coin-flip variance question

[Guy McSucker]

Via a roundabout route, I've been looking at this post by BruceZ containing his own derivation of a risk of ruin formula, using coin-flip games as the basis.

The game is an even-money bet of size b with a biased coin: you win $b with probability p and lose $b with probability q = 1-p.

BruceZ writes this formula for the variance of the game:

variance = p*b^2 + q*(-b)^2 – EV^2

Is that right? I don't know much stats, but I thought variance was computed as the average of the squares of the deviations of actual results from the mean, i.e.

p(b - EV)^2 + q(-b - EV)^2

which doesn't seem to be the same thing.

On the other hand, BruceZ is pretty reliable in these matters.

Can anyone shed any light on this for me?

Guy.