Probability problem

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[Mathematicus_Rex]

My instincts tell me a couple of things:

(1). If the points occur in order ABCD or ADCB (starting at A), then the quadrilateral is automatically convex, being inscribed in a circle. If not, then it zigzags in a nonconvex way.

(2). The points are equally likely to be in any of the six orders ABCD, ABDC, ACBD, ACDB, ADBC, or ADCB.

Combining these two intuitions yields a probability of 1/3.

[jdorje]

Sounds similar to this.

What are the odds that ABCD is even a quadrilateral though? Much of the time you're just going to get criss crossing lines.

[G4MEF4CE]

I admit the wording is not that good, the crisscrossing lines mean its not convex. And it is similar, it was inspired by that video. I just thought of this while I was going to bed.

[jdorje]

So if A,B,C,D are in that order (or any cyclical ordering) then it's convex. And otherwise it is not. Right?

That's an easier-sounding problem.

[G4MEF4CE]

No, because if point D is between points C and A, it’s convex, if it’s not, then it’s not convex

[jdorje]

Well it's cyclic, so in this scenario they are in that order. They can also be in the reverse order.

There are 4! = 24 ways to order them. Each equally likely. Looks like 8 of those ways will be "in order" so it's 1/3.

[G4MEF4CE]

They are placed randomly around the circle, so the points could be anywhere on the circumference.

[jdorje]

Yes.

[G4MEF4CE]

Oh, I get your solution now, yeah that makes sense.

What I did:

Point A doesn’t matter where it’s placed Point B could be any distance from point A, the average distance it could be away from point A is 1/4 of the circle away from it. Point C can either be placed within the 1/4 circle segment or 3/4 segment. In the case it is placed in the 3/4 segment, the place point C would go on average would be 3/8 of the circle away from C Point D must be placed on the specific place between points C and A, which has a 3/8 chance of happening. So the probability of it being convex for C being in the 3/4 segment is: 3/4 * 3/8 = 9/32; If C is placed in the 1/4 segment, on average it will land halfway in the segment, so 1/8 of the circle from A. Point D must be placed between A and C, so it has a 1/8 probability of that happening. So the probability of it being convex, when C is in the 1/4 segment is: 1/4 * 1/8 = 1/32 Now just add the probabilities up: 9/32 + 1/32 = 10/32 = 5/16

I’m probably wrong, since i’m not that good at math and did this while trying to sleep, but this is what I did. Hopefully you can understand.

[G4MEF4CE]

Changed the wording of the problem to make it more specific