[djames]

For x >= 1, the expected result of the hand is always break even at best. For x = 1, the expected loss is exactly zero. As x increases, so does the expected loss. However, the expected loss as a percentage of your starting stack declines.

I can e-mail a spreadsheet whereever you want.

pot to win = 10 + 2(x-1)

Needed win % for E[push]>=0 = (x-1)/(10 + 2(x-1)), or (stack left)/(pot to win)

Min hand range to achieve win % = [(x-1)/(10 + 2(x-1)]^(1/9) or (needed win%)^(1/9) ... call this m

Prob(Push) = 1 - m

Prob(win|push) = integral from m to 1 of y^9 dy, or simply (1/10)*(1-m^10)

E[chips after game] = (1-Prob(Push))(x-1) + P(push)xP(win|push)x(10+2(x-1))

Profit/Loss = E[chips after game] - x

For numerical verification, here are the values for x=2 and x=100.

x: 2
pot to win: 12
needed win%: 0.0833
m: 0.7587
Push %: 0.2413
P(win|push): 0.0937
E[chips]: 1.03
Loss: 0.97
Relative Loss: 48.5%

x: 100
pot to win: 208
needed win%: 0.4760
m: 0.9208
Push %: 0.0792
P(win|push): 0.0562
E[chips]: 92.02
Loss: 7.91
Relative Loss: 7.9%

From this information it would be trivial to express a closed form for any x. Typing this information in a web form was hard enough for me to do, so I'll leave that last step to you.

Interestingly the relative loss stabilizes very quickly. That is, after about 3905 chips, the relative loss stays at 6.6% up to x = a billion. This is because m converges.