The figure shows the shadow of a piece of rope on the ground, and you can't see which part is on which part; Suppose the rope is placed completely randomly. Now tighten the two ends of the rope to the left and right. What is the probability that the rope will be tied into a knot?
Assuming the rope is just a normal rope and the only unknowns are how does it intersect in the three points marked in red in the following picture (i.e. no strange topology, no funny direction changes at the intersections).
There are only 8 possible dispositions, I colored the rope with two different colors (red and green) to show them. I call the dispositions with a sequence of three R (red) or G (green), depending on which part of the rope is above in each of the three intersections.
Assuming that by "completely random" OP means that on each overlapping point there is $1/2$ probability that the red part is above the green part and $1/2$ probability that the green part is above. This will result into a uniform distribution among the 8 possible combinations.
Only configurations RGR and GRG will be tied into a knot, so the requested probability is
$2/8 = 1/4$
Proof that only these combinations will be tied into a knot:
Well, I tried with a physical rope. Sorry I don't have a better proof.
The probability is
There are three points where the ropes cross each other If we start from the left end, and look at that rope, the first three crossings have 8 possible configurations. If we denote "on top" as '1' and "underneath" as '0', these configurations are just like the binary numbers 0 (0b000) through 7 (0b000):
0 0 0
0 0 1
0 1 0
0 1 1
1 0 0
1 0 1
1 1 0
1 1 1
For the rope to end up in a knot, you need either
101- two out of eight possible configurations.