• There are 6 strings clustered together.
• One end of each string is at point Y (the top), and the other is at point X (the bottom).
• First, two of the ends at point X are randomly tied together. Then two more are tied together, and then the last two.
• Next, two ends at point Y are randomly tied together. Then, two more are tied together, and then the last two.
Determine the probability that all the strings will be tied together in one large loop.
Firstly, it doesn't matter what happens at point X, as the result will always be three strings of double the original length, with six untied ends remaining at Y. These can be thought of as three groups of two, each pair being the endpoints of one of the three "doubled" strings.
Then at Y:
Select two ends to tie. There is a 1 in 5 chance you will immediately form a loop, which is not desired. So you have a 4/5 chance of continuing correctly. Assuming you can continue, there are 4 untied ends, with 6 ways of picking two at random. Two of the six picks of two will form an incorrect loop, so there is a 4/6 chance to continue. Assuming you again can continue, there are two ends, and they must be connected to solve the problem.
So, I think the solution is 4/5 * 2/3 = 8/15 chance of forming one big loop. If this is correct, it's larger than I would have intuitively thought it would be.
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Posted by Kenny M
on 2022-10-16 08:44:28 |
Solution?
