Same-Color Neighbors

Question

A standard deck (26 red, 26 black) is shuffled. Of the 51 adjacent pairs of cards, how many are same-colored, in expectation?

Accepted Answer

For the pair at positions i, i+1, condition on the first card: 25 of the remaining 51 cards share its color, so P(textsame) = frac2551, identical for every pair, regardless of position. Linearity: mathbbE[textsame-color pairs] = 51 cdot frac2551 = 25. The 51s cancel: the answer is exactly 25, not approximately. Overlapping pairs are dependent (pair 3 shares a card with pair 4), and that dependence is irrelevant to the expectation, saying this out loud is precisely what the interviewer is listening for. Linearity of expectation needs no independence; that's its entire power. Benchmark against iid: if colors were independent coin flips, the answer would be 51 cdot frac12 = 25.5. Sampling without replacement creates slight negative autocorrelation, each red dealt depletes the reds, shaving off half a pair. That "finite pool mean-reverts relative to iid" effect is a real market phenomenon: flow that consumes resting inventory behaves like the deck, not the coin. Equivalent restatement: color changes average 51 - 25 = 26, so a shuffled deck has about 27 color runs. Counting boundaries with indicators is the same move that counts runs, streaks, and regime switches, one technique, many di

Same-Color Neighbors

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