The monkey who rolls again on a six

Question

A monkey rolls a fair six-sided die. If it shows 1 to 5, the monkey eats that many bananas and stops. If it shows 6, the monkey eats 5 bananas and rolls again, and the same rule applies to every subsequent roll, possibly forever.

What is the expected total number of bananas the monkey eats?

Accepted Answer

Four bananas. Set up the equation. Let E be the expected total. With probability 5/6 the first roll is 1 to 5, and the monkey eats the face value, whose average over those five outcomes is 3. With probability 1/6 the roll is a 6, the monkey eats 5, and then faces exactly the same situation again, expecting E more. So E = (5/6) × 3 + (1/6) × (5 + E). Solve. E = 15/6 + 5/6 + E/6, so E minus E/6 = 20/6, so (5/6)E = 20/6, giving E = 4. A second way to see it. Each roll contributes, on average, (1 + 2 + 3 + 4 + 5 + 5)/6 = 20/6 bananas (a six is worth 5). The number of rolls is geometric: the monke…

The monkey who rolls again on a six

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