a fun little problem

[ArtVandelay]

You are heads-up at the end of a Hold'em tournament (betting structure is irrelevant), and after ridiculously long and even heads-up play you both have T100,000, and the blinds are T50,000 and T100,000. You are in the small blind, and your opponent is all-in on the big blind. Which hands do you call with and which do you fold?

[ArtVandelay]

Oops, that problem is actually boring. The following is not. Exact same set-up except there's only one T100,000 blind (i.e. the SB is zero); now which hands do you play?

[cferejohn]

Any pair. Any Ace. Any King. Any 2 cards ten and up. Any J or Q with a kicker that can make a straight. Any connectors 89 or better. Any suited connectors 67 or better.

[ArtVandelay]

When the SB is T50,000, the answer is you call every hand. The problem is that even 32o wins over 30% of the time against a random hand, but when you fold, you only win the T50,000 back half the time, and you then win again on your SB only half the time, so overall you win only 25% of the time when you fold.

For the more interesting problem, where the SB is 0, the first SB (who clearly has an advantage) wins 53.248% using the following strategy. He should call with all pairs, any hand containing an ace, king, or queen, any suited jack, any offsuit hand with both cards at least 8, any suited hand with both cards at least 7, and the following additional hands: J7, J6, J5, T7, T6s, T5s, 96s.

(In other words, he plays everything but the following: J2-32, J3-43, J4-54, T5-65, T6-76, 97-87, T2s-32s, T3s-43s, T4s-54s, 95s-65s, 86s-76s.)

[Bozeman]

I made one significant assumption to solve this problem analyitically.

P=probability that button wins.
(1-P)=probability that BB wins (I am assuming they use the same strategy)
F=prob. that button folds
W(F)=average prob. that button beats BB's random hand with his non-folded hands.

P=F*(1-P)+W*(1-F), so P=(F+W-FW)/(F+1).

The incorrect approximation I used is that W is linear in F (for holdem, at least, there is some upward curvature). W=aF+b, and b=1/2 if you are facing a random hand.

Given this assumption, P is maximized when
F=sqrt(2+(1-2b)/a) - 1 = sqrt(2) - 1 for b=1/2

This means that you should fold the worst ~41% of your hands.

For hold'em, you should play all Qx or better, all J6 or better, T7 +, 98+, Jxs, T6s or better 3 gappers, 96s + two gappers, 97s one gappers, and 87s connectors.

I suppose one could plug in the winning percentages for all hold'em (or other game) starting hands against a random holding to calculate W(F), though then you need to numerically evaluate P to get the maximum since the numerical value are non-differentiable. Has anyone done this?

Nice problem,
Craig