Quote:
Originally Posted by spadebidder
The first pocket pair at a 10-handed table is
1 - (1-(78/1326))^10 = 45.46%
The second pocket pair of a different rank, given that the first one happened, is
1 - (1 - (72/1225))^9 = 42.03%
The product of those is 19.1%, which is the correct answer. So I have no idea where I got 9.5% in the earlier post. It was wrong.
These are mutual-exclusive approximations but are very accurate. You could use inclusion exclusion to get the exact answer and it would be within about a tenth of a percent or so.
First of all, they are independent approximations, not mutually exclusive. Secondly, if you had used a mutually exclusive approximation, you would have quickly seen that the answer can be no bigger than 15.6%:
C(10,2)*78/C(52,2)*72/C(50,2) =~ 15.6%
Thirdly, if you were going to use an independence approximation, the following one would have given an even tighter upper bound of about 14.4%, though it takes some thought to realize that this is still an UPPER bound:
1 - [1 - 78/C(52,2)*72/C(50,2)]
C(10,2) =~ 14.4%
Fourthly, this turns out to be one case where the independence approximation is not very accurate as the exact answer is about 11.3%.
The exact calculation uses a variation of inclusion-exclusion where the adjustment factors are not just +/- 1, since we are starting with 2 players. Each line of inclusion-exclusion has multiple terms which each require their own adjustment factor which are shown in bold below. Getting these adjustment factors right takes some thought. The rest isn't difficult. Here are the first 4 terms:
C(10,2)*78*72 /C(52,2)/C(50,2) -
C(10,3)*(
2*78*72*66 +
1*78*72*1*3) /C(52,2)/C(50,2)/C(48,2) +
C(10,4)*(
3*78*72*66*60 +
2*78*72*66*1*C(4,2) +
1*78*1*72*1*3) /C(52,2)/C(50,2)/C(48,2)/C(46,2) -
C(10,5)*(
4*78*72*66*60*54 +
3*78*72*66*60*1*C(5,2) +
2*78*72*1*66*1*15) /C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)
=~ 11.3%
The first line is for 2 players, the second line is for 3 players, etc. The first summed term on the second line is for 3 different pairs. These are counted C(3,2) = 3 times by the first line, hence the adjustment factor subtracts 2. The second term is for 3 players holding 2 different pairs. These are counted twice by the first line, hence the adjustment factor subtracts 1.
The first term on each line is for N players having N different pairs, starting with N=2 on the first line, N=3 on the second line, etc. The second term on each line is for N players holding N-1 different pairs starting with N=3 on the second line. Starting with N=4 on the third line, the third term is for N players holding N-2 different pairs.
Finally, the question asked by the OP concerns set over set. For that problem, if you consider the flop first, then the above probability is not needed. It becomes an easy matter to make an exact calculation involving 2 or 3 players holding 2 or 3 specific pairs. See my earlier post in this thread with an example for 3 sets.
Last edited by BruceZ; 03-31-2010 at 07:29 AM.