Theoretical Problem with Practical Implications

[Gabe]

Let’s say you’re playing 20/40 holdem. On the turn the board has no pair. Your opponent bets all in. You know he has either two pair or a set, but he could have any two pair or any set. You have an over-pair. Neither of you has a straight or flush draw. How big must the pot be to call his bet?

[bigpooch]

There are 12 combinations of sets and 54 combinations of
two pairs. In the case of a set, you have a two-outer and
in the last case, you have 8 outs (and you don't even know
6 of them!), all out of 44 cards. Your overall chances are
computed by the weighted average:

1/66 * (12x2/44 + 54x8/44) = 19/121

In other words, the pot must be 6 7/19 bets after you call
to be profitable. If the opponent went all-in for the full
$40 on the turn, the pot size before you call must contain
at least (6 7/19-1) x $40 = $214.74.

Of course, in real life, you probably will need much better
odds, as it is not that likely your opponent will have some
of the two pair combinations! Also, sometimes players
misrepresent their hand as two-pair or better when all they
have is a pair and a four-flush or a four-straight. To top
it all off, your opponent may not always semibluff, so you
basically have to determine how your opponent plays.