[ QUOTE ]

If you were to add all the numbers for everyone in the tournament, would you get 1?

[/ QUOTE ]


Can't believe I didn't check that

If you have a stack of 2Q with a field of n players and they all have equal stacks, then they all have (T-2Q)/(N-1) = Q(N-2)/(N-1). Thus, each has a probability of coming in first of

P = Q(1 +/- ln(e(|Q(N-2)/(N-1)-Q|)/Q)/T
= Q/T * (1 - ln(e/(N-1)) = Q/T*ln(N-1)

Thus, the odds that we come in first should be 1-(N-1)Q/T * ln(N-1)

Ps = Q(1 +/- ln(e(|2Q-Q|)/Q)/T
= Q/T * (2+ln(2)).

I think you need to normalize.