Probability of strings making a complete loop

Question

We got 6 pieces of strings. The top ends of the ropes are randomly paired up and tied together. The same procedure is done with the bottom ends of the strings. What is the probability that, as a result of the process, the 6 pieces of strings will be connected in a single closed loop of string?

My initial attempt is to just multiply the probabilities in each step:

Probability of tying the top of the string in initial procedure is 1. Because any head we tie will not effect the result we are getting.

Second, step if we choose a random string and random bottom end of it, out of our new 3 full strings; only one of the outcomes will not be favorable out of all 5 bottom ends. So favorable probability $\frac{4}{5}$

Third, if we consider one of our new tied strings's bottom ends we have 2 favorable ends out of 3 ends so $\frac{2}{3}$

and finally 1 way to tie the last ends.

Therefore in total;

$1 \times \left(\frac{4}{5}\right) \times \left(\frac{2}{3}\right) \times 1 = \frac{8}{15} $

This was my attempt am I missing something?

Answer 1

We can assign a number to each string from 1 to 6. We get three pairs $(i,j)$ of distinct numbers when choosing the top ends. We will not have a single closed loop if we choose one of these three pairs for the bottom ends. So the probability of having a single closed loop can be decomposed in $\frac{12}{15}$ when choosing the first bottom pair times $\frac{4}{6}$ when choosing the second bottom pair given we didn't choose one of the top three pairs before, which gives $\frac{8}{15}$. $\frac{4}{6}$ comes from the fact that the first bottom pair removes the possibility of choosing two of the three top pairs and we also need to consider the probability of choosing the remaining top pair by default for the last bottom pair.

Answer 2

Here is another approach that yields the same answer...

Note that each configuration of resulting strings is equally likely to occur. So, I proceed by counting $(1)$ all possible ways the six strings can be tied together and $(2)$ all such ways that result in a loop.

As we begin tieing the top ends together, there are ${6 \choose 2} = 15$ ways of selecting two ends of the available six to bind together, each of which leaves four top ends untied. Then, there are ${4 \choose 2} = 6$ ways of selecting two ends of the remaining four to bind together, each of which leaves two top ends untied. Of course, there is only ${2 \choose 2} = 1$ way to tie the remaining two ends together, so there are $ {6 \choose 2} \cdot {4 \choose 2} \cdot {2 \choose 2} = 15 \cdot 6 \cdot 1 = 90$ ways of tieing the top ends together. Now, for each such way, we can repeat this logic to conclude there are $90$ ways of tieing the bottom ends together. Hence, there are $90 \cdot 90 = 8100$ distinct ways of tieing the strings together in the manner you described.

Now for the tricky part... consider that for each of the $90$ ways in which the top ends are tied together, we are always left with the following configuration of strings

$$ \bigcap \bigcap \bigcap$$

where each $\bigcap$ is an elongated string formed by tieing two smaller strings together. As we begin tieing the bottom ends together, there are ${6 \choose 2} = 15$ ways of selecting two ends of the available six to bind together, but there are three ways of doing so that would prevent a loop from being formed between all strings. This would occur if we tied together two ends of the same elongated string. So there are ${6 \choose 2} - 3 = 15 - 3 = 12$ ways of "legally" selecting two ends of the available six to bind together. Now, each such way leaves behind four untied bottom ends coming from two elongated strings. We may repeat the logic just described to conclude there are ${4 \choose 2} - 2 = 6 - 2 = 4$ ways of "legally" selecting two ends of the available four to bind together. Of course, there is only one way to tie the remaining two ends together, so there are

$$ \Big[ {6 \choose 2} - 3 \Big] \cdot \Big[ {4 \choose 2} - 2 \Big] \cdot 1 = \Big[ 15 - 3 \Big] \cdot \Big[ 6 - 2 \Big] \cdot 1 = 12 \cdot 4 \cdot 1 = 48 $$

ways of tieing the bottom ends together that result in a loop. Since this is true for each of $90$ ways of tieing the top ends together, we have $90 \cdot 48 = 4320$ ways of tieing the strings together to form a loop.

Thus, the probability of tieing the strings together in a loop is

$$ \frac{\text{# ways to tie strings in loop}}{\text{# ways to tie strings}}= \frac{4320}{8100} \approx 0.53333 $$