In some televised sports, such as downhill skiing, individual competitors take turns for the best time on a course. After each competitor the announcers will give the current standing. So for example, the first person will always be announced as ranked 1 (though this will likely change), the second person will be announced as either 1 or 2, and so on.
By the end of the competition everyone's final rank will be the same or higher.
a) In an event with n competitors, what is the expected difference between the first announced and final rank of competitor that goes x^{th}?
b) What is the expected average difference for all n competitors?
Note: the order of the competitors is random.
For any competitor, their final rank could be anywhere from 1 to n; on average it will be (n+1)/2.
The xth competitor's first announced rank could be anywhere from 1 to x; on average it will be (x+1)/2.
So the expected difference between the first announced and final rank of the competitor that goes xth will be (n+1)/2  (x+1)/2 = (nx)/2.
The expected average difference for all n competitors would be the sum of (nx)/2 from x = 1 to n, divided by n. So:
[(n1)/2 + (n2)/2 + (n3)/2 + ... + (nn)/2] / n
The numerator will be n*n  the nth triangle number, which simplifies to (n^2  n)/4. Dividing by n to get the average we get (n1)/4.

Posted by tomarken
on 20220223 10:31:21 