#1
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coin-flip variance question
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. |
#2
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Re: coin-flip variance question
[ 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. |
#3
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Re: coin-flip variance question
[ QUOTE ]
EV = pb + q(-b) = b(p - q) [/ QUOTE ] Okay, that's the simplification I missed. Thanks very much! Feeling stupid, Guy. |
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