Esteem the number of fruits that monkey can eat.

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The Mike monkey is under of a tree with many twigs. Each twig contains a fruit, but he is 3 meters higher of the one than the previous twig (the first twig is the 3 meters of the soil). To reach the next twig, Mike gives a jump and has success with probability 1/2. In imperfection case, it it falls in the soil and it has that to start everything of new. It esteem the number of fruits that Mike obtains to eat n after jumps. It justifies.
 

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Hmm,
Again lets do some encoding,
Whenever the monkey is successfull in its jump, encode that as a 1 and if it fails encode that as a 0. Hence the sequence of jumps made by the monkey are encoded as a n-bit binary string.
Then the required expectation is the expected longest sequence of 1's in the binary string. This expectation can be calculated from first principles, again a bit tedious.
E[longest sequence of 1's]
= 0 * P(longest sequence of 1's = 0)
+ 1 * P(longest sequence of 1's = 1)
+ 2 * P(longest sequence of 1's = 2)
+...+ n * P(longest sequence of 1's = n)

The probabilities can be calculated by looking that 2^n possible binary representations.

-- AI
 
log(n+1)/log(2) fruits
 
Last edited:
log(n+1)/log(2) fruits
 

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