• 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.
Amswer is 8/15.
My thoughts: There are only two opportunities to go wrong.
After the X's are tied together we effectively have 3 strings with 6 ends. No problem yet.
Pick a Y for the first tie. There are now 5 ends that it can be tied to. The string will be tied to itself with probability 1/5 (thus preventing a single big loop) or to another string with probability 4/5.
If we have surmounted the first obstacle, then we effectively have two strings with 4 ends. Pick a Y for the second tie. There are now 3 ends that it can be tied to. The string will be tied to itself with probability 1/3 (thus preventing a single big loop) or to the other string with probability 2/3 (in which case nothing can go wrong).
The probability of a single loop is therefore (4/5)*(2/3) = 8/15.
There may be a knot in the loop, but it is still a single loop.
A knotty problem (spoiler)
