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The decisive straw (Posted on 2014-07-25) Difficulty: 3 of 5
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.

No Solution Yet Submitted by Ady TZIDON    
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Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts re(2): Analytic for 2 straws; spreadsheet | Comment 4 of 6 |
(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:

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

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