2 evenly matched teams play a series that ends as soon as one team has 3 wins. Each game is won by either team with probability 1/2, independently, with no draws. What is the expected number of games played? Give a fraction a/b in lowest terms.

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Length of a first-to- $3$ series

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$2$ evenly matched teams play a series that ends as soon as one team has $3$ wins. Each game is won by either team with probability $\frac{1}{2}$ , independently, with no draws. What is the expected number of games played? Give a fraction a/b in lowest terms.

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Hint

Find the probability that the series lasts exactly $3$ , $4$ or $5$ games.

Solution

The series lasts $3$ games if one team wins all of them: probability $2 \times \frac{1}{8} = \frac{1}{4}$ .
It lasts $4$ games if the winner takes the last game and exactly $2$ of the first $3$ : probability $2 \times \frac{( 2 3 )}{16} = \frac{6}{16} = \frac{3}{8}$ .
It lasts $5$ games with probability $2 \times \frac{( 2 4 )}{32} = \frac{12}{32} = \frac{3}{8}$ ; the probabilities sum to $1$ .
The expected length is $3 \times \frac{1}{4} + 4 \times \frac{3}{8} + 5 \times \frac{3}{8} = \frac{6}{8} + \frac{12}{8} + \frac{15}{8}$ .
This equals $\frac{33}{8}$ .
The answer is $\frac{33}{8}$ .

Answer: $\frac{33}{8}$

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