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

Home > Probability
Shoelaces (Posted on 2005-04-05) Difficulty: 3 of 5
You have a drawer with n shoelaces in it. You reach into the drawer and select two lace ends at random, and tie them together. Repeat this until there are no more untied lace ends left.

If the two ends are from the same lace, or from the same group of laces already tied at their ends, it will form a loop, and therefore no longer have any free ends.

Continue this until all laces are parts of loops, of either 1 lace, 2 laces, etc.

When you're finished, a certain number of loops will have been formed in this fashion, consisting of one or more laces.

What is the expected number of loops?

See The Solution Submitted by Charlie    
Rating: 3.8333 (6 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Intuitive trash talk | Comment 5 of 7 |
(In reply to re(2): Recursive definition by Richard)

Af first it seems a bit counterintuitive (at least to me) that one expects 1.7 loops for 4 shoelaces, while for 100 shoelaces one only expects 3.2 loops.

But seeing that you are all right I have rethought my initial impression.

So "revising my intuitive thought" I would  say that while it is true that the more shoelaces there are there is more potential for loops to be created, it is more difficult to actually close them since you need to grab a particular opposite end from the many ends you can choose from.

Both effects kind of cancel out leaving the tiny logaritmic growth of the number of expected loops with the number of shoelaces.

Edited on April 6, 2005, 7:22 pm
  Posted by ajosin on 2005-04-06 19:21:10

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


Search:
Search body:
Forums (2)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (5)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2020 by Animus Pactum Consulting. All rights reserved. Privacy Information