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