[Steve]

1) You're right, the solution I gave assumes they are assigned a random real number between 0 and 1, and the lower number wins.

Like I said, I came up with the solution mathematically and then tested it with a high speed simulation. The math got kind of annoying but it was stuff like this:

Given x is the percentage of legitimate hands player A should bet, and y is the percentage of hands player B should call with. Assuming y>x and y<(1-x/2), A's profit is

P = x[(1-y)($1) + y[(x/2y)(-$1)+(1-(x/2y))($2)]] + x/2[(1-y)($1)+y($-1)] + (1-(3x/2))[(x/2)($1)+(1-(3x/2))($.50)]

Where:
x is the legitimate hands A bets
(1-y)($1) is the times B folds those and A wins $1
y[(x/2y)(-$1)+(1-(x/2y))($2)] is the times B calls, and they show down
x/2 are the times A bluffs
(1-y)($1) are the times B folds when A bluffs
y($-1) are the times B calls As bluff
(1-(3x/2)) are the hands A checks
(x/2)($1) are the times B has a hand inferior to any hand A would check
(1-(3x/2))($.50) are the times A and B have hands both in the range [x, (1-(x/2))]

simplifies to:

P = x/2 - (9/8)x^2 + .50

Which reaches a maximum at x=2/9, P = 5/9


2) I haven't read Jerrod Ankenman's recent posts but I will go check that out... thanks.