Six identical pieces of string run through an opaque tube; there is no way to identify which ends form one piece of string.
At one end of the tube tie three knots, each knot joining two of the pieces of string.
Then tie three knots at the other end of the tube, again each knot joins two pieces of string.
What is the probability that you have made one continuous loop from the six pieces of string?
The three knots at the first end of the tube have no effect on the probability, as you will always produce, in effect, three paired pieces of string looped through the tube.
At the other end of the tube choose an end of string. If you tie this to produce a loop of two pieces then you can't make one continuous loop of six pieces. You can tie the end you have chosen to any of the five other ends - one makes the loop of two pieces, and the other four do not. So the probability of not failing at this stage is 4/5.
For the remaining four ends the probability of not failing is 2/3, and the other two ends must be tied together.
So the probability of one continuous loop is 4/5 x 2/3 or 8/15 - better than 50%!