Posted 14 Mar 2014 Get two random numbers between 0 and 1. Subtract the smaller from the bigger. What is the probability that the result is <0.3? Is it bigger than the probability it is > or equal to 0.3? For what number the probability for both cases is the same? 0 Share this post Link to post Share on other sites

0 Posted 14 Mar 2014 The probability that the difference is <0.3 is 72%. Equal probability is achieved for the difference of (3-sqrt(5))/4 or approx. 0.191 0 Share this post Link to post Share on other sites

0 Posted 15 Mar 2014 How'd you come to that answer?If you pick one number N, then the area from N - 0.3 to N + 0.3 will be 0.6, so the probability that a second point M would fall within that range (to make the difference between N and M be <0.3) can be no higher than 60%. 0 Share this post Link to post Share on other sites

0 Posted 16 Mar 2014 Probability is very close to 51%. Mean length is 1/3. Median length is .2925. Here are 2000 random distances plotted in ascending order. 0 Share this post Link to post Share on other sites

0 Posted 16 Mar 2014 (edited) Consider placing one point N in the range from 0 to 1, and then placing a second point M in the range from 0 to 1.If N is in the range from 0.3 to 0.7, then there will be a total area of 0.6 around N in which M could be placed to be within 0.3 units of N.Probability (N is within 0.3 to 0.7) and (M is within 0.3 units of N, given that N is within 0.3 to 0.7)= 4/10 x 6/10 = 24/100If N is in the range from 0 to 0.3, or 0.7 to 1.0, then some of the area within 0.3 units of N will be outside of the 0-1 range and inaccessible to M. At either extreme (at 0 or at 1.0), there would be 0.3 units in which M could fall that are within 0.3 units of N. That accessible area within 0.3 units of N increases linearly from 0.3 to 0.6 as N moves from 0 to 0.3 (or as N moves from 1.0 to 0.7). If N is within 0 to 0.3 or 0.7 to 1.0, then on average, M would have 0.45 units of area that would put it within 0.3 units of N.Probability (N is within either 0 to 0.3 or 0.7 to 1) and (M is within 0.3 units of N)= 6/10 x 4.5/10 = 27/100The sum of those two is 51/100. So slightly higher than 50/50 odds that M is within 0.3 units of N.Now to find the distance D for which the odds are 50/50. Those equations would become:Probability (N is not within D of 0 or 1) and (M is within D of N)= (1 - 2D) x 2D= 2D - 4D^{2}Probability (N is within D of 0 or 1) and (M is within D of N)= 2D x 1.5D= 3D^{2}The sum of those two is2D - D^{2}Confirming this works with the previous answer, 2 x 0.3 - 0.3^{2} = 0.51Solving for a 50/50 chance:2D - D^{2} = 0.5D^{2} - 2D + 0.5 = 0Using the quadratic formulaD = [2 +/- sqrt(4 - 4x1x0.5)] / 2= [2 +/- sqrt(2)] / 2The + solution would produce a number greater than one which would not be meaningful for this problem, so the - solution is the only one that makes senseD = [2 - sqrt(2)] / 2 = 1 - sqrt(2)/2roughly 0.29289 Edited 16 Mar 2014 by plasmid 1 Share this post Link to post Share on other sites

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