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?

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    $\begingroup$ Once again you've chosen tags that have no relevance to this puzzle. $\endgroup$ Commented Jul 14, 2022 at 14:13
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    $\begingroup$ I think there won't be an answer unless "completely randomly" is fully defined. In mathemacial terms, you need to specify your probability space, e.g. does every crossing has equal and independent probability of being over- or under-crossing. $\endgroup$
    – WhatsUp
    Commented Jul 14, 2022 at 16:43
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    $\begingroup$ @WhatsUp If "completely randomly" in this case does not mean "every crossing has equal and independent probability of being over- or under-crossing", then what else could it possibly mean? $\endgroup$
    – JLee
    Commented Jul 14, 2022 at 17:08
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    $\begingroup$ @JLee A lot of possibilities. E.g. the rope could be interpreted as an eulerian path from one end to the other; or it could make turns at random on each vertex (thus a conditional probability); it could also be the trace of a brownian movement on that "shadow", which in the end covers the whole shadow. I'm not saying what is probably the intended meaning, just saying that the question is not well defined. $\endgroup$
    – WhatsUp
    Commented Jul 14, 2022 at 18:04
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    $\begingroup$ @MrMonkey where did you find the image? $\endgroup$
    – bobble
    Commented Jul 15, 2022 at 13:45

3 Answers 3


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.

  • $\begingroup$ @JLee no problem. I actually cheated a little bit by posting the text of the answer and adding the images and the "proof" later. I think you can post your answer too, looking at the timestamps most user will know that we were trying to solve this puzzle independently. $\endgroup$
    – melfnt
    Commented Jul 14, 2022 at 18:55
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    $\begingroup$ LOVE the proof! $\endgroup$ Commented Jul 15, 2022 at 6:15
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    $\begingroup$ Love it! Nice and obvious. $\endgroup$
    – Mr Monkey
    Commented Jul 15, 2022 at 7:01

The probability is


The reason:

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 010 or 101 - two out of eight possible configurations.





— Unless the two end crossings are different from the center crossing, it's not a knot.
— So there's a 50% × 50% = 25% chance that it is a knot.

  • $\begingroup$ ... and for completeness, if they are both different then we have a standard knot. $\endgroup$
    – Florian F
    Commented Jul 15, 2022 at 6:52

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