I am curious if there is a way to approximate the variance of the sum of cards out of a 52 card deck.

Assume A = 1, J = 11, Q = 12 and K = 13

For example the variance of one card is 14. The variance of the sum of two cards however will already become more instensive if you were to calculate it by hand/head.

The minimum value is AA = 2 and the maximum value is KK = 26. The mean is (2)(7) = 14. Do you know of any way to approximate th variance with and/or without replacement without writing it all down? If not, how would you do it?

Thank you!

  • $\begingroup$ if this is HW please add the self-study tag $\endgroup$
    – Antoine
    Sep 22, 2016 at 8:26
  • 1
    $\begingroup$ Hi, thank you. This is not homework, I am just curious about it. $\endgroup$ Sep 22, 2016 at 9:12

2 Answers 2


Let $X_i$ be the value of the $i$'th card. You're interested in $S=\sum_{i=1}^nX_i$. Specifically, you want the variance which is, by linearity:

$\mbox{Var}(S)=\sum_{i=1}^n \mbox{Var}(X_i)+2\sum_{i<j}\mbox{Cov}(X_i,X_j).$

The individual variances are easy if you realize that even though $X_i$ are not independent, their joint distribution is invariant under permutations. This implies:


The covariances are similar, but a bit annoying. You need to calculate $E[X_iX_j]-E[X_j]E[X_j]$. The only difficult term is the first one. Fix $X_i$. Then

$E[X_iX_j]=E[X_i E[X_j|X_i]]=E\left[X_i\left[\frac{4}{51}(1+\cdots+13)-\frac{1}{51}X_i\right]\right],$

where we accounted for using up card $X_i$. Now use linearity and the individual variance results from above.

  • $\begingroup$ "their joint distribution is invariant under permutations" — Or in other word, exchangeable. $\endgroup$ Sep 22, 2016 at 19:14

The actual question is rather: what is the pdf f(x) for your chosen event, at least for the exact answer. Once you have the pdf, the calculation of the variance is an exercise in summing. The variance is the second central moment of a distribution of variable x, where x in your case is the sum of n randomly drawn cards: $$x = \sum_1^n card_n$$ $$\sigma^2 = E[(x - \mu)^2]$$

Getting a single expression for each possible sum for (1, 2, 3, 4, ..., 51, 52) card(s) might not be easy or pretty.

Concerning approximate variances I would much rather write a small program that does the random summing for you and histogram the result -- and extract the variance from the histogram. Quoting an exact error for the result might also become a bit more involved, but for a good guess just make sure that the histogram is "smooth" and the individual errors for each bin (== outcome) are small -- take enough samples.


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