[GrunchCan]

I don't know if you're missing anything; I could be wrong.

But let's try an example with 2 short stacks.

Hero: 100 BB
MP: 10 BB
LP 10 BB

Preflop: Hero is UTG with 2[img]/images/graemlins/club.gif[/img] 2[img]/images/graemlins/diamond.gif[/img]. Hero calls 1 BB. Some folds. MP raises to 4 BB. Some folds. LP calls 4 BB. Blinds fold. Hero ... ?

Pot is 10.5 BB, offering 3:10.5 expressed. For simplicity, assume that if Hero calls he will win 10% of the time (I claim this is a valid assumption -- readers are invited to confirm it). Also assume that MP will get is alll in the middle 100% of the time, and LP will also get it in 25% of the time.

EV = {0.10 * [10.5 + 6 + (.25 * 6)]} + (0.90 * -3) = 1.8 - 2.7 = -0.9

Which makes this call -EV.

Assuming that the PFR will always get all-in no matter how many opponents he has (not a valid assumption), and coldcalling opponents will get all-in 25% of the time, how many coldcalling opponents do we need for our call to be breakeven?

EV = 0.10 * [10.5 + 6 + (n * .25 * 6)] + (0.90 * -3)

...setting EV to 0 and solving for n...

0 = .1(16.5 + 1.5n) - 2.7
2.7 = 1.65 + .15n
2.7 - 1.65 = .15n
(1.05/0.15) = n
n = 7

So if the opponents all have an average stack of 10BB at the beginning of the hand and PF coldcallers will get all-in 25% of the time postflop, we need 7 coldcallers to make this call +EV.

This is kind of an extreme example, but it illustrates the point that I was trying to make: there is no always in poker. The formula could be generalized to make the chance that the coldcallers will get all-in and the average stack variables, but I have to get back to work. [img]/images/graemlins/smile.gif[/img]