 Solution

Represent the removed half as the first $m=500$ cards in a uniformly random ordering. Every unordered half is then equally likely. Let $X_1,\dots ,X_{500}$ be their values. Each has mean zero and variance one, and every distinct pair has covariance $-1/999$ by the same ordered-pair argument as in (ii).

The general covariance expansion from (b), applied without imposing independence, givesThis is the finite population correction for a half-sample; treating the draws as independent would give the wrong variance.

Finally $\{Z\ge 50\}\subseteq \{|Z|\ge 50\}$, so Chebyshev's inequality yieldsThe cards returned before this sampling step restore the original population; no conditioning on the earlier two observed values is imposed.