Thread
:
At least two Aces?
View Single Post
#
3
10-05-2002, 07:45 AM
BruceZ
Senior Member
Join Date: Sep 2002
Posts: 1,636
Correction to 1st part
In part 1 I computed the probability of getting 3 or more aces given that you got 2 aces. We want the probability that we have 2 aces given 1 ace which is just P(2 ace)/P(1 ace) = 26%/48% = 54%, so you are a favorite here too.
BruceZ
View Public Profile
Find all posts by BruceZ