Esteem the number of fruits that monkey can eat.

  • Context: Undergrad 
  • Thread starter Thread starter Alexsandro
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around estimating the number of fruits a monkey can eat based on its jumping success rate. Participants explore the problem through probabilistic modeling and encoding of jump outcomes, focusing on the expected longest sequence of successful jumps represented as a binary string.

Discussion Character

  • Exploratory
  • Mathematical reasoning

Main Points Raised

  • One participant describes the scenario where the monkey jumps to reach fruits on twigs, with a success probability of 1/2 for each jump.
  • Another participant proposes encoding successful jumps as a binary string, suggesting that the expectation of the longest sequence of 1's can be calculated using probabilities associated with different lengths of sequences.
  • Two participants provide a formula for estimating the number of fruits as log(n+1)/log(2), although it is unclear if this is universally accepted or derived from the previous discussions.

Areas of Agreement / Disagreement

There appears to be no consensus on the final estimation of the number of fruits, as multiple approaches and interpretations are presented, particularly regarding the mathematical formulation and its justification.

Contextual Notes

The discussion includes assumptions about the monkey's jumping behavior and the probabilistic model used to estimate outcomes. The derivation of the formula log(n+1)/log(2) is not fully explained, leaving some steps and dependencies unresolved.

Alexsandro
Messages
49
Reaction score
0
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.
 
Physics news on Phys.org
Hmm,
Again let's 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
 

Similar threads

  • · Replies 34 ·
2
Replies
34
Views
3K
  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 12 ·
Replies
12
Views
5K
  • · Replies 93 ·
4
Replies
93
Views
22K
  • · Replies 11 ·
Replies
11
Views
4K
Replies
7
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 0 ·
Replies
0
Views
4K
  • · Replies 13 ·
Replies
13
Views
7K
  • · Replies 11 ·
Replies
11
Views
2K