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.

[BruceZ]

[ QUOTE ]
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.
Guy.

[/ QUOTE ]

Sure it is. Expand it out.

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

p + q = 1
EV = pb + q(-b) = b(p - q)
-2bEV(p - q) = -2EV^2

so we have

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

= pb^2 + q(-b)^2 - EV^2

Actually this form is just

variance = E(x^2) - [E(x)]^2

where E means expected value.

[Guy McSucker]

[ QUOTE ]

EV = pb + q(-b) = b(p - q)


[/ QUOTE ]

Okay, that's the simplification I missed.

Thanks very much!

Feeling stupid,

Guy.