There is a circle where you put 3 points randomly on it as shown below:

enter image description here

What is the chance of these randomly chosen three points passing on an half circle?

Reference: Bilim Teknik Dergisi 2018-07

  • $\begingroup$ I agree with Bass. This looks like the same question. $\endgroup$ – hexomino Jul 6 '18 at 14:29
  • $\begingroup$ Indeed. While that other question uses three points inside the circle instead of on the perimeter, that makes no difference to the answer. $\endgroup$ – Jaap Scherphuis Jul 6 '18 at 14:33
  • $\begingroup$ @Jaap I concur. First paragraph of accepted answer there says it all: "First, observe that the triangle does not contain the center if and only if all three points are contained in a half-circle. So, only the polar angles of the points matter, not their distance to the center." $\endgroup$ – xhienne Jul 6 '18 at 14:40
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    $\begingroup$ I don't agree that this question is duplicate. It can have the same implications, yes, but I think the formation of the questions changes the puzzle dramatically. This question seems a lot simpler and intuitive, which is why the solutions are more diagram/simple probability. The original question has done a lot more things with the "triangle" mentioned (in almost all the answers except for the top-voted one.) $\endgroup$ – Aryaman Jul 7 '18 at 11:26
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    $\begingroup$ And more importantly, the questions are obviously asking for the opposite thing. The answer to this question is, atleast according to the OP in the comments, is meant to be the "probability-inverse" of the answer given by @CakeMaster (forgetting what they're called - basically 100 - that answer) $\endgroup$ – Aryaman Jul 7 '18 at 11:30


The first dot can be considered a single position, because we can always rotate the circle.

The second dot can always be considered as not crossing the half line, because we will simply redefine which half we're talking about depending on where it falls.

Thus the second dot can fall anywhere within half the arc of the circle. The third dot must fall within the other half (50%), AND also opposite to the arc between the first 2 dots (another 50% on average). If it falls on the opposite half but outside the opposite of the arc, then the half can just be redefined to contain all 3 dots again.

Multiplying those probabilities gets us 25%.

  • $\begingroup$ It would be interesting to know why this answer, which match all the answers of the allegedly duplicate question, was downvoted. Surprisingly, at the time of writing, the only other answer which proposes 50% was upvoted. You should really comment your downvotes, guys. $\endgroup$ – xhienne Jul 6 '18 at 14:57
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    $\begingroup$ the answer is 75% $\endgroup$ – Oray Jul 6 '18 at 15:00
  • $\begingroup$ Probably because I didn't include a picture, or explain it very well. I did some napkin math here to prove it to a coworker as well. Basically it boils down to the integral of x/2pi on the interval from 0 -> pi, and then divide out pi to find the average. $\endgroup$ – CakeMaster Jul 6 '18 at 15:01
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    $\begingroup$ @Oray, Oh, I see, you are right, this is the exact opposite of the other question. $\endgroup$ – xhienne Jul 6 '18 at 15:02
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    $\begingroup$ @xhienne exactly $\endgroup$ – Oray Jul 6 '18 at 15:03

50%. The first two points will always be on a half-circle, the chance that the last point is on the same half-circle is 1/2 or 50%.

enter image description here
The image above illustrates the possibilities: in stage #1, it really doesn't matter where you place the first point, it just becomes the reference for the following points.

For stage 2, point #2 will end up on a half-circle with point #1, pretty much by definition. It has to be on the same half circle. So two points down and still at 100% probability.

So it all comes down to the last point. There are two major cases, counting clockwise: 1, 3, 2 or 1, 2, 3. No matter the order, as long as point #3 ends up on the same side of the circle as point #2 (with reference to point #1) (depicted as red vs. pink in the diagram), all three points will end up on the same half circle.

Now, if the points are close together, the probability actually increases, approaching 100% as points #1 and #2 approach each other. This can be envisioned by swapping points #1 and #2. Nevertheless, the minimum probability is 50%.

  • $\begingroup$ What if the first two points are close together? Would you still say 50% in this scenario? $\endgroup$ – hexomino Jul 6 '18 at 14:21
  • $\begingroup$ -creating graphic to illustrate $\endgroup$ – IronEagle Jul 6 '18 at 14:25
  • $\begingroup$ I don't think this is right. The first 2 points will rarely fully define the half-circle, so in most cases the 3rd point has a more than 50% chance of being in it. $\endgroup$ – Cain Jul 6 '18 at 14:58
  • $\begingroup$ Point 3 can fall just to the left of point 1 and still lie in the same half circle as point 2 (and 1). $\endgroup$ – Jaap Scherphuis Jul 6 '18 at 15:01
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    $\begingroup$ And the maximum probability is 100%, which gives an average of 75% which is the actual probability of 3 randomly chosen points to be in the same half. Strictly speaking you have to integrate over all possibilities of point 2, but since things are very linear, it is the same as taking the average of the minimum and maximum. $\endgroup$ – Jaap Scherphuis Jul 6 '18 at 15:14

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