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A game played by B and K involves indepenent rounds. In each round if B wins they receive 1 dollar from K, if K wins they receive 2 dollars from B, and in the event of a draw no money is given. K knows they will win a round with probability 0.25 and will lose a round with probability 0.5. What is the standard deviation of B's winnings over four rounds?

I solved this in two ways and got two different answers. Please help me understand which one is correct & why!

In the first method I found E(X) = 0 and E(X^2) = 1.5. Thus I was able to calculate the variance as 1.5 and the SD(X) = sqrt{1.5} and therefore for a four game series SD = 4*sqrt{1.5}

In the second method I found E(X) = 0 and E(X^2) = 1.5. Then I determined that variance of X must be equal to 1.5 and thus the variance of a four game series = 6. Thus the SD for a four game series SD = sqrt{6}

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    – joriki
    Mar 23, 2023 at 13:12

1 Answer 1

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The second method is correct; when summing independent variables, variances add, but standard deviations do not.

Formally, let $B_i$ represent $B$'s winnings from round $i$. Then $B$'s total winnings are $\sum_{i=1}^4 B_i$, the variance of which is

$$\text{Var}\left(\sum_{i=1}^4 B_i\right)=\sum_{i=1}^4 \text{Var}\left(B_i\right)=4\text{Var}(B_1),$$

the final equality coming from the assumption that the earnings from the different rounds are independent and identically distributed.

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