We simply use a general situation to prove it. Suppose there are always n lots, of which m are "medium". The probability of the first person drawing is obviously m/n, so how to calculate the probability of the second person winning?
We know that there are n(n- 1) methods to randomly select two of the n signatures, which is our total sample space. In these arrangements, to ensure that the second person wins the lottery, he has m kinds of lottery methods; In this way, the first person can choose from the remaining n- 1, and then there are m(n- 1) ways to ensure that the second person can draw. Therefore, the "probability of the second person drawing" is m(n- 1)/n(n- 1), which is still equal to m/n.