One by one straws are placed on a mathematical camelís back.
The straws have random weight, being drawn independently from the uniform distribution 0 to 1.
The camelís back breaks when the total weight of the straws exceeds 1.
Whatís the expected weight of the straw that breaks the camelís back?
Source: abridged from "Mind your decisions" blog.
(In reply to re: Analytic for 2 straws; spreadsheet
by Steve Herman)
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:
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