[JocK]

[ QUOTE ]
Pot size = P
Your chance of winning once = x
Expected value of running it once = Px

Your chance of winning twice = 0 (x^2 in most cases, but 0 now because you only have one out)
Your chance of winning once and losing once to split the pot = 2(1-x)x
Expected value of running it twice = 0*P + 2(1-x)x*(P/2) = Px - Px^2.

Is the analysis correct?

[/ QUOTE ]
No, it is not. If you take card elimination effects into account for the chance of hitting twice, you should do the same for the chance of hitting once and losing once:

Chance of winning first and then losing is x

Chance of losing first and then winning is (1-x) times x/(1-x) = x