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Rolling past a threshold

ProbabilityHard

A fair $6$ -sided die is rolled repeatedly and independently, and the results are added up. You stop as soon as the running total is at least $7$ . What is the expected number of rolls? Give an exact fraction in lowest terms.

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Hint

Let $f (j)$ be the expected number of rolls when the total is $6 - j$ , and find a relation between $f (j)$ and $f (j - 1)$ .

Solution

Let $f (j)$ be the expected number of further rolls when the total is $6 - j$ . At total $6$ every roll ends the game, so $f (0) = 1$ .
From total $6 - j$ , a roll of $i$ leads to total $6 - j + i$ . It ends the game if $i \geq j$ , and otherwise moves to state $j - i$ . So $f (j) = 1 + (\frac{1}{6}) \sum f (i)$ over $i = 0.. j - 1$ .
Subtracting the same relation for $j - 1$ gives $f (j) - f (j - 1) = \frac{f ( j - 1 )}{6}$ , so $f (j) = (\frac{7}{6}) f (j - 1)$ .
Therefore $f (j) = (\frac{7}{6})^{j}$ , and we start at total $0$ , which is $j = 6$ .
The expected number of rolls is $(\frac{7}{6})^{6} = \frac{117649}{46656}$ .

Answer: $\frac{117649}{46656}$

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