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Breaking the back of a camel

Posted on: April 7th, 2014 by

A camel is loaded with straws until it's back breaks. Each straw has a weight uniformly distributed between 0 and 1, independent of other straws. The camel's back breaks as soon as the total weight of all the straws exceeds 1. Find the expected weight of the last straw that breaks the camel's back. (knowledge of probability required)

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3 Responses to Breaking the back of a camel

  1. Lizard had this to say about that:

    Let f(x) be the expected weight of the back-breaking straw given that the camel is already carrying weight x. Integrating over the weight y of the next straw, we have: f(x) = integral [y=0 to 1-x] of f(x+y) dy + integral [y=1-x to 1] of y dy. Differentiating this with respect to x, we have: f'(x)=-f(x)+1-x. Solving this differential equation with initial condition f(1)=1/2 gives f(x)=2-x-exp(1-x)/2, so f(0) = 2-e/2.

    • Aditya Mittal had this to say about that:

      I think you're missing the uniform independence. When you integrate from y=0 to 1-x you assume, the back breaking straw has a weight between 0 and what's already on the camel. That's not true the straw has a weight from 0 to 1 independent of what's on the camel.

      I think the answer is just 0.5. The expected weight of a straw is 0.5 The expected number of straws it would take to break the back is 2. The expected weight of the second straw that breaks the back is also independently 0.5 as it has nothing to do with any other straw.

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