Semicircle Probability

Three points are chosen randomly from the circumference of a circle. What is the probability that they lie on a common semicircle?


Bonus : Generalize this to n n points.

0.75 0.67 0.5 0.25

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7 solutions

I'll look at the general problem of finding the probability that n n points chosen at random on the circumference of a circle lie on a common semicircle.

First choose any one of the n n points. Now for each of the other ( n 1 ) (n - 1) points the probability that they are within π \pi radians going clockwise of the chosen point is 1 2 , \frac{1}{2}, and so the probability that all the remaining ( n 1 ) (n - 1) points lie on the (clockwise) semicircle that has the initially chosen point as the 'leftmost' point is ( 1 2 ) n 1 . (\frac{1}{2})^{n-1}.

Since the choice of the initial point was arbitrary, this result holds for each of the n n points on the circle, and since these 'events' are mutually disjoint, (i.e., we can't have two different points serving as the leftmost point on the clockwise semicircle on which all n n points lie), we can simply sum these n n probabilities to find that the probability that all points lie on the same semicircle is n × ( 1 2 ) n 1 . n \times (\frac{1}{2})^{n-1}.

In this case we have n = 3 , n = 3, which gives us the solution 3 × ( 1 2 ) 2 = 3 4 = 0.75 . 3 \times (\frac{1}{2})^{2} = \frac{3}{4} = \boxed{0.75}.

I understand the adding of mutually disjoint probabilities, but I didn't think it could be applied in this way. Took a while to convince myself that this was actually valid.

What a brilliant solution.

Bufang Liang - 5 years, 7 months ago

Well, as a reversal of roles, let me be the one to post the calculus solution this time :3. Lol, I used averaging probabilities (I don't know what it's even called).

Don't know when it will be up as my round of golf is about to start.

Trevor Arashiro - 6 years, 2 months ago

Exactly as solved.

Prakash Chandra Rai - 6 years, 2 months ago

I don't get this problem at ALL.

Can someone explain to me where I go wrong? Perhaps I'm misunderstanding the question. Fixing two points, the chance of a third can be 100% ( If they are across from each other or on top of each other, there can be a semicircle regardless of the placement of the third point). There are definitely positions with two points of 100%, and therefore, to have a probability of 75%, there must be a setup where there is a chance of a semicircle under 75%. But no matter where I fix the two points, I can't reach it.

This problem has been bugging me for WEEKS, especially since this solution makes sense to me as well. Can someone PLEASE explain how I'm wrong? Just a picture of a position with two points where a third point placed on the circle at random would not fall on a semicircle under 25% of the time and a description. A below solution claims that, when the two points are in the same place, this is true, but it is clearly not since you can move the semicircle around those points to encompass the whole circle.

Alex Li - 5 years, 2 months ago

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Picture a clock face, and without loss of generality place one point on the circumference at 12 o'clock. If the second point is placed at 3 o'clock then any third point placed between 6 and 9 o'clock will be outside any semicircle that contains the first two points. In this case we would get the 75% you mention.

But now place the second point at, say, 5 o'clock. Then any third point placed between 6 and 11 o'clock would be outside any semicircle that contains the first two points. In this case we would only have a 58.33% probability. So as the second point makes its way around clockwise (or counterclockwise) from 12 to 6, the probability decreases linearly from 100% to 50%, (with a point of discontinuity at 6 o'clock where the probability spikes to 100%). This results in an overall probability that is the mean of 100% and 50%, i.e., 75%.

Brian Charlesworth - 5 years, 2 months ago

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Okay, thank's a lot, the fact that 6 o'clock didn't work made me not consider the points directly next to it. This cleared things up (and makes me feel kind of dumb, but I'm just glad I understand now)

Alex Li - 5 years, 2 months ago

Would the probability at just before 6 o'clock be exactly 50% though? Because I am thinking that you are looking at an angle created by first the two points where the third point cannot be placed, and the percentage of all 3 being there is 360 θ 360 \frac{360-\theta}{360} . But if θ = 180 \theta=180 then you'd be back to 100% chance. If you consider the points prior to θ = 180 \theta=180 would it be enough to say that θ = 179.999 = 180 \theta=179.999\ldots=180 ? giving exactly 50%?

