March 22nd, 2010, 9:52 pm
not sure about the exact "graphic way" DevonFangs has in mind, but one geometrical interpretation of the problem is to calculate the area of a strip along the diagonal in a unit square (defined by |y-x|<=w, where w is the waiting time as a fraction of the total time. w=1/6 here).one way to get the area is to subtract two triangles from the square which gives 1-2*(1-w)^2/2=2*w-w^2. w=1/6 results 11/36 as FritzJacob gave.[one could generalize to n people, all waiting for w, and consider the volume of a region along the diagonal of the hypercube defined by |max-min|<=w. the result is n*w^(n-1)-(n-1)*w^n]if the waiting time for each person could be different, the strip is not symmetric w.r.t. the diagonal any more. still simple for two person, but very messy for the general case.for dunrewpp's question, FritzJacob's answer is correct, and one could easily see the geometrical meaning of each term in p. the final result will be slightly larger than 1/6*2*60=20mins (exact result is w=[3/2-2/sqrt(3)]*60~20.7min), as w^2 is convex, and has a slightly larger average than the average squared.----- ----- ----- ----- -----there is a very nice discussion of the general case
Last edited by
wileysw on March 23rd, 2010, 11:00 pm, edited 1 time in total.