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MrDannimal
07-28-2003, 05:15 PM
In a ten-handed Hold 'Em table, what's the probability that 3 people (one each) will be dealt AA, KK, QQ?

Copernicus
07-29-2003, 03:05 PM
I get about .0014%, or 1 time out of 70,689 hands. To estimate it, the probability of any particular pair being dealt to 1 player is about .45%. You need three particular pairs which would be .45%^3. However you have C(10,3) players that can draw those 3 hands, so the total is 120*.0045^3, or .0011%. Its actually a little higher because after one of the players gets one of the pairs the probability of another player getting one of the other pairs increases slightly.

MrDannimal
07-30-2003, 03:08 PM
Much thanks!

BruceZ
07-31-2003, 01:12 AM
You would have to multiply your probability by 6 since there are 3! ways the first 3 guys can hold the 3 pairs. This would give .0066%, but to take into account the hands not being independent, instead of .0045^3*C(10,3) we would use (18/1326)*(12/1225)*(6/1128)*C(10,3) = .00849%. This is accurate, but for an exact solution we would also have to take into account the fact that C(10,3) overcounts times we get more than 3 of these pairs. We can use inclusion-exclusion:

Probability of 3 particular guys having these 3 pairs:
P3 = (18/1326)*(12/1225)*(6/1128)

Probability of 4 particular guys having these 3 pairs:
P4 = (18/1326)*(12/1225)*(6/1128)*(3/1035)*3

Probability of 5 particular guys having these 3 pairs:
P5 = (18/1326)*(12/1225)*(6/1128)*(3/1035)*(2/946)*C(5,2)

Probability of 6 particular guys having these 3 pairs:
P6 = (18/1326)*(12/1225)*(6/1128)*(3/1035)*(2/946)*(1/861)*C(6,3)

P(AA,KK,QQ out with 10 players) =
C(10,3)*P3 - C(10,4)*P4 + 3*C(10,5)*P5 - 17*C(10,6)*P6
= <font color="red">.00836%</font> or <font color="red">1 in 11,959.</font>

This is a slight twist on the normal inclusion-exclusion. When we subtract off the times we have 4 pairs, the times we have 5 pairs get subtracted off 4 times since 5 pairs count as 4 pairs 4 ways, so we add back 3 times the number of 5 pairs. Then the number of 6 pairs get added 6*3 = 18 times, so we subtract off 17 times the number of 6 pairs. The inclusion-exclusion step doesn't change the final answer much.

Copernicus
07-31-2003, 05:10 PM
My bad on the factor of 6. I'm too used to thinking order can be ignored with combinatorials!

Thats why I let programs do the math, and my staff do the heavy calculating, while I do the heavy lunch and golfing!

BruceZ
07-31-2003, 08:54 PM
My bad on the factor of 6. I'm too used to thinking order can be ignored with combinatorials!

Usually. This case is interesting since there there is a permutation group within a single success.

Thats why I let programs do the math, and my staff do the heavy calculating, while I do the heavy lunch and golfing!

Actuary? Engineer? Got any contract work to farm out? You can email me a problem in the morning, and I'll send you back the solution with an invoice.

BruceZ
08-06-2003, 05:52 AM
Irchans has found the exact portion of this calculation to be in error. The approximate result is OK. Here is a correction to the exact calculation. We are confident that this is now correct since P3-P6 were used to obtain the same answer to a related problem 2 different ways.

Probability of 3 particular guys having these 3 pairs:
P3 = 18/C(52,2)*12/C(50,2)*6/C(48,2)

Probability of 4 particular guys having these 3 pairs:
P4 = 6/C(52,2)*6/C(50,2)*6/C(48,2)*1/C(46,2)*3*4!/2

Probability of 5 particular guys having these 3 pairs:
P5 = 6/C(52,2)*6/C(50,2)*6/C(48,2)*1/C(46,2)*1/C(44,2)*3*5!/2^2

Probability of 6 particular guys having these 3 pairs:
P6 = 6/C(52,2)*6/C(50,2)*6/C(48,2)*1/C(46,2)*1/C(44,2)*1/C(42,2)*6!/2^3

In P4 and P5, the 3 represents the number of ways to pick the duplicated pair. In P4, 4!/2 represents the permutations of the 4 cards with 1 duplicate pair. In P5 there are 2 duplicate pairs, and in P6 there are 3 duplicate pairs.

P(AA,KK,QQ out with 10 players) =
C(10,3)*P3 - C(10,4)*P4 + C(10,5)*P5 - C(10,6)*P6
= <font color="red">0.0084020111%</font> or <font color="red">1 in 11,902</font>

Amazingly, the coefficients are +/-1 just like in the normal inclusion-exclusion case. In this case we want to count 3,4,5, and 6 pairs exactly once which is the way inclusion-exclusion is normally used.

Here is a summary of the counting. Each row is for the number of pairs being counted, and the columns are how many of each number of pairs get counted for each one of the number of pairs for that row. For example, when we count the 3 pairs, we get 2 four pairs, 4 five pairs, and 8 six pairs.

<font class="small">Code:</font><hr /><pre>
3 4 5 6
------------------
3: 1 2 4 8

4: 1 4 12

5: 1 6

6: 1

</pre><hr />

This next table shows the accumulated number of each number of pairs counted for each term of the proposed solution. For example, after the second term we have counted the 3 pairs and 4 pairs once, the 5 pairs 0 times, and the 6 pairs -4 times. The goal is to count each one exactly once.

<font class="small">Code:</font><hr /><pre>
3 4 5 6
------------------
3: 1 2 4 8

4: 1 1 0 -4

5: 1 1 1 2

6: 1 1 1 1
</pre><hr />

irchans
08-06-2003, 12:09 PM
Bruce gives the probability that at least one player gets QQ, at least one player gets KK, and at least one player gets AA.

To get the probability that exactly one player gets AA, exactly 1 gets KK, and exactly 1 gets QQ, we can use two methods:

1) Modified inclusion exclusion formula using Bruce's P3, P4, P5, and P6 numbers

P3exact = c[10,3]*P3 -c[10,4]*2*P4 +c[10,5]*4*P5 -c[10,6]*8*P6
= 0.000083167198468

2) Sum up the probability that 3 specific people get AA, KK, QQ in any of the six orders AND the other 7 do not get AA or KK or QQ. This comes out to

method2 = c[10, 3]* (18/1326*12/1225*6/1128)*(prob noone else get AA,KK, or QQ)

Use Inclusion Exclusion to get that last probability:

(prob no one else get AA or KK or QQ)
=(1 - 7*partic + c[7, 2]*partic2 - c[7, 3]*partic3);

where

partic = 3/c[52 - 6, 2];
partic2 = 6/c[52 - 6, 2]/c[52 - 8, 2];
partic3 = 6/c[52 - 6, 2]/c[52 - 8, 2]/c[52 - 10, 2];

Then I get

method2 = 0.000083167198468