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Another version, slightly faster (8 sec) Code:sim=:3 : 0 ((>./)(<./))(?3$0) ) (+/%#) (%3)>((sim "0) 1000000$0)
Hi bobbym,
Hi gAr;
Hi, Code:sim=:monad define a=. ?3$0 (>./a)(<./a) ) (+/%#)(%3)>(sim&+) 1000000$0 ≈0.259516
Okay, see you tomorrow.
Okay.
Thanks gAr, I appreciate you saying that. You will understand it or quickly find the hole in it.
Hi bobbym,
Since this is a thread devoted to computer math programs let's see how mathematica can be used to solve the problem experimentally. We can form a table of n subdivisions and solutions: The table means that if we have the set, {0,1/20,2/20,3/20,...1} there are 245 ways. A finite set with 21 elements representing points on the number line is a poor substitute for the unit interval with its infinite number of points. We need a relationship between the solutions and n. We ask mathematica to find one. It comes up with the following difference equation: We run the recurrence in the forward direction to get n = 2 000 000. The number of ways to pick three from the 2 000 001 points is 345677975308901235. The probability is: This only disagrees with 7 / 27 by less than .00000077, so we are on the right track. Can we do better? We could, if we could solve that recurrence but it will not even fit on a piece of paper. We need a trick. Let's study the solutions again starting at 6. It is well behaved at 7 and 8 and then takes a big jump at 9. Well behaved again at 10 and 11 and then another big jump. This disproportionate jump occurs at every third one. Let's see if we can understand it. The first tool to understanding any sequence is a difference table. Old timers like Newton,Leibniz, Gauss, Euler, Legendre and Lagrange lived off of them. They are much less used and understood today. Let's form a difference table of every third one. That is what we want a last line of constant differences. That means there is a polynomial fit for every third entry in the table. Now we find that polynomial. So you see hiding inside the major sequence was another smaller pattern. One much easier to deal with. This is sometimes the case with recurrences and you should look for it. Now to finish up. The more points we use in the interval 0 to 1, the closer we approximate the continuous unit interval. We will use the calculus to get an infinite number of them. The probability of drawing three numbers from the unit interval with the maximum  minimum < 1/3 is: And we are done!
The only way you will understand what I have done there is if you buy my brain. It is on auction on ebay for 19.95. I had it removed right before I wrote post #1.
The sad truth about the above tremendous solution is that it is only true for one set of numbers. Any attempt to generalize that method to any other number of points and any other distance fails miserably.
hi bobbym,
Hi Bob;
Good morning bobbym,
This question popped up in another thread. The question was well answered by TheDude over there so the only purpose of this post is to show that a nice neat drawing can provide much insight into solving a problem.
The expected value of the maximum point out of 3 points is 3 / 4 but that is not important. I placed the maximum there, just as a start. Figure 2. chance of one point getting in the interval. For two points to get in, For anywhere we place a point, p the probability of the other two getting within the interval is, When the maximum is at 1 / 3 or less, figure 1, then the probability that both points will be inside it is 1 this can happen 1 / 27 = ( 1/3 *1 / 3 *1/3 )of the time. We hold on to this number. Since we want the probability for any point p that the two points ( one will be the minimum ) will be inside the interval we just have to sum for all the points p. This is done by integration. So the answer is 7 / 27 which agrees with simulations. Drawings done with geogebra and screen captor. 