I used 50% because it is easy to work with, and it shows that 50% isn't a magical breakeven point - which would have to exist if you agree that the expected ret

I used 50% because it is easy to work with, and it shows that 50% isn't a magical breakeven point - which would have to exist if you agree that the expected return is a continuous function and you believe there is a point where your expected return is greater than 0. > You don't have a 50% chance of winning 48%, you have a >50% chance of winning 98%. You only win 48% in my model because the other 50% is your own money that you're getting back. Anyway, let's do it generically. You have put in x of the money (more than 0.5 in your scenario, but it doesn't matter). Then there are two ways to formulate your expected return. First, what I was doing before - if you lose, you lose x and if you win you win (.98-x). You have x chance of winning, so expected return is: (1-x) chance of losing x + x chance of winning (.98-x) (1-x) * -x + (.98-x) * x (x-1 + .98-x) * x -.02x Alternatively, you initially give up x and then have x chance of winning the whole pot, .98: give up x + x chance of winning .98 -x + .98x -.02x So either way, no matter what portion of the pot is yours, your expected return is to lose 2% of the money you put in.