[Monte314]

"Three points are selected at random on a circle. What is the probability that they all lie on the same side of some diameter?"

(Keep in mind that a diameter must pass through the center of the circle.)

Be careful: this problem does not ask, "What is the probability that three randomly selected points fall on the same side of a given diameter." The answer to that question is 1/4 (use the Binomial Probability Distribution). This question asks about finding any diameter that puts all three points on one side.

[HIDE="Here is a hint. This problem is EQUIVALENT to this one... do you see why?"]
"Randomly select three points in the plane. What is the probability they form an obtuse triangle?"
[/HIDE]


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[Syntax]

[HIDE="answer"]
.5

the first two points always lay on the same side(since it can be arbitrarily chosen)...then there's a 50% chance the last point will fall on that side.[/HIDE]

[EDIT] WRONG!

[SirJac]

( Show Spoiler )

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Once two points are determined, the region that the third point can lie in so that there is an obtuse angle is shaded in green. From this we can see that the region where it is not possible to find a diameter that all three points lie on (shaded in red) is equal to the angle the first two points make with respect to the center of the circle, measured in radians. From this we get that the probability that the third point will lie in the green region is:

P=(2pi-theta)/2pi

The largest angle the first two points can make is pi radians, and there is equal probability for all angles between 0 and pi to be made. So repeated over many times, we would expect an average value of pi/2 as the angle made by the first two points. Using this expected average in the probability function above we get an expected value of P=0.75, or there is a 75% probability that 3 random points will fall on one side of some diameter.

[Monte314]

SirJac's answer is correct (no surprises there), and his proof is very neat... completely different from what I did.

Anyone else?

[Latro]

SirJac's proof reminds me of a result we ran into in my linear algebra class recently: as you increase the number of dimensions, the probability that a given vector is orthogonal to the coordinate axes increases, and in fact it tends to 1. For example:
(1 1 ....)
is, in the limit, orthogonal to
(1 0 ...)
because the angle between them is cos^(-1)(1/sqrt(n)), where n is the number of components in the vectors. If you send n to infinity, you get cos^(-1)(0) = pi/2.

But no, I don't really have anything to add to the actual problem; having not taken a probability class I'm fuzzy on setting up a distribution function like that, and having read SirJac's proof it would just wind up being a spin on that anyway.

[ArtistTyrant]

bah, i really need to study mathematics, its undoubtedly my weakest field, i had a high school understanding at like 6 and stopped studying it :/ i should have kept going -.-

um, the answer i had was 0.5^3=12.5%, but i think that is wrong because i'm preconceiving the circle already being sliced in half and the points all placed simultaneously or something O_o idk :/

[Latro]

	Originally Posted by ArtistTyrant To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	bah, i really need to study mathematics, its undoubtedly my weakest field, i had a high school understanding at like 6 and stopped studying it :/ i should have kept going -.- um, the answer i had was 0.5^3=12.5%, but i think that is wrong because i'm preconceiving the circle already being sliced in half and the points all placed simultaneously or something O_o idk :/

Part of the problem with that approach is that any two points MUST be on the same side of some diameter, because you can just choose a diameter that would work. The interesting part is whether there is a diameter such that the third point is on the same side of it as the other two.

[Monte314]

Yes, Latro, this is strange but true.

Here is a slide I created for a seminar I gave a few years back that shows this fact:


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[ArtistTyrant]

oh, so the question is asking whether three random points will be in the same side of the circle, and the circle can be cut AFTER...idk what the hell i was thinking, i was assuming that the circle was already cut, idk =.=

[broken]

My initial guess was wrong--I snuck a peak at Jac's answer and rethought my solution. I'm just starting out as a Mathematics major and tried to keep the explanation intuitive--I'd welcome a critique. I wish it wasn't as wordy, but it first came from a few drawings.

