One by one straws are placed on a mathematical camel’s back.
The straws have random weight, being drawn independently from the uniform distribution 0 to 1.
The camel’s back breaks when the total weight of the straws exceeds 1.

What’s the expected weight of the straw that breaks the camel’s back?

Source: abridged from "Mind your decisions" blog.

Here are some more expected values:
the ultimate straw is              (2-e/2)  ~ 0.6409
the penultimate total weight is (e-2)    ~ 0.71828
the ultimate total weight is      (e/2)    ~ 1.3591

Posted by Larry on 2014-07-30 17:48:45