Assuming the rope is only a regular rope and the one unknowns are how does it intersect within the three factors marked in crimson within the following image (i.e. no unusual topology, no humorous path modifications on the intersections).
There are solely 8 attainable inclinations, I coloured the rope with two totally different colours (crimson and inexperienced) to point out them.
I name the inclinations with a sequence of three R (crimson) or G (inexperienced), relying on which a part of the rope is above in every of the three intersections.
Assuming that by “utterly random” OP implies that on every overlapping level there may be $1/2$ chance that the crimson half is above the inexperienced half and $1/2$ chance that the inexperienced half is above. This may end result right into a uniform distribution among the many 8 attainable mixtures.
Solely configurations RGR and GRG can be tied right into a knot, so the requested chance is
$2/8 = 1/4$
Proof that solely these mixtures can be tied right into a knot:
Properly, I attempted with a bodily rope. Sorry I haven’t got a greater proof.