[elindauer]

See my response to your post titled "Cute Applicable Math Question" in which I derive the formula for the probability of player 1 doubling up when facing this situation over and over is:

p^2/(1-2p(1-p))

Setting this equal to .6, we find that p=.55.


This is a fantastic question, since when you are calling all-in, your call needs to be a 60% favorite (in most tournament payout schemes) to actually be positive EV in real dollars. I believe Mr. Sklansky is driving at a fundamental question in tournament poker... how do you apply this fact to situations when you are not calling all-in?

We've found part of the answer... if you're calling for half your stack, you'll need to be a 55% favorite just to be breaking even in real payout dollars. I suspect that as the percentage of your stack gets smaller, the amount you need to be favored by will drop perpetually closer to 50%.


Now an open question... you almost never face an even money call in real tournament poker, you'll always getting odds on your money. How does this factor into the equation?

For example, you are faced with calling for half your stack on the river, and the pot is laying you 2:1 odds. What win rate do you need to guarantee a double up probability of 60%? We know you'd need a 55% win rate if it was an even money bet... do you just need the same 10% expected return on your money (translating to 11/30 win rate if the pot lays you 2:1), or is the answer more complex?