Probability of lottery problem

Equal, no matter who draws first, it is fair.

use a general case to prove it. Suppose there are always n lots, and m of them are "middle". The chance for the first person to draw is obviously m/n. We randomly select two of the n labels in sequence, and there are n(n-1) methods in one * *, which is our total sample space. In these arrangements, to ensure that the second person wins the lottery, he has m kinds of drawing methods.

In this way, the first person can choose from the remaining n-1 lots at will, so there are m(n-1) ways to ensure that the second person draws. So the "probability of the second person drawing", that is, m(n-1)/n(n-1), is still equal to m/n.

The order of drawing lots has nothing to do with the results

Using similar methods can prove that everyone's chances of winning lots are m/n from now on. In fact, there is a simpler idea for this problem. No matter how these people draw lots, the result they finally draw is nothing more than a permutation and combination of n lots.

In this permutation and combination, no position is more special than others, so the possibility of winning lots in each position must be equal. Lottery selection is a fair selection method, and the lottery sequence will not affect the winning probability if the results are not announced.