I think this is actually a complicated problem because of the possible combinations that 4 hands can wield. I can calculate the odds of AA hitting two times in a row fairly easily.
((4/52 * 3/51)^2) == 2.04746013 × 10^-5 which is pretty damn rare.
However, there are more combinations involving AA in 4 hands.
I will label the starting hands as A - B - C - D where A is the first hand dealt. I will assume that the letters beginning with ! are the ones that got dealt the AA. Meaning a Hit.
So according to my little chart, with 4 hands there are 6 different ways the hole cards can land with AA twice.
Case 1 !A - !B - C - D
Case 2 !A - B - !C - D
Case 3 !A - B - C - !D
Case 4 A - !B - !C - D
Case 5 A - !B - C - !D
Case 6 A - B - !C - !D
Out of 4 sets of hole cards how many possible combinations are there?
(52*51)^4.
Well in 1 set there are 52*51 combinations. So 52/51 ^ 4 are how many combinations that exist in 4 hands.
So out of 4.94645519 × 10^13 only 6 wield combinations with AA.
But there are 12 ways to be dealt an AA. So 12 * 6 = 48.
Possible AA hands:
As Ah Ac Ah Ad As Ah Ad
As Ac Ac As Ad Ah Ah As
As Ad Ac Ad Ad Ac Ah Ac
Putting it altogether...
48 / 4.94645519 × 10^13 == 9.7039189 × 10^-13
Intuitively speaking, in 1 hole card deal there are 2652 possible hole card combinations. And 12 of them are AA. Raise this to the 4th power because you have 4 combinations, and your sample set will be even smaller than this due to the ordering.Statistics: Posted by Rhapsodus — Sun Jul 10, 2005 6:15 pm
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