Nathaniel Bronson - 2 years, 4 months ago

Can you please explain. Here just found out the sum of all the probabilities which were disjoint. But in the undermentioned calculus solution, he found the average of the probability. In another level 2 problem, which you must have solved, of getting the probability of getting head while tossing a coin on an even numbered flip, the solution was to find just the sum not the mean. So what do you do, if the probabilities are disjoint. Find the sum or the mean . I am confused, please elucidate.

Prayas Rautray - 3 years, 11 months ago

Just divide the circumference into 4 equal parts. It is rest assured that the 2 points both lie on the same part. Now for the last point it should lie on one of the 2 adjacent parts or within the part where the 2 fixed points are located. Since the last point can only lie on the 3 different parts out of 4 total parts then the answer is 0.75.

All thanks to 3blue1brown. I have watched one of his videos that contains this kind of problem. That's why I solved this instantaneously😎

Hans Duran - 10 months, 2 weeks ago
Trevor Arashiro
Apr 5, 2015

First, fix a point on the circumference of the circle, call it P P now, draw a diameter through the circles center C C and call the other end point of the diameter Q Q .

Now, place a second point S S anywhere on the circle and draw a diameter with S S as one end point and T T as the other.

Label angle S C P \angle SCP as θ \theta . Hence T C Q \angle TCQ also equals θ \theta .

Now, the third point can't be on minor arc T Q TQ . So the probability that it's not on that arc is

2 π θ 2 π = 1 θ 2 π \dfrac{2\pi-\theta}{2\pi}=1-\dfrac{\theta}{2\pi} .

Note that theta varies between 0 θ < π 0\leq \theta <\pi

Using probability distribution function, we have

1 π 0 π 1 θ 2 π = 3 4 \dfrac{1}{\pi} \displaystyle \int_0^{\pi} 1-\dfrac{\theta}{2\pi}=\frac{3}{4}

Nice approach. One typo to mention: I think that you mean "minor arc T Q TQ ". And the "whatever-it's-called" is a probability distribution function, (pdf). :)

So how would you generalize this method for n n points?

Brian Charlesworth - 6 years, 2 months ago

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Ya, I did mean minor arc T Q TQ . Thanks for pointing that out.

I'm new to this probability distribution function thing :3 lol. I'm not sure how you would use it however, because after picking two points, the probability of picking the nth point will either be the same as that of the (n-1)th point or less. This is because if the (n-1)th point is chosen between the two most exterior points, the probability of the nth point being in that same range won't change.

Trevor Arashiro - 6 years, 2 months ago

What is a probability distribution function..?

Soumak Bhattacharjee - 6 years, 2 months ago

This was out of the box ... fascinated ...

Ujjwal Mani Tripathi - 6 years, 2 months ago

I guess I don't understand this very well...shouldn't you be able to integrate the pdf with the appropriate constant and get 1? I integrated the pdf between 0 and pi with and without the constant between 0 and pi and couldn't get 1.

Scott Shannon - 6 years, 1 month ago

Outstanding solution.

Swapnil Das - 5 years, 3 months ago
David Vaccaro
Apr 9, 2015

A nice geometric argument is this....let x,y,z be the proportions of the circle lying between the three points.

If you plot ( x , y , z (x,y,z in 3-space you get a point uniformly distributed in the region x + y + z = 1 x+y+z=1 with x , y , z 0 x,y,z\geq0 which geometrically is an equilateral triangle with vertices at i, j and k.

The three points do not lie on the same semicircle if x , y , z < 1 2 x,y,z<\frac{1}{2} which is the equilateral triangle with vertices i 2 , j 2 , k 2 \frac{i}{2},\frac{j}{2},\frac{k}{2} which has area one quarter of the original triangle. (Easy to see in a diagram, it's just the middle quarter if you split original triangle by joining midpoints of adjacent sides)

Good proof. But wouldn't it have been to visualise if there were more than 3 points? Maybe you'd require more than 3-d

Led Tasso - 6 years, 2 months ago

Let there be x x points on a semicircle.

