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Posted 23 March 2013 - 12:21 AM
I did find the number, as RG gave it, from simulation, not calculation. Uniform point-picking, necessary to form uniformly distributed triangles was an issue. Equally likely radius and angle gives preference to central points.
What was interesting to me when I simulated was that I was off by a factor of four that the four points would not be convex. That is, and should be 4 time the probability the 4th dart lands inside the triangle. There are three other areas the 4th dart can land that make a non-convex configuration, and those areas, on average, are equal to the triangle size. Not an intuitive result until you realize any of the darts can be taken to be the 4th dart.
So my last question now does not need to be asked, namely, to compare the probabilities of the two cases in the OP. also, that question has been answered in this thread.
Also, the mean triangle area was (to me) surprisingly small.
- Bertrand Russell
Posted 28 February 2014 - 12:33 AM
Since question 1 was not a part of that problem posted in 2008, let me provide a solution to that one real quick.Spoiler for probability of 4th dart inside
Now, that question 1 is solved, I am inclined to yield the opportunity of solving question 2 to others.
I think the proof is fine as it is, but I believe there are a few typos. See below
Question for bonanova: about question 1- the probability that the 4th point falls within the triangle,
Tardy answering your question ... better late than never.
Sextuple integrals are indeed the solution. Nasty, but thankfully someone has already done it.
- Bertrand Russell
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