There is a common situation in life that requires probability knowledge: less things should be distributed to more people, such as giving three movie tickets to five people. Because the points are not enough, we have to draw lots to allocate them. The obvious question is: are the chances of winning the lottery in the first lottery and the last lottery equal? The answer is: equality, no matter who smokes first, it is fair.
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.
The order of drawing lots has nothing to do with the result.
In a similar way, it can be proved that from now on, everyone's chance of winning the lottery is m/n.
In fact, this problem has a simpler idea. No matter how these people draw lots, the final result is nothing more than the arrangement and combination of n lots. In this arrangement, no position is more special than other positions, so the possibility of winning the ticket in each position must be equal.