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Matching neighbours in a row

StatisticsHard

A row of $10$ cards, $5$ red and $5$ black, is shuffled so that all $252$ colour patterns are equally likely. Let $S$ be the number of the $9$ neighbouring position pairs whose cards have the same colour. What is $Var (S)$ ?

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Hint

Write S as a sum of indicators and compare the covariances of neighbouring and separated pairs.

Solution

Each indicator equals $1$ with probability $\frac{4}{9}$ , so its variance is $\frac{4}{9} \times \frac{5}{9} = \frac{20}{81}$ , and the $9$ variances sum to $\frac{20}{9}$ .
For $8$ overlapping pairs the covariance is $\frac{1}{6} - \frac{16}{81} = - \frac{5}{162}$ , since $3$ cards in a row share a colour with probability $\frac{1}{6}$ .
For the other $28$ pairs the probability that both are same-coloured is $\frac{13}{63}$ , so the covariance is $\frac{13}{63} - \frac{16}{81} = \frac{5}{567}$ .
The covariance total is $2 \times (8 \times (- \frac{5}{162}) + 28 \times \frac{5}{567}) = 2 \times (- \frac{20}{81} + \frac{20}{81}) = 0$ , so $Var (S) = \frac{20}{9}$ .

Answer: $\frac{20}{9}$

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