Four cards are drawn from a standard set of 52. What is the expected value of the square of the sum of the cards? Does neglecting the covariance result in an overestimate or underestimate? How would your answer change if you flipped over a card and it read $5$?
I assume A, J, Q, K are $1$, $11$, $12$ and $13$. Ignoring the covariance of the draws and assuming they are independent, I think it is probably safest to use linearity of expectation such that $$\mathbb{E}[(X_1+X_2+X_3+X_4)^2]=\mathbb{E}(X_1^2+...+X_4^2+2X_1X_2+...+2X_3X_4)=4^2\mathbb{E}(X)^2=$$ $$=16\times49=784.$$
I suspect the covariance should cause the result to drop, but I’m unsure how. Could someone prod me in the right direction? Thanks!
You can evaluate $E(X_1X_2)$ in the following way: $$\ E(X_1X_2) = \sum_i\sum_jijPr(X_1=i, X_2=j)= \sum_i\sum_jijPr(X_1=i)Pr(X_2=j|X_1=i)\\ Pr(X_2=j|X_1=i)=4/51 \, \text{ } \forall \, i\neq j\\ Pr(X_2=j|X_1=i)=3/51 \, \text{ } \forall \, i=j\\ \implies E(X_1X_2) \approx 48.7255\\ E(X_2)=E(E(X_2|X_1=x))\\ E(X_2|X_1=x)=\sum_ii\times Pr(X_2=i)=\sum_{i\neq x}i\times {4\over 52} + \sum_{i= x}i\times {3\over 52}={364-x \over 51}\\ \implies E(X_2)=7\\ \implies Cov(X_1,X_2)<0 $$
Hence, Covariance causes the result to drop. Try to do it for four cards.