If you had AA and you knew your opponent was set mining what stack-to-pot ratio would you have to have to 'go with the hand' or i.e. to make it so that you never have to fold postflop. It turns out that it is actually less than 12. Let's do 100 trials of the hand.
Flop comes out, you bet full pot with your AA
Opponent will flop a set 11.7 times, he can't call a full pot bet here without a set as he isn't getting the right price so he folds 88.3 times.
+88.3 pots
He's flopped a set 11.7 times but of that you will have flopped a better set 8.4% of those times which is 11.7% X 8.4%= .98% of the time we flop set over set, in other words we will flop set over set about 1 time during our 100 trials and when we do, we win 90% of his stack (95% equity that we win – 5% equity that we lose = 90% of his stack). Therefore 90% of .98 X 12 pots (his stack size) = 10.584 pots
+10.584 pots
He's flopped a set and is not set under set 10.7 times, we have 10% equity in these hands, so we will win 1.07 times by sucking out. 1.07 X 12 pots (his stack size) = 12.84 pots
+12.84 pots
We won 88.3 on the flop, we won 1 by set over set and we won one by sucking out which means that we won 90.35 times, so he won the other 9.65 times. And we are going to just stack off for our entire stack each time. 9.65 X 12 = 115.8
-115.8 pots
If you total this up, we come out slightly down, about 4 pots.
So what is the exact number you need to break even in this scenario?
88.3+.882P +1.07P – 9.65P = 0
Where P is the stack-to-pot ratio
P = 11.47
Keep in mind that people won't play their underpairs perfectly, if they ever call a flop bet without a set then you will show a profit here. And the overpair will never play it incorrectly, ever because he just plays the line bet-bet-bet-get it in.
The implications are that you need at a minimum of 12 stack-to-pot ratio to break even playing the underpair and even then, theoretically, it is possible to show a small profit but in practice, you have absolutely no chance.