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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

enter image description here

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

1 Answer 1

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

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

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