Probability of Tying Grass Together

[HKfish]

Homework Statement

Each of three indistinguishable blades of grass are bent roughly at their midpoints and clasped by these midpoints so that an observer can't match any of the ends in any way. In other words, all six ends are just dangling and appear completely unrelated. Suppose you are asked to tie the ends together (which, given the stipulations, can only be done randomly by selecting one pair at a time). First, determine the probability, as a reduced fraction, that a large circular loop results, and then, generalize for n blades of grass.

Homework Equations

The Attempt at a Solution

I have no idea what angle I should attempt this question at. If someone could just decipher this question for me it is much appreciated!

 

[Ray Vickson]

Homework Statement

Each of three indistinguishable blades of grass are bent roughly at their midpoints and clasped by these midpoints so that an observer can't match any of the ends in any way. In other words, all six ends are just dangling and appear completely unrelated. Suppose you are asked to tie the ends together (which, given the stipulations, can only be done randomly by selecting one pair at a time). First, determine the probability, as a reduced fraction, that a large circular loop results, and then, generalize for n blades of grass.

Homework Equations

The Attempt at a Solution

I have no idea what angle I should attempt this question at. If someone could just decipher this question for me it is much appreciated!

It is asking for the probability that one end of blade A is tied to one end of blade B, the other end of B is tied to one end of C and the other end of C is tied to the other end of A---so you form a big loop. Of course, the order could instead be ACB, and either end of each blade can be chosen each time.

 

[HKfish]

It is asking for the probability that one end of blade A is tied to one end of blade B, the other end of B is tied to one end of C and the other end of C is tied to the other end of A---so you form a big loop. Of course, the order could instead be ACB, and either end of each blade can be chosen each time.

Thanks for the reply! So, I attempted the question and figured out there is 6! ways of tying the grass (total). Now the problem is how do I figure out the factors to take out of the total? Thanks for your reply!

 

[Gatsby88]

Rather than simplifying it and then accounting for various things, it might be clearer what you need to do if you label the ends

A1 A2 B1 B2 C1 C2

And consider how many different ways there are to arrange this such that A1 is next to A2, B1 is next to B2 and C2 is next C3

And remember that A1A2B1B2C1C2 is different to A1A2B1B2C2C1

Once you start writing these out you should see ways of simplifying the problem

 

[Ray Vickson]

Thanks for the reply! So, I attempted the question and figured out there is 6! ways of tying the grass (total). Now the problem is how do I figure out the factors to take out of the total? Thanks for your reply!

You might also think about what it must mean if the *complement* of the desired event were to occur; that is, think of what must happen if the grasses do NOT form a large loop.

 

[haruspex]

figured out there is 6! ways of tying the grass (total).

Rather less. It's just a matter of arranging the 6 ends into pairs.