All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Probability
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    
No Rating

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:
(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

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (7)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information