Re: Prob. dealt AA back to back losing/pushing both times?
P(AA) = C(4,2)/C(52,2) = 1/221 ~ 0.005
(or ~0.5% of the time)
So P(AA) twice is (1/221)*(1/221) = 0.00002
(or ~0.002% of the time)
The probability for losing both is harder to do, since it depends on your opponents. If we assign them both random hands, you'll win ~85% of the time for each, so we can multiply P(AA) by P(losing) for each.
P(AA twice and losing both) = (1/221)*(0.15)*(1/221)*(0.15) = 0.000 000 5
(or about 0.00005% of the time)
I think that's all right. Who knows. ;-)
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