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.

(In reply to re: Analytic for 2 straws; spreadsheet by Steve Herman)

Awesome, Steve.  
I looked at the pattern of Steve's fractions to predict the straw weight and the data fits very closely. 
Steve's pattern is  (n+2)/(2*(n+1))
Here is another pattern:  the probability that the final straw will be the n-th straw looks like it might be:   (n-1) / n!

So the weighted average ought to be the exact value, summing from 2 to infinity:
(n-1)*(n+2)/(2*(n+1)*n!)

And doing that results in:  0.640859 (truncated)

Edited on July 28, 2014, 1:15 am

Posted by Larry on 2014-07-28 01:10:42