|
|
#1
08-13-2004, 01:00 PM
|
|||
|
|||
Splitting AA
This one may have been answered before, but its happened to me twice in about 50-60 AA hands and I was curious. If you're holding AA (or any pocket pair). What are the odds that someone else is too?
|
|
#2
08-13-2004, 01:22 PM
|
|||
|
|||
|
Re: Splitting AA
[ QUOTE ]
If you're holding AA (or any pocket pair). What are the odds that someone else is too? [/ QUOTE ] The probability that any particular opponent has AA given that you have AA is 1/(50 choose 2)=1/1225. The events are disjoint, so the total probability is the sum of the probability for each opponent, 9/1225 at a full table, or about 135.1:1. The probability that some pair of players is dealt all 4 aces is (10 choose 2)/(52 choose 4) = 45/270725.
|
 |
|
|
Linear Mode
