View Full Version : AA dealt 4 times in 3 hands

09-15-2005, 11:46 PM
I'm not sure how to calculate it so here it is.
I was playing in a $50 buy in home tourney when this happened. Down to 2 tables with 8 at my table. Players A and B go to showdown and split the pot, each with AA. Next hand goes to showdown with me and player A (same player as in first hand) who shows AA again (takes down pot). Next hand player A (same player again!) raises preflop and everyone folds. After several comments about how he probably has AA again he shows his third consecutive pocket aces!
What are the odds?!?!?!
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09-16-2005, 12:01 AM
Quick and dirty...

3 consecutive A-A hands by itself: 10793860 to 1.

2 consecutive, full ring A-A vs. A-A hands:
[C(10,2) * 1/C(52,4)]^2 = 36193592 to 1

2 consecutive, full ring A-A vs. A-A hands, followed by one particular guy getting A-A:
[C(10,2) * 1/C(52,4)]^2 * 1/221 = 7998784030 to 1

If you'd like to calculate it as if it were heads up, just take the C(10,2)s out.

09-23-2005, 09:27 AM
The odds of someone getting aces 3-times in a row in a home game are infinitely higher than the chances someone stacked the deck /images/graemlins/laugh.gif