[BruceZ]

[ QUOTE ]
I know the probability of getting AA on any individual hand is 1/221. My question is, what is the probability that this will happen 5 times out of 60 hands? I'm not sure which formulas would be helpful in this situation. Thanks for your help!

(Hopefully this doesnt get jammed in the trash with all the "OMG AA VS KK" posts)

[/ QUOTE ]

Exactly 5 AA:

C(60,5)*(1/221)^5*(220/221)^55 =~ 1 in 123,873

or in Excel =BINOMDIST(5,60,1/221,FALSE) =~ 1 in 123,873


For 5 or more:

1 -
[C(60,0)*(1/221)^0*(220/221)^60 +
C(60,1)*(1/221)^1*(220/221)^59 +
C(60,2)*(1/221)^2*(220/221)^58 +
C(60,3)*(1/221)^3*(220/221)^57 +
C(60,4)*(1/221)^4*(220/221)^56 ]

=~ 1 in 118,746

or =1 - BINOMDIST(4,60,1/221,TRUE) =~ 1 in 118,746