Now we have to find out the probability to have all the points on a semicircle which will be equal to 1 - {probability of having no three points in same semicircle}

Probability of selecting 1 point on the semicircle = 1/2

Probability of selecting 2 points on another semicircle = x ( x 1 ) / 2 x ( 2 x 1 ) x(x-1)/2x(2x-1)

This selection can be done in two ways so the probability of selecting three points that do not lie on a semicircle = 2 / 2 x ( x 1 ) / 2 x ( 2 x 1 ) = x 1 / 2 x 1 2/2 * x(x-1)/2x(2x-1) = x-1/2x-1

There fore the probability of having all the 3 points on a semicircle = (1- (x-1/2x-1) = 3x-1/4x-1 Now as x tends to infinity the limit comes to be 0.75.

Dan Lawson
Apr 9, 2015

Fix the first point and choose the second point and then find the probability that the third is on a semi-circle.

Extreme Case I: The first two points are opposite each other. In this case the probability that the third point is on a semi-circle with the first two points is 1.

Extreme Case II: The first two points are the same point. In this case the probability that the third point is on a semi-circle with the first two points is 1 2 \frac{1}{2} .

Assuming the probability is linearly distributed, the overall probability is the mean of 1 and 1 2 \frac{1}{2} , which is 3 4 \frac{3}{4} .

Dan, I used the same approach as you and had (almost) exactly the same graph. However, I think that for Extreme Case II the probability would be 1 once again, as you could use the common first two points as the endpoint of a semi-circle that could proceed either clockwise or counter-clockwise around the circle, and thus encompass the entire circle. This would represent a point of discontinuity for your function, as the rest should align with the graph you posted, and wouldn't actually impact the answer.

Michael Hunt - 6 years, 1 month ago

This is how I pictured it.

Figure Figure

Dan Lawson - 6 years, 2 months ago

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This graph is backwards. The probability is 1 at 0 degrees and approaches 1/2 as theta approaches 180 degrees. There is a discontinuity at 180 degrees, but that's not important for the integral.

Richard Desper - 4 years ago

Since the points are randomly chosen, how can you fix two points? As far as I can say you can fix only one point.

Prakash Chandra Rai - 6 years, 2 months ago

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Wherever you place the second point, it will be a part of either one of the two semi-circles formed by the first point alone (A circle is, essentially, 2 semicircle joined end to end). It doesn't matter until the third point comes in, in which case you need to consider the probability of the third point being a part of the same semi-circle containing both the points.

Vishnu Bhagyanath - 5 years, 10 months ago

It's not useful here to look at the Extreme cases, since the options open to the antiipodal point are nothing like the options open to any other point near it. Why? Because the initial point and the antipode create no minor arcs.
Also, you've made an error with Extreme Case II. If your first two points are identical, then the probability that the third point is on a semi-circle is also 1.
As other solutions show, if your second point is theta radians away from the first point, for 0 < theta < pi, the odds that the third point will be on some semi-circle with the first two is (1 - theta/2 pi), as the excluded region will be the minor arc directly opposite the arc between the first two points.

Richard Desper - 4 years ago

Thats how I did it. But I do not know whether this method follows from the axioms of probability. That is we may need to prove our algorithm.

Siddharth Iyer - 6 years, 2 months ago

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Check out Geometric Probability .

Calvin Lin Staff - 5 years, 7 months ago

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Was it ok to assume that it was linear?

Eric Kim - 4 years, 10 months ago

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@Eric Kim Of course not. That has to be proved (and we also have to take the underlying probability distribution into account) Look at the other solutions.

Calvin Lin Staff - 4 years, 10 months ago

Consider the circle having a circumference of 2 2 , then let's put the first to point two points having a distance k k between them with k k between 0 0 and 1 1 . Then for k k chosen randomly between 0 0 and 1 1 the probability that the other point is in the same semicircle is ( 1 + k ) / 2 = 3 / 4 (1+k)/2=3/4 .

Rob Bradley
Jan 6, 2018

This is the complementary problem to the following: break a stick in two places - what is the probability that you can form a triangle with the three pieces that result?

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