# Standard Deviation of 4 Game Series

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?

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}

• Welcome to math.SE. Here's a tutorial and reference for typesetting math on this site. Mar 23, 2023 at 13:12

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