Tying ends of spaghetti probablility

Question

A dish contains n strands of cooked spaghetti. Two ends of spaghetti are chosen at random and tied together. Two other ends are selected and tied together. If this continues until there are no more loose ends,

a) What is the probability that all n strands form one big loop?

b) What is the probability that there will be n loops each consisting of one strand?

Answer 1

Start with simplest case n=1. Then both probabilities are 1 (you get one "big" loop of 1 strand and n=1 loops of 1 strand).

So take n=2. You have 4 ends to choose from. Pick one, you have three others. You have a 1/3 chance of picking the one on the same strand. That will commit you to tying the other strand in a singular loop. So probability of b) is 1/3. Pick one on the other strand and now you are committed to making one big loop of 2 strands, so probability of a) is 2/3.

Now take n=3. You have 6 ends to choose from. So you have a 1/5 chance of picking the same strand, and 4/5 of picking a different strand. Either way, on the next tie you are effectively in the n=2 case. So probability of a) is 2/3*4/5 and b) is 1/3*1/5.

In general, probability of b) is

n

∏ 1/(2i-1)

i=2

and of a) is

n

∏ (2i-2)/(2i-1)

i=2