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Runs in a colour sequence

Expected valueHard

$4$ red balls and $6$ blue balls are arranged in a row, with every distinct colour arrangement equally likely. A run is a maximal block of consecutive balls of the same colour. What is the expected number of runs? Give a fraction a/b in lowest terms.

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Hint

The number of runs equals $1$ plus the number of adjacent pairs with different colours.

Solution

Each boundary between runs is an adjacent pair with different colours, so runs $= 1 +$ number of such pairs.
The row has $9$ adjacent pairs. For one pair, the probability of red then blue is $\frac{4}{10} \times \frac{6}{9}$ , and the same for blue then red.
So a pair differs with probability $2 \times \frac{4}{10} \times \frac{6}{9} = \frac{8}{15}$ .
By linearity the expected number of differing pairs is $9 \times \frac{8}{15} = \frac{24}{5}$ .
Adding $1$ gives $\frac{29}{5}$ .

Answer: $\frac{29}{5}$

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