[eastbay]

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At the point of maximum edge for the small blind, the optimal strategies are:

sb push hands: 22+,A2+,K2+,Q2+,J2+,T5o+,T2s+,96o+,93s+,86o+,84s+, 76o,74s+,64s+,54s (0.737557)

bb call hands: 22+,A2+,K2+,Q2+,J2+,T5o+,T2s+,96o+,93s+,86o+,84s+, 76o,74s+,64s+,53s+ (0.740573)

Again, there are no mixed strategies, and moreover, within one hand they are playing the same strategy! I suspect this may point to a deeper principle of some kind, but I'm not sure what it might be.

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I thought this might be some kind of interesting mathematical result, but now I can't see how it could be true that SB and BB play almost the same hands. I'll do some quick computations by rounding the numbers, and ignoring ties.

We're talking about when the stacks are around 4BB (I'll assume that includes the posted chips). Your strategies have the players playing roughly 3/4 of their hands.

Let's look at the BB's "worst" hand. Whatever it is, he should be indifferent between folding and calling the all-in when his pot odds leave him indifferent. This means he should have a 3/8 chance (37.5%) of winning across SB's range of hands, with that worst hand.

Look at SB's worst hand (whatever it is). He should be indifferent between pushing all-in and folding. Since BB folds 1/4 of the time, one can verify that his worst hand must have around a 11/24 chance (45.8%) of winning (based on elementary EV calculations).

Those two probabilities are so far apart, I have a hard time believing that both players would use the same range of hands in equilibrium. Which of us made a mistake somewhere?

alThor

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Using blinds 1,2 and stacks 8:

For the BB:

EV_fold = EV_call
-2 = 8*pwin +(1-pwin)(-8)
pwin = 3/8

For the SB:

EV_fold = EV_push
-1 = 3/4[ 8*pwin +(1-pwin)(-8)] +1/4(2)
-1 = 3/4(16*pwin - 8) +1/2
-1 = 12*pwin -6 +1/2
pwin = 9/24 = 3/8

Did you maybe set the LHS for SB to zero rather than -1?

Moreover, 76o which I eyeballed as likely to be BB's weakest hand against SB's range, has an equity of .376.

eastbay