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 nth straw looks like it might be: (n1) / n!
So the weighted average ought to be the exact value, summing from 2 to infinity:
(n1)*(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 20140728 01:10:42 