Let's say the blinds are such that folding is better than calling with a worse hand. Also, our opponent is Super Man and because of his x-ray vision he will only call with a hand that dominates ours. Then we can work backwards from the perfect play number to guess what the number might be if a worse hand calls.

Here is my math for when it's break even against perfect play:

hands dominated by
AA - 3 ways
22 - 3 ways
KK - 33 - 6 ways each (11) for 66 ways
AK - A3 - 12 ways each (11) for 132 ways

132 + 66 + 6 = 204

total hands left
50 * 49 / 2 = 1225

0.167 better hand
0.833 worse hand

We beat a better hand 25% (conservitive)


Let X be the big blind
Let Y be the short stack

EV is
(best hand * amount won on fold) + (worst hand * (amount won - cost))
(0.833 * 1.5x) + (0.167 * ((0.25 * 2 * y) - (y - 0.5X) ))
(0.833 * 1.5x) + (0.167 * ((0.5y - (y - 0.5X)))
(0.833 * 1.5x) + (0.167 * (-0.5y + 0.5X))

Solve for an EV of 0
(0.833 * 1.5x) + (0.167 * (-0.5y + 0.5X)) = 0
(0.833 * 1.5x) + (0.167 * (-0.5y + 0.5X)) = 0
1.2495x - 0.0835y + 0.0835x = 0
1.333x = 0.0835y
y = 15.96x