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You have a deck of 52 cards. You draw 4, what is the expected value of the sum of those 4 cards that you draw?

Attempt:

Basically, the possible sum ranges from 4 to 52. Pr{sum = 4} = Pr{ 4 ace} = 4/52*3/51 *2/50*1/49

And then, you repeat this for sum =5, 6, 7, .... , 52. Then.

$E[Sum] =\sum_{i=4}^{53}i*Pr(i)$

I am wondering is there better/quicker approach besides manually finding each probability?

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    $\begingroup$ Use linearity of the expectation operator: $$E(X_1 +X_2 +X_3+X_4) = E(X_1) + E(X_2) +E(X_3) +E(X_4)$$ and $E(X_i) = 7$. $\endgroup$ – mlc Mar 26 '17 at 6:38
  • $\begingroup$ I was thinking that, if you let $X_2$ be the random variable of the 2nd draw, does $X_2$ be dependent on the first draw, that's m confusion. $\endgroup$ – wrek Mar 26 '17 at 17:15
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    $\begingroup$ The linearity of the expectation operator holds regardless of dependence. $\endgroup$ – mlc Mar 26 '17 at 17:21
  • $\begingroup$ I agree, but I was thinking that $E[X_1] \neq E[X_2]$, because of some dependency relationship $\endgroup$ – wrek Mar 26 '17 at 17:47
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Elaborating on the comment by mlc, let $X_1,\ldots,X_4$ denote the four chosen cards, and let $\text{rank}(X_i)$ denote the rank (i.e., the face value, starting with $1$ for ace and ending with $13$ for king) of the card $X_i$ for $i=1,2,3,4$. The tuple $(X_1,X_2,X_3,X_4)$ is uniformly random from the set of all tuples of four distinct cards. Thus by symmetry, $X_1$ is uniformly random from the set of all $52$ cards (no card is favored more than the others), and therefore $$\mathbb E\ \text{rank}(X_1)=\frac{1+2+\cdots+13}{13}=7.$$ By the same argument, $\text{rank}(X_2),\text{rank}(X_3),$ and $\text{rank}(X_4)$ also have expected rank $7$.

Thus, by linearity of expectation, $$ \mathbb E(\text{rank}(X_1)+\text{rank}(X_2)+\text{rank}(X_3)+\text{rank}(X_4))=7+7+7+7=28. $$

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  • $\begingroup$ I was thinking that, if you let $X_2$ be the random variable of the 2nd draw, does $X_2$ be dependent on the first draw, that's m confusion. $\endgroup$ – wrek Mar 26 '17 at 17:15
  • $\begingroup$ Good observation. Even though it is not the case that $X_1$ and $X_2$ are independent, we may still use linearity of expectation: it works whether or not the random variables in question are independent. That is the beauty of it! $\endgroup$ – pre-kidney Mar 26 '17 at 21:58
  • $\begingroup$ I agree, but I was wondering how come $ E[X1]=E[X2]$ ? $\endgroup$ – wrek Mar 27 '17 at 0:35
  • $\begingroup$ By symmetry, $X_1$ and $X_2$ have the same distribution, so their expectations are equal. $\endgroup$ – pre-kidney Apr 16 '17 at 23:42

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