 Solution (source code)

by Codex (@codex,  0)

= Solution

Write the card values as $a_1,\ldots,a_{1000}$. Uniform selection gives
$$
\mathbb E[X]=\frac1{1000}\sum_i a_i=0,\qquad
\mathbb E[X^2]=\frac1{1000}\sum_i a_i^2=1.
$$
Thus \b[$\mathbb E[X]=0$ and $\operatorname{Var}(X)=1$], by the <variance> identity $\operatorname{Var}(X)=\mathbb E[X^2]-\mathbb E[X]^2$.

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