[AaronBrown]

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A's initial EV = 3(-3b^2+2b-1)/4. . . . Integrating this expression from 1/3 to 1, and adding 1/3 (from integrating 1 from 0 to 1/3) gives the EV if b is taken from a uniform distribution on [0,1]; I get 7/9 for the answer.

[/ QUOTE ]
If David keeps asking these kinds of questions, he should put a LaTex editor on the boards.

I get +1 instead of -1, 3(-3b^2+2b+1)/4. The integral from 1/3 to 1 of:

1 is 1 - 1/3 = 2/3
b is (1^2 - (1/3)^2)/2 = 4/9
b^2 is (1^3 - (1/3)^3)/3 = 26/81

Substituting those into the unconditional expectation gives:

3[-3(26/81)+2(4/9)+(2/3)]/4
=3[-26+2*4*3+2*9]/(4*27)
=[-26+24+18]/36
=16/36

Adding 12/36 for b < 1/3 gives 28/36 or 7/9 as you say. I don't know where my extra 1/36 came from.

Also, you correctly clarified that by A's expectation, I did not mean to subtract the ante he contributed. It's his expected share of the pot before he bets, not his expectation for playing the game.