Variance of a small arcade game's net token payout

Question

An arcade game costs a token to play and returns a net result $X$ (in tokens) each round:

Net tokens	-1	1	2
Chance	40%	40%	20%

Here $-1$ means you came out one token behind that round.

What is the variance of the net token result?

Accepted Answer

Negative values are no problem: variance still measures how spread out the outcomes are around the average. Use \mathrm{Var}(X) = E[X^2] - \big(E[X]\big)^2. Mean. E[X] = (-1)(0.4) + (1)(0.4) + (2)(0.2) = -0.4 + 0.4 + 0.4 = 0.4. On average you net +0.4 tokens per round. Average of the squares. Squaring makes every term positive, so the losing outcome contributes just as much as a win of the same size: E[X^2] = (-1)^2(0.4) + (1)^2(0.4) + (2)^2(0.2) = 0.4 + 0.4 + 0.8 = 1.6. Subtract. \mathrm{Var}(X) = 1.6 - (0.4)^2 = 1.6 - 0.16 = 1.44. The standard deviation is \sqrt{1.44} = 1.2 tokens. So a typ…

Variance of a small arcade game's net token payout

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