by shanen » Wed Jan 13, 2021 8:00 am
The comic is
https://www.smbc-comics.com/comic/percent
The mathematics is called the Secretary Problem.
https://en.wikipedia.org/wiki/Secretary_problem for reference. I was just looking at it because I wanted to figure out the probability that you wind up with #2 or #3 or some other lower ranked choice. Obviously #2 is much less than 37%, but how much less?
There are two solution approaches. I can't remember the direct analytic solution, but the other approach is based on permutations. For example, it is (if I remember correctly) also true that the #1 choice is within the first n/e choices 1/e percent of the time (again about 37%), in which case you are screwed and cannot get #1 since you discarded those candidates. that leaves 63% to be tested, but you subtract the 37% for success at #1, which leaves about 26%. The cases in that 26% where you are going to get the #2 option are the permutations where #2 was not in the first n/e but appears before #1 in the remainder. Basically the same for #3 and down.
So does anyone know the solution?
The comic is https://www.smbc-comics.com/comic/percent
The mathematics is called the Secretary Problem. https://en.wikipedia.org/wiki/Secretary_problem for reference. I was just looking at it because I wanted to figure out the probability that you wind up with #2 or #3 or some other lower ranked choice. Obviously #2 is much less than 37%, but how much less?
There are two solution approaches. I can't remember the direct analytic solution, but the other approach is based on permutations. For example, it is (if I remember correctly) also true that the #1 choice is within the first n/e choices 1/e percent of the time (again about 37%), in which case you are screwed and cannot get #1 since you discarded those candidates. that leaves 63% to be tested, but you subtract the 37% for success at #1, which leaves about 26%. The cases in that 26% where you are going to get the #2 option are the permutations where #2 was not in the first n/e but appears before #1 in the remainder. Basically the same for #3 and down.
So does anyone know the solution?