[cbloom]

Ah yes, good catch.

An interesting thing about this game is that if your opponent is playing "optimally" you can't beat him just by betting every time.

Consider heads up, n=2, m=1. I have card X, my opponent has card Y. He will bet only if Y > c, with c = b/(a+b)

for this EV computation I consider that the antes are already part of the pot.

if Y < c
you win 2a
if Y > c
if X < c
you lose b
if X > c
you win (2a+b) half the time
and you lose b the other half

So your EV is :

c * (2a) + (1-c) * ( c * (-b) + (1-c) * ( 1/2 * (2a+b) - 1/2 b ) )
c * (2a) + (1-c) * ( c * (-b) + (1-c) * a )

plug in for c and simplify -

b/(a+b) * (2a) + (a/(a+b)) * ( b * (-b) + a * a )/(a+b)

(2ab*(a+b) + a * ( a^2 - b^2 ))/(a+b)^2
(2ab + a * ( a - b ))/(a+b)
(2ab + a^2 - ab )/(a+b)
(ab + a^2 )/(a+b)
a

your EV is exactly the ante, so his is too, and this is just a break-even game. That is, you can't beat him if he's playing optimally, the best you can do is break even.

In fact, if your opponent is playing optimially, you MUST bet if your card is X > c , but if it is less it doesn't matter if you bet or fold. c has been set so that the EV is indifferent to whether you bet or fold when X < c.

Let's solve for what c needs to be to make this indifference -

fold -
if Y > c
EV is 0
if Y < c he folds too
EV is a (push)

bet -
if Y > c
EV is -b
if Y < c
EV is 2a

EV of fold is :

c*a

EV of call is :

c*(2a) - (1-c)*b


c*(2a) - (1-c)*b = c*a

c*(a) + c*b = b
c = b/(a+b)

which is just what 'c' is.