Prob. dealt AA back to back losing/pushing both times?

[Money2Burn]

Just curious, I was playing at a private game last night and was dealt AA in consecutive hands 9 handed. Was then HU to the flop both times and lost the first one and pushed the second. It really had me steamed because the second hand was a +10k pot. Anyway I'm not too good with statistics and i was wondering if anyone could calculate this for me. thanks.

[numeri]

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. ;-)

[Money2Burn]

Sounds plausible. Thanks. I'm going to go drink away my depression now.