[thechainsaw]

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Before we answer the problem with 10 hands , we should find the solution for 2 hands and then proceed from there .

Since you are forced to shove on the second hand , you must take the gamble if your first hand is better than 50 % .

Here is my conjecture . If you play this game with n trials , then you should wait to shove with a hand in the top 1/(n-1)*100% of all hands . For n=10 , you should expect to see a hand in the top 11.11% of hands on your 9th trial .

top 11.1% of hands includes 66+ ,a7S,k10s+ a10+ kq

However , the hands you decide to shove all in with changes with each hand dealt . You have to loosen your shoving range .

[/ QUOTE ]

I like the conjecture.

I think it can be proved inductively.

let n be the number of trials before a shove and P(n) your probability of winning.

if n = 0, P(0) = 50%

P(1) = .50*.50 + .50*.75 = .50(1.25) = 62.5%

and so on - i think it's possible to find a formula - basically if P(n-1) is favourable to P(n), fold it, otherwise shove