Earn pips on every roll until a six appears

Question

You roll a fair six-sided die over and over, stopping the moment a $6$ appears. For every roll before that first $6$ , you earn the number of pips it shows (so $1, 2, 3, 4,$ or $5$ ).

What total should you expect to earn?

Accepted Answer

Let N be the number of rolls before the first 6. Both the count N and each roll's value are random, so condition on N and use the tower rule: E[	ext{total}] = E\big[\,E[	ext{total} \mid N]\,\big]. Step 1, freeze the count. Each roll that counts is a roll that was not a 6, so it is equally likely to be 1, 2, 3, 4, or 5, averaging \frac{1+2+3+4+5}{5} = 3. If N = n, E[	ext{total} \mid N = n] = 3\,n. Step 2, find the expected count. A 6 stops you (probability 	frac16), any other face lets you continue (probability 	frac56). The expected number of non-six rolls before the first six is E[N] = …

Earn pips on every roll until a six appears

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