[cbloom]

The solution is : bet all your hands X > c with

c = ( b / (n*a + b) ) ^ (1/(n-1))

For your example case, a = 1, b = 2, n = 3 this is

c = 0.63246

How do you get this? Principle of indifference. When your hand is X = c, you should have the same EV for betting or folding. Consider your ante as already being in the pot. If you fold, your EV is zero. If you bet, I assume that all other players are using the same strategy. They will only call if they have hands X >= c. The chance of a tie is zero, so that means if you have any callers, they will beat you. Thus, you only win the pot if noone calls. Thus, your chance of winning is just the chance that everyone else has a hand < c :

P = c ^ (n-1)

And your EV is :

P * n*a + (1-P) * (-b)

That is, when you win (P chance) you win the antes (n*a). When you lose (1-P chance) you lose your bet (-b). This EV should be equal to the EV of folding, which is zero.

P * n*a + (1-P) * (-b) = 0

P * (n*a + b) - b = 0

P = b / (n*a + b)
c^(n-1) = b / (n*a + b)

c = ( b / (n*a + b) ) ^ (1/(n-1))

And really this game's not very interesting at all.

For the case of a=1,b=2, n = 2 (heads up) , this is:

c = 0.4

it's correct to bet slightly more than half your hands.