• 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.
Several people have gotten the answer but we can go further. Lets generalize 6 into 2N strings; with their X ends randomly tied in pairs as stated in the problem. So now to the Y ends.
In general, we will have 2K ends to choose from during an arbitrary point in the process of tying the Y ends. Pick a string end then there are 2K-1 available ends to choose from.
Exactly one other string end is the complementary end of the string we first chose. We don't want that so we have 2K-2 valid choices to grab.
Then the probability that we have a desired pair of ends to tie together is (2K-2)/(2K-1).
Lets iterate this on all N-1 times we need to draw at random. (The last draw is just the two remaining ends that will finish the large loop).
Then the probability that we will end up with a single large loop is (2N-2)/(2N-1) * (2N-3)/(2N-4) * .... * 4/5 * 2/3 = (2N-2)!!/(2N-1)!!
For the problem as stated we have N=3 then 4!!/5!! = 8/15.
Solution
