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Old 04-01-2005, 12:51 AM
gaming_mouse gaming_mouse is offline
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Default Re: simple Q for you math geniuses

[ QUOTE ]
Before getting into this, I would like to have somebody critique the logic I'm using here. I feel it's a very good approximation, but I could be missing something.

Here is the way I would calculate the probabilty: The chance of any one person being dealt a pocket pair is 13 * (4C2)/52C2 (mathematically, this means 13 times "4 Choose 2" divided by "52 Choose 2". I get this to be 5.88%.

If you model this as a Bernoulli trial with a binomial distribution (and I think it's a good approximation), this means a "success" {i.e., pocket pair} will occur 5.88% on any one particular hand and a "failure" {i.e. a non-pair} will occur 94.12% of the time on any one particular hand.

Using this assumption, I get the following results at a nine-person table:

Chance of No Pocket Pairs: (0.9412)^9 = 58.0%

Chance of Exactly 1 Pocket Pair:
(9C1) * (.9412)^8 * (.0588) = 32.6%

Chance of Exactly 2 Pocket Pairs:
(9C2) * (.9412)^7 * (.0588)^2 = 8.1%

Chance of Exactly 3 Pocket Pairs:
(9C3) * (.9412)^6 * (.0588)^3 = 1.2%

Chance of Exactly 4 Pocket Pairs:
(9C4) * (.9412)^5 * (.0588)^4 = 0.1%


John <font color="black"> </font> <font color="blue"> </font> <font color="blue"> </font> <font color="black"> </font>

[/ QUOTE ]

The approximation will be fairly accurate. I'll do the second term of inclusion exclusion to compare the approximation for 2 pocket pairs. It has:

((9 choose 2) choose 2) = 630 terms.

Of those, (9 choose 4)*3 = 378 are non-overlapping (ie, they represent 4 people all having pocket pairs)

The chance of 3 particular people all having a pocket pair is (using a very slight and negligible cheat by assuming that all 4 are of different ranks):

(13*6/(52 choose 2))*((12*6+1)/(50 choose 2))*((11*6+2)/(48 choose 2)) = 0.00021131857

And for 4:

0.00021131857*(10*6+3)/(46 choose 2) = .0000128628695

The second term of inclusion exclusion is therefore:

378*.0000128628695 + (630 -378)*0.00021131857= 0.0581144443

Which means that the 1 term approximation is not very accurate after all. The 2nd term approximation is:

12.6 - 5.8 = 6.8

but is only guarenteed to be accurate within 5.8%. I don't feel like going after the 3rd term, as it only get messier. But it looks like the binomial answer will be pretty close after we add back on the 3rd term.
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