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Expected weighings on a balance

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There are $5$ coins and exactly $1$ is heavier, chosen uniformly at random; the others weigh the same. You use a balance scale with this fixed strategy: weigh $2$ coins against $2$ coins. If the pans are level, the fifth coin is heavy and you stop. Otherwise, weigh the $2$ coins from the heavier pan against each other. What is the expected number of weighings?

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Hint

Find the probability that the first weighing is level, and the number of weighings in each case.

Solution

The first weighing is level only if the heavy coin is the fifth one, which has probability $\frac{1}{5}$ ; then the total is $1$ weighing.
With probability $\frac{4}{5}$ the heavy coin is among the weighed $4$ , and a second weighing is needed, so the total is $2$ weighings.
The expected value is $1 \times \frac{1}{5} + 2 \times \frac{4}{5}$ .
This equals $\frac{1}{5} + \frac{8}{5} = \frac{9}{5}$ .
So the answer is $\frac{9}{5}$ .

Answer: $\frac{9}{5}$

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