Represent the removed half as the first cards in a uniformly random ordering. Every unordered half is then equally likely. Let be their values. Each has mean zero and variance one, and every distinct pair has covariance 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 , 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.
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