Tying Up Loose Ends to Create Different Size Loops

Date: 10/3/95 at 1:10:43
From: Margaret 
Subject: Tying Up Loose Ends

    Wandelina has six identical pieces of string held firmly in 
her fist, with one end of each sticking out above her fist, and 
the other end of each sticking out below her fist.

    Geribaldi comes along and ties two of the ends above her fist 
together.  Then he ties two other ends above her fist together.  
Finally, he ties the last two ends above her fist together.

    There are still six loose ends hanging out below Wandelina's 
fist.  

    Now Katalana comes along.  She has no idea which end below 
belongs to which end above.

    She randomly picks two of the ends below and ties them 
together; then two more; and finally, the last two.

    So Wandelina now has 6 strings in her fist, with three knots 
above and three knots below.

HERE ARE THE QUESTIONS:

1) When Wandelina opens her fist and looks at the strings, what 
combinations of different size loops might there be?

2) What is the probability of each possible combination occurring?

Date: 10/3/95 at 2:56:21
From: Doctor Andrew
Subject: Re: Tying Up Loose Ends

Neat question!  Well, I'd look at it this way.  When Geribaldi is 
done there are 3 pieces of string of length 2; no probability 
involved.  When connecting the pieces there are 3 possible loop 
sizes: 2, 4 and 6.

There are (6 choose 2) = 6 * 5 / 2 = 15 choices for knots among 6 
string ends.  There are (4 choose 2) = 4 * 3 / 2 = 6 choices for 
knots among  4 string ends.  And there is one choice among 2.

3 of the choices on 6 string ends produce a loop of length 2.  So, 
on  the first pick, 3/15 of the time a loop will be created.  
Within that  3/15 of a time, 2/6 (two pairs among six left) of the 
time another loop  of length 2 will be chosen.  So, 3/15 * 2/6 = 
1/15 of the time three  strings of length 2 will be created.  3/15 
* 4/6 = 2/15 of the time first a length 2 is chosen and then a 
length 4.  Now let's go back to the 12/15 of the time when a 
length 4 string is created.  Only two choices will produce two 
strings; the two that would create a loop.  So,  12/15 * 2/6 = 
4/15 of the time a length 4 loop is created and then a length 2.
  
So in conclusion, we have a 1/15 probability of three 2's, 6/15  
probability of a 4 and a 2, and the remaining 8/15 = 2/5 
probability of a single 6.  I suppose it isn't intuitive that the 
chance of a getting a single string is so much higher than the 
chance of getting 3 separate strings.  Pretty cool.

I hope this explanation works for you.  If you'd like something 
clarified or want to point out an error, please write back.

-Doctor Andrew,  The Geometry Forum