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Forced reroll on lowest face

Expected valueEasy

A fair $6$ -sided die is rolled. If it shows $1$ , you must roll it once more and take the new result, even if it is $1$ again. Otherwise you keep the first result. What is the expected value of the final result?

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Hint

Split by whether the first face is $1$ or not, and use the mean of a fresh roll in the first case.

Solution

Faces $2$ to $6$ are kept, each with probability $\frac{1}{6}$ , so they contribute $\frac{2 + 3 + 4 + 5 + 6}{6} = \frac{20}{6}$ .
A first face of $1$ has probability $\frac{1}{6}$ and is replaced by a fresh roll worth $3.5$ .
That case contributes $\frac{3.5}{6} = \frac{7}{12}$ , and the parts sum as $\frac{10}{3} + \frac{7}{12}$ .
Adding gives $\frac{47}{12}$ .

Answer: $\frac{47}{12}$

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