[fnord_too]

It's been a while since I've played around with this type of game theory, but I think there is some subtlety here.

Looking at the one where B can call or raise, but A can only call or fold:

I am going to make the simplifying assumption that A knows B's strategy, and plays optimally against it.

If B raises every time he has a .5 or better (an no other times), then A's best play is to call if he has .625 or better other wise fold. Since he knows that B has somewhere between .5 and 1, 1/4 of the time he will have between .5 and .625 so he is getting break even odds to call here. So before investigating further let me compute the value of the game. Over 16 plays, B will fold 8 and raise 8. Of the 8 he raises, he will win 5 out right for +5, and will be called on 3. Of the three he is called on, A will have between .625 and 1 or an average of .8125 to B's average of .75. So B will gain (.75/.8125) * 2 - (.8125/.75)*2 = ~ -.32 per hand, so ~-.96 for the three hands or just over +4 units for the 8 hands he plays. So that is JUST barely over calling with .5 or above.

I'm going to run through this again with slightly tighter raising standards because I believe I will see a way to solve this in closed form without bluffing. Adding in bluffing will be somewhat trickier.

Say B calls with .5-.75 and raises with .75-1

Then, A should call a raise with .8125 or greater.

So 16 hands, B folds 8, calls 6, raises 2. On the 6 he calls, he wins .75 * 6 = 4.5 units.

On the two he raises, his average hand will be .875. .8125 he wins outright.
of the .1875 he gets called, A will have an average hand of (1.8125/2) = .90625, so B wins (.875/.90625) * 2 - (.90625/.875) * 2 = ~1.931 - 2.071 =~ -.14 per hand or ~ -.28.

So, all in all, he wins 4.5 + 2*(.8125*2 + .1875*(-.14)) =~ 7.6975 units over 16 hands or about .48 per hand. Nice improvement.

I think I am ready to write this out abstractly:
We at least call with anything .5 and above
R = raising point.
C = calling point for A = R + .25*(1-R)
Note: Probability of Raise = 1-R. Probability of call = .5 - (1-R) = R - .5

E(x) = P(folding)*E(folding) + P(calling)*E(calling) + P(raising)*E(raising)
= (0.5)*(0) + (R-0.5)(0.75) + (1-R)E(raising)

E(raising) = P(A Folding) * 1 + P(A calling)*E(A calls) = C + (1-C) E(A calls)

E(A calls) = (((1+R)/2)/((1+C)/2)-((1+C)/2)/((1+R)/2)) * 2

so, deep breath, substituting in R + .25*(1-R) for C and expanding everything out, we need to maximize:

E(x) = (0.5)*(0) + (R-0.5)(0.75) + (1-R)*((R + .25*(1-R)) + ((1-R + .25*(1-R))* (((1+R)/2)/((1+R + .25*(1-R))/2)-((1+R + .25*(1-R))/2)/((1+R)/2)) * 2))

I may have some parens screwed up, but that is about it. I need to go off line and write this out. I will simplify and maximize and post my result for the published, no bluffing, optimal strategy shortly.

Regards,

Eric