[ QUOTE ]
[ QUOTE ]

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


[/ QUOTE ]

Err, check it over. ln(e(|2Q-Q|)/Q) = ln(eQ/Q) = 1

Ps thus evaluates to 2Q/T = 2Pq.

Your earlier calculation is correct, however.

I'll think about how to normalize it, but in the meantime, ponder whether it's really necessary. This doesn't have to be a actual probabilistic calculation to be useful. It only needs to model the relationship between probabilities of different stacks well enough.

If you come up with a normalized version, by the way, I'd love to see it.

[/ QUOTE ]

It's pretty necessary, as the example that I gave returns negative probabilities at N>4. It's pretty cool that you got a formula that fits 3 of the four criteria though. I'll try and normalize it when I get a chance (it's parents' weekend at school right now.)