Eight Strings

As they are tied first, the top ends don't matter, the result of tying them in any order will be 4 single strings, each with 2 ends.

Choose one end of a string. There are 7 other ends, of which 6 are ok to tie it to (the 7th would result in a single loop.)

You can consider what you have now as 3 single strings.

Choose one end of a string. There are 5 other ends, of which 4 are ok to tie it to (the 5th would result in a single loop.)

You can consider what you have now as 2 single strings.

Choose one end of a string. There are 3 other ends, of which 2 are ok to tie it to (the 3rd would result in a single loop.)

You can consider what you have now as 1 single string. Go ahead and tie it.

Total probability: $\frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3} = \frac{48}{105} \approx \boxed{0.4571428571}$

Tie the top ends together however you like.

Once this is done, the number of ways the bottom ends can be tied together is $7 \cdot 5 \cdot 3 \cdot 1 = 105$

The number of ways that result in a loop is: $6 \cdot 4 \cdot 2 \cdot 1 = 48$

$\dfrac{48}{105} = \boxed{0.457}$

Another way of looking at Geoff's explanation: Tie the tops of the strings together any way. Now tie a bottom end to one of the other ends. There are 7 ends to choose from. Do not choose the string already connected at the top. The probability of choosing suitably is 6/7. Continue in this way. The probabilities that the other three knots are tied suitably are 4/5, 2/3 and 1/1. Combined probability: 6/7.4/5.2/3 = 0.457.

$\frac{6}{7}\times\frac{4}{5}\times\frac{2}{3}\times\frac{1}{1}=\frac{48}{105}\approx\boxed{0.457}$

Suppose we matched threads (ab) (cd) (ef) (gh) at the top, but of course we do not know which is which.

Start with any thread (say it is a) connect it to another one. Only b is wrong, so 6 out of 7 belong to a new pair, say it is c. At some point, we will connect d. 4 out of 5 belong to a new pair. If we do that right (say we chose e), we have to match f. 2 out of 3 are right. Now connecting the remaining 2 threads completes the circle. Altogether we have probability $\frac{6}{7} \frac{4}{5} \frac{2}{3} = \frac{48}{105}$ to do it right

I started by labeling 8 strings A through H. Pair them as AB, CD, EF, GH.

Start by tying A to any of the 7 other strings. You have reduced your 8 untied strings to 6. Choose any untied string, there are 5 choices to tie it to. You are left with 4 untied strings. Choose one, there are 3 choices of untied strings. Your last 2 untied strings must be tied together. Total choices are $7 \times 5 \times 3 \times 1 = 105$

Which choices create only one loop?

Start with A. You cannot tie A to its partner B, but any of the other 6 will be OK. Call your choice C.

B cannot be tied to D, C's partner, otherwise we create a loop of 4 string segments, but any of the other 4 will be OK. Call your choice E.

D cannot be tied to F, E's partner, otherwise we create a loop of 6 string segments. D must tie to G or H. F ties to the other one.

$6 \times\ 4 \times\ 2 \times\ 1 = 48$

$48 / 105 = 0.45714...$

The solution can be easily found out for k number of strings by using the general expression (k-1)! / {(k-1)!!}^2 i.e factorial of one less than k , k-1 divided by square of DOUBLE factorial of one less than k, k-1.

For the given question, k is 8 Probability of single loop =(8-1)!/ ( (8-1)!!)^2

= 7! / (7!! X 7!!)

Now 7! = 5040 and 7 !! = 7x (7-2) X(5-2) x(3-2)= 105

So it is 5040 / (105 x105)

= 0.457142857

Once the top ends are joined, there are:

7×5×3×1=105

ways to join the lower ends.

To find the number of favourable cases, fix one loop and its orientation, for example in the figure of the loop in the question, fix the top left loop alongwith its orientation.

Then there are 3!=6 ways to arrange the three other loops and 2 ways each to orient them.

Therefore, 6×2×2×2=48 favourable cases.

Probablity=48/105=16/35=0.4571...

The Answer can be extended to 2n strings to show that the probability that the pieces of string are all joined in one loop is

(2n-1)!/[(2n-1)!!]^2

For n=4,

Answer=16/35=0.4571

Blue $=$ completed node.
Yellow $=$ current node.
Green $=$ valid node to connect to.
Red $=$ invalid node to connect to.

The fractions are the probabilities that the yellow node will connect to a green node, so the total probability for a single loop is $\frac67 \cdot \frac45 \cdot \frac23 = \frac{16}{35} \approx0.45714285714$ .