# Probability of forming a circle with 6 ribbons

Tradition tells us that in certain rural areas of Russia the marriage of a young woman was determined as follows: The young woman held in her hand 6 ribbons by the middle, so that the tips were above and below the hand. The young suitor had to tie by pairs the 6 tips that went up, and then tie the 6 tips below, also by pairs. If the young man tied the 6 ribbons in a single circle, then the wedding would take place in less than a year.

a) What is the probability of forming a single circle if the ribbons were tied randomly?

Can someone please explain to me why?

I am using the formula for combinations, nCr = n! / (r! * (n - r)!), where n represents the number of items, and r represents the number of items being chosen at a time

• What do you mean by 6C2 etc.? – pendermath Feb 9 at 17:27
• 6 choose 2 I guess – pendermath Feb 9 at 17:28
• nCr (combinations of 2 ribbons in 6 without order) – Lollipop Feb 9 at 17:28

I can confirm that the answer is $${8\over15}$$ but I don't understand how the answer with combinations is arrived at. I would do it this way:
After the tops have been tied, we have $$3$$ ribbons, with $$6$$ ends. The man picks one end, then ties it to one of the $$5$$ remaining ends. Four of the five don't belong to the same ribbon, and are okay. If he succeeds in the first step, there are now $$2$$ ribbons, and by the same argument he has probability of $${2\over3}$$ of success. If he succeeds at the the second step there is only one ribbon left, so he must succeed at the third step.
This gives a probability of success of $$\frac45\cdot\frac23={8\over15}$$