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We have a square. we will opt three random point from inside of this square and name it $p_{1},p_{2},p_{3}$ then opt another random point $p_{4}$. what is the probability of that $p_{4}$ lies in triangle $p_{1}p_{2}p_{3}$ ?

I write program and test it 10 times and in each case test it for $10^{7}$ points and the results are:

TEST CASE # 1 = 0.076566
TEST CASE # 2 = 0.076430
TEST CASE # 3 = 0.076308
TEST CASE # 4 = 0.076378
TEST CASE # 5 = 0.076433
TEST CASE # 6 = 0.076340
TEST CASE # 7 = 0.076289
TEST CASE # 8 = 0.076382
TEST CASE # 9 = 0.076402
TEST CASE #10 = 0.076260
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  • $\begingroup$ I believe this is sufficient to answer your question. people.missouristate.edu/lesreid/Adv41.html Once you have the expected area, the probability $p_{4}$ lies in your triangle should follow. $\endgroup$ – JessicaK Jun 1 '14 at 9:23
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    $\begingroup$ @JessicaK. This looks like magics to me ! Thanks for the link. $\endgroup$ – Claude Leibovici Jun 1 '14 at 9:27
  • $\begingroup$ @JessicaK: Link not available, are there some regional restrictions on that site? Other link to and partial answer at rqna.net/qna/… $\endgroup$ – Dr. Lutz Lehmann Jun 1 '14 at 11:57
  • $\begingroup$ Link's not working for me either. Looks like a related question was asked before, which has that link. $\endgroup$ – gar Jun 1 '14 at 12:04
  • $\begingroup$ Are you looking for an exact answer, in a formula, or just an approximation? What did the original question ask for, or was it just curiosity? $\endgroup$ – Henno Brandsma Jun 1 '14 at 13:01
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A mathworld link and a paper answers the question.

$$ \mathbb{P} = \frac{11}{144} \approx 0.0763889 $$

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