100 people in a contest top 50 move on to the next round. I am currently in 20th place with 40 people already finished. What are my chances of staying in the top 50?

Answer 1

There are 100! ways that the 100 people can finish, and in 1/40 of these, you finish in each possible place. In particular, in 1/40 of these you finish in 20th. Thus, there are 100!/40 ways that you finish 20th

Then, constrained on your finishing 20th of 40, we need to add the number of ways you can finish between 20th and 50th

There are C(39,19) ways we can select the particular 19 ahead and 20 behind of those that have already finished.

Let's say you finish in pth place, p between 20 and 50.

We will be choosing p - 20 of the 60 who have not yet finished to also finish ahead, C(60,p - 20)

Then, there are p-1! ways to select the p-1 ahead of you and 100-p! ways to select the 100-p behind you.

Thus, the count is C(39,19)C(60,p-20) p-1! 100-p! p = 20 to 50

Dividing by 100!/40, we get

We add 40 * C(39,19)C(60,p-20) p-1! 100-p! /100! =

40 * C(39,19)*C(60,p-20)/(C(100,p)p)

Adding from 20 to 50, I got 0.580791678039851

I added these from p = 20 to 80 and am happy to say that it added to 1.

There are simpler ways to model this, and the SAT is probably using one of these, I would guess the following. It is equally likely for each of the people finishing to be ahead or behind you. Then, note that P(29 or less) equals P(31 or more), and P(29 or less) + P(30) + P(31 or more) equals 1, so

P(29 or less) = (1 - P(30))/2, and P(30 or less) = 1/2 + P(30)/2

P(30) = C(60,30)/2^60, so 1/2 + P(30)/2 = 1/2 + P(60,30)/2^61 = 0.551289086504285

Note the assumption that it is equally likely for each of the remaining to finish ahead or behind you, a faulty assumption, I would say.

You could also model this as there are 41 possible positions among the first 40 finishers that each of the 60 remaining could finish, ahead of person 1, between person i and person i+1, i = 1 to 39, and behind person 40. Then, 20/41 of the positions are ahead of you, so the probability that 30 or less finish ahead of you is, from the binomial distribution, P(X <= 30), p = 20/41, n = 60, which we get from Excel as binom.dist(30,60,20/41,1) = 0.625009335148428.

These two approximations bracket what I believe is the right answer.