Dear fa Shin,

I'm sorry to report that both algorithms appear to be incorrect.  The first one solves a different problem than the one stated.  You have chosen two particular semicircles.  The probability that three random points of a circle lie on one of these two particular semicircles is much smaller than the probability that they lie on any semicircle.

Three points of a circle lie on the same semicircle if and only if the interior of the triangle they determine excludes the center of the circle (or a side of the triangle contains the center).  However, you can ignore the case where the center of the circle lies on a side of the triangle because that event has probability 0.

Using the above observation, below is a Matlab code similar to your first code that I believe solves the stated problem.

N = 1000;

P = zeros(N,1);

j = sqrt(-1);

p = [0,0];

for i = 1:N

  count = 0;

  for k = 1:i

    a = rand(3,1)*2*pi;

    e = exp(j*a);

    node = [real(e),imag(e)];

    if ~inpoly(p,node)

      count = count + 1;

    end

  end

  P(i) = count/i;

end

figure

plot(P);

I'm not sure what you're trying to do with your second code, because it is not as well documented as the first.  However, I believe that it gives an incorrect result because the above code converges to a probability near 0.75.

Sincerely,

Greg.

Dear fa Shin,

I'm sorry to report that both algorithms appear to be incorrect.  The first one solves a different problem than the one stated.  You have chosen two particular semicircles.  The probability that three random points of a circle lie on one of these two particular semicircles is much smaller than the probability that they lie on any semicircle.

Three points of a circle lie on the same semicircle if and only if the interior of the triangle they determine excludes the center of the circle (or a side of the triangle contains the center).  However, you can ignore the case where the center of the circle lies on a side of the triangle because that event has probability 0.

Using the above observation, below is a Matlab code similar to your first code that I believe solves the stated problem.

N = 1000;

P = zeros(N,1);

j = sqrt(-1);

p = [0,0];

for i = 1:N

  count = 0;

  for k = 1:i

    a = rand(3,1)*2*pi;

    e = exp(j*a);

    node = [real(e),imag(e)];

    if ~inpoly(p,node)

      count = count + 1;

    end

  end

  P(i) = count/i;

end

figure

plot(P);

I'm not sure what you're trying to do with your second code, because it is not as well documented as the first.  However, I believe that it gives an incorrect result because the above code converges to a probability near 0.75.

Sincerely,

Greg.

Thank you for your insights, it is much more clear now!! :)

If you could still give me your insights, do you think the numbers of trials are not going to change the percentages of probabilities?

If I understand your question, I think you're asking if the probability (call it p) of interest changes as a function of the number of trials.  The probability p doesn't change; however our computed estimates of it change.  That's the short answer, but there is a longer answer as well.

Here is how I think of it. In the above code, the integer i represents the number of trials and k represents a specific trial. For each i, P(i) represents a random variable estimate of p.  That is, for a given i, P(i) changes every time we repeat the underlying experiment.  This becomes evident when you run the above code twice in succession and get somewhat different plots each time.  However, the larger i becomes, the smaller the variance in P(i) and the closer it gets to p.

We can demonstrate the variance of P(i) getting smaller as the number of trials increase by running a different code (given below).  When you run the code given below, you will see the sample variance of P(i) approximately decrease as a function of i.  The reason it isn't strictly decreasing is because the sample variance of P(i) is only an approximation of the variance of P(i).

clear all

n = 60; % Sample size of random variable P(i)

Pi = zeros(n,1); % Instances of (data for) P(i)

j = sqrt(-1); % Complex i

N = 40; % Number of sample variances

S2 = zeros(N,1); % Sample variances

for i = 1:N

  for L = 1:n

    count = 0;

    for k = 1:i

      a = rand(3,1)*2*pi;

      e = exp(j*a);

      node = [real(e),imag(e)];

      if ~inpoly([0,0],node)

        count = count + 1;

      end

    end

    Pi(L) = count/i;  % An instance of P(i)

  end

  S2(i) = var(Pi); % Sample variance of P(i)

end

figure

plot(S2);

xlabel('Number of trials (i)')

ylabel('Sample variance of P(i)')

Thank you very much Dr. Wilson with your insight helped me much! I appreciate it!

