[schubes]

Playing poker all night, now I can't seem to sleep [img]/images/graemlins/ooo.gif[/img]
I know I'm talking to myself here, but this is the exact answer to the problem using the fact that our hand is 22.

In situation A, hand 1 does not contain a 2, prob. is (36/47)*(34/46)
In B, hand 1 does contain a 2, prob. 2*(2/47)*(36/46)

a) If the 1st card of hand 2 doesn't pair flop or hand 1's cards, and isnt a 2:
For A, prob. is (28/45)*(32/44)
For B, prob. is (32/45)*(32/44)

b) 1st card pairs a card in hand 1, isn't a 2:
A, (6/45)*(33/44)
B, (3/45)*(33/44)

c) 1st card is a 2:
A, (2/45)*(34/44)
B, (1/45)*(35/44)

P =
P(A)*(P(a|A)+P(b|A)+P(c|A)) + P(B)*(P(a|B)+P(b|B)+P(c|B))
= (43!/47!)*( 36*34*(28*32+6*33+2*34) + 2*2*36*(32*32+3*33+35) )

=37.12%

Which is slightly more than using the easier way. I predicted it would be less: if you have a pocket pair shouldn't it be more likely that your opponents have pairs?

Anyone want to speculate why having a pair makes it slightly less likely that your opponents do? [img]/images/graemlins/confused.gif[/img]