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Esteem the number of fruits that monkey can eat.

  1. Aug 28, 2005 #1
    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.
  2. jcsd
  3. Aug 29, 2005 #2
    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
  4. Aug 30, 2005 #3
    log(n+1)/log(2) fruits
    Last edited: Aug 30, 2005
  5. Aug 30, 2005 #4
    log(n+1)/log(2) fruits
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