= Solution
Represent the removed half as the first $m=500$ cards in a uniformly random ordering. Every unordered half is then equally likely. Let $X_1,\ldots,X_{500}$ 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>, gives
$$
\boxed{\mathbb E[Z]=0,\qquad
\operatorname{Var}(Z)=500-\frac{500\cdot499}{999}
=\frac{250000}{999}.}
$$
This is the <finite population correction> for a half-sample; treating the draws as <independent> would give the wrong <variance>.
Finally $\{Z\ge50\}\subseteq\{|Z|\ge50\}$, so <Chebyshev's inequality> yields
$$
\boxed{\mathbb P(Z\ge50)\le
\frac{\operatorname{Var}(Z)}{50^2}
=\frac{250000}{999\cdot2500}
=\frac{100}{999}.}
$$
The cards returned before this sampling step restore the original population; no conditioning on the earlier two observed values is imposed.
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