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
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