Dear fa Shin,

I couldn't sleep last night so I thought about your problem a little more. It turns out that it's not hard to solve it using probability theory (no programming).

Given three randomly placed points on a unit circle, we can always choose a coordinate system so that the first point is (1,0) and the second point is (cos(a),sin(a)) for some angle a with 0

Thank you, Dr. Wilson. Would you elaborate Pr(~E/a)? I am assuming its negation of Expectation value. Thank you again for your insight. Dr. Wilson :)

Your welcome. :)

E denotes the event that three randomly selected points on a unit circle lie on a semicircle.  It is not expected value in this case.

Pr(~E|a) is notation for a conditional probability.  It is the probability that the event ~E occurs (i.e. E does not occur) given a.

In general, if A and B are events, Pr(A|B) = probability of A occurring given that event B has occurred.  In our case, Pr(~E|a) is the probability that event ~E (not E) occurs given that a particular value of the angle a has resulted from the selection of three random numbers on a unit circle.

You should do some reading about conditional probability.  In particular,

Pr(A|B) = Pr(A&B)/Pr(B)   so that   Pr(A|B)Pr(B) = Pr(A&B).

This latter equation is the basic justification of the analysis in my previous answer. Once you get used to working with conditional probabilities, you'll find them useful for solving problems similar to the one you posed.

Another relevant topic is probability distribution functions (PDF).  In my previous post, Pr(a) = da/pi comes from the assumption that a is uniformly distributed over the interval [0,pi]. That means the PDF of a is that of the uniform distribution over [0,pi].

Greg.

Thank you Dr. Wilson for the thoughtful insights. :)

The problem can be treated as a reconstruction for a well-known one. I mean a “meeting problem”. Two agreed a meeting at some point at any time between 12 a. m. and 1 p. m. Each should wait for other for only 15 minutes, then go. What the probability of success {which means they can see each other}? This is a classic example on “geometric probability”. To resolute the problem, we only need plotting a square-form “probability space”, i. e. [0,1] x [0,1], and calculate the “good” area which is a hexagon {(x,y): 0≤x≤1, 0≤y≤1, |x-y|≤1/4}. The answer is 7/16.

Now, on that “3 points” problem. Let us set the first point to the α₁=π position (meaning 0≤α≤2π). Then we should take |α₂ – α₃| ≤π for the other two so this problem is analytically equivalent to the first one. I believe the correct answer is 0.75.

And, I will read those codes later, the problem is interesting. Note the contradiction You picked is impossible, as well as I know.

I have modernized Gregory L Wilson's code {see it in his answer above} a little. This is for Scilab (a free analog for the proprietary MatLab program.) Note I can't give You 100% warranty this is correct since I've used my  own idea which You can read on in my previous post.

I think it will be of no problem to correct it so it's usable in MatLab.

N = 1000;

P = zeros(N,1);

p = [0,0];

for i = 1:N

count = 0;

for k = 1:i

a = rand(2,1)*2*%pi;

d_12 = abs(a(2) - a(1))

if d_12

 Thank you, Dr. Bagmanov :) I will take a close look at it! This would help endeavoring for the better knowledge!  

The following is an excerpt from an answer byAndrey

" Now, on that “3 points” problem. Let us set the first point to the α₁=π position (meaning 0≤α≤2π). Then we should take |α₂ – α₃| ≤π for the other two so this problem is analytically equivalent to the first one. I believe the correct answer is 0.75."

this is not a subject of believe, It is just true according to the algorith recalled by Andrey:

[ (2\pi)^2 - (2 \pi - \pi)^2] / (2\pi)^2 = 3/4 :)