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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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Analytic for 2 straws; spreadsheet | Comment 2 of 6 |
First I tried an analytic approach, and began with just cases where the second straw broke the back.  The probablity density function of the sum of two random variables, each being uniform (0,1) is:
 f(s)=2-s (where 1<=s<=2) or f(s)=s where 0<=s<=1).
We are just concerned with s>1 or f(s)=2-s,  since that's when the camel breaks.

So all cases where the second straw broke the back occur in the triangular region between s=1 and s=2 and f(s)=2-s.  Weighted by area, the expected value of the sum, s, should be at s=4/3, or one third the way between 1 and 2, since this is a triangle.  And since there are only two straws in this simplified example, and it is equally likely that each could have been first or second, the expected value of the straw that did the breaking should be 2/3.  This is in the ballpark of Charlie's programatic finding of about .64


Then I made a spreadsheet where each of 20,000 rows was a trial, and 10 columns represented 10 successive straws.  
The next set of 10 columns were the cumulative sums.  
In the next 10 columns was an identifier to determine in which column of the previous set did the sum exceed 1 for the first time, i.e. which straw did the breaking.  
The next set of 10 columns showed weight of the straw that did the breaking, and averaged out about .64 as Charlie found.

Straw Numbers:
The breaking straw was the second straw about half the time, and the average weight of these straws was very near 2/3 which happily agrees with the non computer analysis above of the oversimplified situation where there are just two straws.
The 3rd straw did the breaking about 1/3 of the time, with average weight about .625.
The 4th straw did the damage about 1/8 of the time with average weight about .6

By pressing F9 many times, many trials could be run (although I'm not sure of the precision of Excel's RAND() function when being called 200,000 times with each press of F9).

I find it curious that the higher the straw number, the lower the weight of the final straw.

Notice:  no camels were harmed in the running of this simulation

Edited on July 27, 2014, 7:31 pm
  Posted by Larry on 2014-07-27 19:29:41

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