# Expected Value multiple dice throws

I am solving this particular question :

A very innocent monkey throws a fair die. The monkey will eat as many bananas as are shown on the die, from 1 to 5. But if the die shows '6', the monkey will eat 5 bananas and throw the die again. This may continue indefinitely. What is the expected number of bananas the monkey will eat?

The correct answer is 4, but I am getting 3. Cannot figure out what is wrong with this approach:

$$E[x] = \Sigma P[x]x$$

$$E[x] = (1/6)(1+2+3+4+5) + 1/6(5 + 1/6(1 +2 +3+4+5)....)$$

$$E[x] = 5/2 + 1/6(E[x] + 5)$$

Solving this I get 3 but the correct answer as shown in the image is 4

• Solving what you wrote on the third line, I get $4$...
– md5
Oct 25, 2019 at 13:23

Not sure where your equations come from.

Two things can happen on the first toss. Either he throws $$≤5$$ and the game ends (after some bananas are consumed), or he throws a $$6$$ and the game restarts (again, after some eating).

Thus $$E=\frac 16\times 15+\frac 16\times (5+E)\implies E=4$$

• He has the same equation. He just did an error when solving it apparently. Oct 25, 2019 at 13:24
• @nicomezi Yes, that makes sense. It's the second equation that puzzles me...though perhaps it is just intended to rewrite the third.
– lulu
Oct 25, 2019 at 13:25
• I agree this is not a very rigourous way to derive the equation. Oct 25, 2019 at 13:27
• I did an error while solving, this is embarrassing. Oct 25, 2019 at 13:29
• @DhruvMahajan Oh, no need to apologize for that. We all make mistakes of that form.
– lulu
Oct 25, 2019 at 13:32