I am reading "Probability and Computing" by M.Mitzenmacher, E.Upfal and I have problems understanding how the probability of comparison of two elements is calculated. Input: the list (y1,y2,...,YN) of numbers. We are looking for pivot element. Question: what is probability that two elements yi and yj (ji) will be compared? Answer (from book): yi and yj will be compared if either yi or yj will be selected as pivot in first draw from sequence (yi,yi+1,...,yj-1,yj). So the probablity is: 2/(y-i+1). The problem for me is initial claim: for example, picking up yi in the first draw from the whole list will cause the comparison with yj (and vice-versa) and the probability is 2/n. So, rather the "reverse" claim is true -- none of the (yi+1,...,yj-1) elements can be selected beforeyi or yj, but the "pool" size is not fixed (in first draw it is n for sure, but on the second it is smaller). Could someone please explain this, how the authors come up with such simplified conclusion? Thank you in advance