<t>QuoteOriginally posted by: quantystHi QuantR,After all, despite my 'unwarrented' assumption about uniformity and independence of all X(1), X(2), ..., on the same set {1,2, ...,N}, it turns out that although X(i) is dependent on all X(j) with j<i, we have P{X(i) < a} = a/N. To see this, begin with...