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Check my math, please

Posted By: Chuck Bower
Date: Friday, 27 July 2007, at 9:35 p.m.

In Response To: Rare events (Stick)

I think this is right, but someone should verify, please.

Assume a session of 100 deals with six players at the table for all 100 deals.

P(AA) = 1/221.

Probability that a specific player never gets aces in that session = (220/221)^100 = 0.635390.

P(exactly one AA) = (100)*(1/221)*(220/221)^99 = 0.288814.

P(exactly two AA) = (100*99/2)*(1/221)^2*(220/221)^98 = 0.064983.

P(exactly three AA) = (100*99*98/3/2)*(1/221)^3*(220/221)^97 = 0.009649.

P(exactly four AA) = (100*99*98*97/4/3/2)*(1/221)^4*(220/221)^96 = 0.001064.

P(zero through four AA) = sum of above five numbers = 0.999900. (Wow, THAT looks pretty bizarre in itself! But I diverge.)

P(five or more AA) = 1 - sum = 0.000100 or one time in 10,000 sessions.

We assumed six players so if we were wondering how likely it would be for any of the six to get five (or more) AA we multiply by 6 and get 0.000600 or one time in 1660 sessions.

Fact is that because this is elimination, there won't always be six players at the table. Say on average it's four (or pick your own number). Then 1/(4*0.000100) = one time in 2500.

So, Michael, do you think the chances of winning the lottery are one in a few thousand?

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