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Waiting for the first red

ProbabilityHard

An urn holds $6$ red and $4$ blue balls. You draw balls one at a time at random without replacement and stop as soon as a red ball appears. What is the expected value of the number of draws, counting the final red draw? Give a fraction in lowest terms.

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Hint

Use linearity: for each blue ball, find the chance it comes before every red ball.

Solution

The number of draws equals $1$ plus the number of blue balls drawn before the first red ball.
Consider one particular blue ball together with the $6$ red balls: all $7$ orderings of their relative positions are equally likely.
So this blue ball precedes all red balls with probability $\frac{1}{7}$ .
By linearity, the expected number of blue balls before the first red is $4 \times \frac{1}{7} = \frac{4}{7}$ .
The expected number of draws is $1 + \frac{4}{7}$ , which equals $\frac{11}{7}$ .

Answer: $\frac{11}{7}$

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