[HIDE="answer"]
Let’s say P1 falls at angle 0 degrees. P2 is sure to fall within 180 degrees of P1, forward or backward, so there is ~100% chance that P1 and p2 share a diameter. There’s an equal likelihood that P2 will fall in an adjacent quadrant to P1 or not. If angle_1 = angle_2, P3 will always be within 180 degrees of P1 and P2, and so it will be 100% chance of success. But this is the extreme, unlikely case. From 0 < angle_2 < 90, P3 can fall between 75% and 100% of the circle. From 90 < angle_2 < 180, P3 can fall between 50 and 75% of the circle. The space that P3 can occupy as a function of angle_2 is inversely proportional and linear. It’s equally likely for P2 to fall between 0-180 degrees, so we can take the average of the two areas and say there’s a 75% chance P3 falls on the same side as P1 and P2.[/HIDE]

[Monte314]

	Originally Posted by broken To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	I'm just starting out as a Mathematics major and tried to keep the explanation intuitive--I'd welcome a critique. I wish it wasn't as wordy, but it first came from a few drawings.

You clearly have the knack for mathematical thinking. One of the things you will learn in college is how to formalize this so that hidden assumptions and other sources of error can't sneek in and ruin things... further, the abstraction of good formalism will help you generalize your ideas, recognize which conditions are essential, and make your thoughts unambiguously accessible to others.

I'm an analyst, so I often start with a picture, too. But the picture is merely a way of compactly representing the fundamentals of a problem, and symbology is introduced almost immediately.

[Syntax]

	Originally Posted by Monte314 To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	...One of the things you will learn in college is how to formalize this so that hidden assumptions and other sources of error can't sneek in and ruin things...

Or he'll end up like me and oversimplify concepts and give answers like the one I did! X0



...in my defense, I formulated my answer within about 60 seconds. Arrogance was to blame.

[Monte314]

	Originally Posted by Syntax To view links or images in this forum your post count must be 2 or greater. You currently have 0 posts.
	Or he'll end up like me and oversimplify concepts and give answers like the one I did! X0 ...in my defense, I formulated my answer within about 60 seconds. Arrogance was to blame.

This has always been a weakness of mine as well. In graduate school, my advisor said the faculty felt I was "glib", which I've taken to mean that I tend to jump to conclusions and express ideas before they are mature. I think this assessment is accurate.

To address this shortcoming, I've trained myself to stop and carefully think through each step in my reasoning, reserving judgement and trying to make sure "I know what I know" before proceeding. It doesn't always work, and my research notebook is full of entire pages that have a big "X" across them; but I usually arrive at good results eventually...

...and every once in a while, one of those leaps turns out to be something I wouldn't have gotten to any other way.

[Malle]

( Show Spoiler )

Pick two points p1, p2.
Orient your circle such that p1 is at the left-most point and p2 is in the top half of the circle (if your circle is centred at (0,0) and has radius 1, p1 would be at (-1,0)).
The part of the circle that can be selected for the criterion to be fulfilled is then 1/2 + alpha / 2*pi, where alpha is the angle from the centre of the circle to p2.

Integrate this over alpha from 0 to pi (enough due to the symmetry in how we align our circle and points) and average out by dividing by the range of the alpha integration and we have our answer:

integrate(1/2 + alpha/(2*pi), for alpha = 0 to pi) / pi = (1/pi)(pi/2 + pi/4) = 1/2 + 1/4 = 3/4

The probability that 3 random points (presuming random selection here means uniformly distributed angle selection) are on some same semi circle is 3/4 or 75%

[Irrational84]

Nice problem! I wouldn't even know where to begin attempting to figure that out. Hopefully that will change as I continue my education.

[acyckowski]

Infinitesimally less than 1/2.

[HIDE="Spoiler"]The first point has 100% chance of laying on the diameter. Call the first point 0 degrees.

The second point has infinitesimally less than 100% chance of laying at some point that is not exactly zero or 180 degrees.

The third point has infinitesimally less than 50% chance of laying on the same side of the diameter as the second point and infinitesimally less than 100% chance of not landing on points one or two.

This answer depends on my understanding of the problem statement that the circle is a given, and cannot be circumscribed after the fact based on points one and two.[/HIDE]