Variance when playing a game with a fair coin

Question

I am having a hard time with this question for some reason.

You and a friend play a game where you each toss a balanced coin. If the upper faces on the coins are both tails, you win \$1; if the faces are both heads, you win \$2; if the coins do not match (one shows head and the other tail), you lose \$1. Calculate the expected value and standard deviation for your total winnings from this game if you play 50 times.

PMF Values: \begin{array}{c|c} $& p\\\hline +$1 & .25\\ +$2 & .25\\ -$1 & .50 \end{array}

I have calculated the expectation as $$1(.25)+2(.25)+(-1)(.5) = .25,$$ so $$E(50X) = 50\cdot.25 = \$12.5,$$ which I have confirmed is correct.

I know I need to get $\operatorname{Var}(50X)$, but doing a standard variance calculation and then using the formula $a^2\operatorname{Var}(X)$ is not giving me the correct value.

What step am I missing?

Answer 1

Variance is the mean of the squares minus the square of the mean. $$ 0.25\cdot1^2+0.25\cdot2^2+0.5\cdot(-1)^2-0.25^2=1.6875 $$ For independent events, the variance of a sum is the sum of the variances, so the variance for $50$ events is $$ 50\cdot1.6875=84.375 $$

Answer 2

$Variance= 50 Var(X) = 50.[{.25(1-.25)^2 + .25(2-.25)^2 +.5*(-1-.25)^2}]=84.375$

Answer 3

You are confusing the distribution of $50X_1$ and $\sum_{k=1}^{50}X_k$ when ${(X_k)}_{k=1}^n$ is a sequence of independent and identically distributed random variables.

It is true that $\mathsf E(50X_1)=50\mathsf E(X_1)$ and $\mathsf{Var}(50X_1)=2500\mathsf {Var}(X_1)$.   However, that is not what you are dealing with.

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Due to the Linearity of Expectations, the expectation of the series is the series of expectations.   Its because the distributions are identical that this series is equal to $50$ times an individual expectation .

$$\begin{align}\mathsf E(\sum_{k=1}^{50}X_k) ~&=~ \sum_{k=1}^{50}\mathsf E(X_k) & \text{Linearity of Expectation} \\[1ex] & =~ 50\,\mathsf E(X_1) & \text{Indentical Distributions}\end{align}$$

Similar result, different reasoning.

(Note: We have not use independence at thi point.)

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The distinction becomes apparent in dealing with the variance.

When it comes the the variance of the series, we have to employ the Bilinearity of Covariance.

$$\begin{align}\mathsf {Var}(\sum_{k=1}^{50} X_k) ~&=~ \mathsf {Cov}(\sum_{k=1}^{50}X_k,\sum_{j=1}^{50}X_j) \\ &=~ \sum_{k=1}^{50}\sum_{j=1}^{50}\mathsf{Cov}(X_k,X_j) &&\text{Bilinearity of Covariance} \\ &=~ \sum_{k=1}^{50}\mathsf {Cov}(X_k,X_k) ~+~ 0 &&\text{Independence: }\mathsf{Cov}(X_j,X_k)=0\text{ when }j\neq k \\[1ex] &=~ 50\mathsf {Var}(X_1) && \text{Identical distributions} \end{align}$$