If the opponent folds their overpair 14.92% of the time then pushin all-in is break even.

EV of a call with overpair
(.39 * 362 + .61 * -274) = -25.96
EV of a fold
+148

X=call %
Y=fold %

X+Y = 100%
X(-25.96) + Y(148) = 0
Solve for Y
Y= 14.92%

UPDATe: This calculation is incorrect

Your EV of a fold is wrong
depending on the exact starting situation u need between 20-25% fold here to break even

Yea your right. It is only +108. Also the EV of a call is wrong as well. I over looked the fact that Kenny had not yet put in the $40 to call the re-raise.

Correct Calculations:

EV of a call with overpair
(.39 * 362 + .61 * -314) = -50.36
EV of a fold
+108

X=call %
Y=fold %

X+Y = 100%
X(-50.36) + Y(108) = 0
Solve for Y
Y= 31.8%

You need a fold 31.8% of the time to break even. therefore, I would say this is a bad all-in ... unless it was against a rock(who would fold overpair) or a LAG (who might just have overs therefore fold)

I hope this doesn't come off as sarcastic, but can you guys really do that math on the fly during the time you have to make up your mind whether to push, call, or fold?

NO but by doing the math afterwards you can get to where you judge the odds fairly well by just estimating.

For example, To generalize the math in this hand

The Pot Odds for the all-in is 1:3
The odds you can outdraw an Overpair is 40%
Therefore, you need a fold from an overpair 30% of the time.

Your calculations are wrong again.

There is $108 in the pot when Kenny decides to push. If villian calls, he has to call $326 more.

So when Kenny wins, he wins $434....not $362. You have to figure what he wins as compared to what he wouldnt have if he folds. Part of that $108 used to be Kennys, but its still included since its not his if he folds.