,4) / C(52,4) +
C(9,3)*C(40,6) / C(52,6) -
...
+ C(34,12) / C(52,12)

= 99.99781063 %

The convergence with PIE was much slower this time, requiring almost all the steps to start getting close.

 02-02-2017, 02:22 PM #12 heehaww Pooh-Bah     Join Date: Aug 2011 Location: It was an attractive goat in AR Posts: 4,182 Re: How often is a pair being dealt in a 9 handed game? Actually for David's problem, it's probably easier to calculate P(no pairs) and subtract.
 02-02-2017, 04:04 PM #13 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 3,789 Re: How often is a pair being dealt in a 9 handed game? Here is a first attempt at the "easy" way of doing David's problem. Prob(At least one pair) = 1 - Prob(No Pairs) Prob(No Pairs) = [[Sum[N=0,18] {Prod[K=1,N] (18-2(K-1))/(18-(K-1)) * C(40,N)*C(12,18-N)}]] / [[Sum[M=0,18] C(40,M)*C(12,18-M)]] I coded this in R and got: = 99.99780480%
 02-03-2017, 03:43 AM #14 heehaww Pooh-Bah     Join Date: Aug 2011 Location: It was an attractive goat in AR Posts: 4,182 Re: How often is a pair being dealt in a 9 handed game? What I had in mind was, for P(no pair), we only need to focus on where the non-aces are placed in the deck and among the players. 1 - [C(34,3)*29 + C(34,2)*9*28 + 34*C(9,2)*27 + C(9,3)*26] / C(52,12) = 99.99781063% (same as my PIE answer) There are 29 ways for one non-ace to go to each player, and C(34,3) ways to put the other 3 in the deck, so the product of those is the #ways to deal 9 non-aces and give one to each player. With 10 dealt it's the same thing except there are 9 possibilities for who gets zero aces. And so on.
 02-03-2017, 04:15 AM #15 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 3,789 Re: How often is a pair being dealt in a 9 handed game? My formula in post #13 purports to calc the prob by placing the aces so that no pair results. Theoretically, the two formulas should lead to the same result. Clearly I made either a conceptual, forumulaic, or execution error somewhere. And I am not really interested in knowing which or where.
 02-04-2017, 12:36 AM #16 David Sklansky Administrator     Join Date: Aug 2002 Posts: 13,891 Re: How often is a pair being dealt in a 9 handed game? You guys are so funny. I posed the problem as a cute way to illustrate that there are cases where if the first hand dealt is not a pair, the chances the next one is not, go down instead of up. I didn't actually want it answered.
 02-04-2017, 01:13 AM #17 AllInNTheDark old hand     Join Date: Jun 2011 Location: Pacific coast Posts: 1,483 Re: How often is a pair being dealt in a 9 handed game? My guesstimate is ~42.1%.
 02-04-2017, 10:44 PM #18 AllInNTheDark old hand     Join Date: Jun 2011 Location: Pacific coast Posts: 1,483 Re: How often is a pair being dealt in a 9 handed game? 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.
02-05-2017, 01:31 PM   #19
heehaww
Pooh-Bah

Join Date: Aug 2011
Location: It was an attractive goat in AR
Posts: 4,182
Re: How often is a pair being dealt in a 9 handed game?

Quote:
 Originally Posted by AllInNTheDark 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 %

 02-05-2017, 04:47 PM #20 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 3,789 Re: How often is a pair being dealt in a 9 handed game? 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.
02-05-2017, 05:09 PM   #21
AllInNTheDark
old hand

Join Date: Jun 2011
Location: Pacific coast
Posts: 1,483
Re: How often is a pair being dealt in a 9 handed game?

Quote:
 Originally Posted by heehaww 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 %
Yes, sorry for the confusion.

02-06-2017, 05:59 AM   #22
whosnext
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Join Date: Mar 2009
Location: California
Posts: 3,789
Re: How often is a pair being dealt in a 9 handed game?

Quote:
 Originally Posted by AllInNTheDark 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.
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.

02-06-2017, 03:43 PM   #23
AllInNTheDark
old hand

Join Date: Jun 2011
Location: Pacific coast
Posts: 1,483
Re: How often is a pair being dealt in a 9 handed game?

Quote:
 Originally Posted by whosnext 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.

 02-06-2017, 04:01 PM #24 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 3,789 Re: How often is a pair being dealt in a 9 handed game? 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.
 02-08-2017, 12:23 AM #25 heehaww Pooh-Bah     Join Date: Aug 2011 Location: It was an attractive goat in AR Posts: 4,182 Re: How often is a pair being dealt in a 9 handed game? 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). Last edited by heehaww; 02-08-2017 at 12:32 AM.

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