Probability in a cone, numerical or theoretical answers welcome?

Well, I simulated over 10^8 experiments, using the rand() function in C. Each experiment independently generated 3 points on the unit disk according to your procedure (first finding them uniformly on the 3d cone, then projecting down). I got:

Pr[No intersections among the 3 points] = 0.348883.

This seems higher than I would have guessed. I don't know if my program had bugs or not. The same simulation showed that the probability the first 2 points did not intersect was 0.642930.

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Analytically, it may be useful to note that if we let (A, theta, r) be the parameters associated with a disc in the 2-d unit circle, where A is the distance from the origin, theta is the angle, and r is the radius of the disc, so that the disc has radius r and origin (A*cos(theta), A*sin(theta)), and we assume (A, theta, r) are formed from your procedure, then:

A has probability density f(a) = 6a(1-a) for a in [0, 1].

Given A, the random variable r is uniformly distributed in [0, 1-A].

The random variable theta is independent of (A, r) and uniform over [0, 2*pi).

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So given 2 independent points with parameters (A1, theta1, r1), (A2, theta2, r2), the probability they do not intersect is:

Pr[(r1+r2)^2 < (A1*cos(theta1) - A2*cos(theta2))^2 + (A1*sin(theta1) - A2*sin(theta2))^2 ]

which is:

Pr[(r1+r2)^2 < A1^2 + A2^2 - 2A1*A2*cos(theta1 - theta2) ]

which in principle could be computed by conditioning on A1, A2, cos(theta1-theta2) (which are mutuallly independent with known marginal PDFs). Hopefully the exact answer is close to my experimental number 0.642930.

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@Gianlino: Sorry. I did the simulation over looking for the number you ask:

Pr[{1 and 2 intersect} AND {1 and 3 intersect} AND {2 and 3 intersect}] = 0.114925

I don´t know how to do this theoretically, although I wouldn´t mind being told. I am running a small simulation now. I´m running just 2 disks now, in batches of 10000 and for that get approximately:

P(no intersection) = 0.644, P(intersection) = 0.356 with standard deviation 0.004

That is with 23 batches of 10000 each. I will look at the 3 disk situation later today.

edit:

Also, I´m generating x,y uniformly on [-1,1] and z uniformly on [0,1] and rejecting (x,y,z) if it is outside the cone. That is how I am getting a random point in C. Also, if the three disks are A,B,E then my understanding of pairwise non-empty intersections is that A intersect B is non-empty and B intersect E is non-empty and E intersect A is non-empty.

edit2: Using 30 batches of 10000 each I get:

P(3 disks intersections pairwise non-empty) = 0.1148 with standard deviation 0.0030.

This jibes well with Michael´s result.

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@gianlino: Thanks for sharing the write-up. It looks pretty complicated, but with a very nice answer. I just saw your second comment to me. I´ll think about the A intersect B intersect E non-empty thing. It does seem more complicated.

theoretical danger = favorable outcomes/complete possible outcomes = 5/(3+4+5) = 5/12 experimental danger = genuine form of cases the favorable outcomes takes position / complete form of experiments = 40 8/one hundred twenty they're very close. in case you run the try 1200 cases, 12000 cases, then you definitely can anticipate both opportunities are severe closer and closer.