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.
For an event with n competitors there will then be a sequence of announced ranks. What is the expected average of these ranks?
In long track speed skating, 2n competitors go on the same track but in pairs. So the first pair will get announced ranks of 1 and 2, the next pair will have two ranks from {1,2,3,4}, and so on.
For an event with n pairs, what's the expected average of the ranks?
(In reply to
solution by Charlie)
I may be misunderstanding the question but I came up with something different for the first part (haven't extended to the 2nd part yet).
Take the case n = 2. The first person's announced rank will always be 1. The second person's announced rank will either be 1 or 2 (with equal probability, I'm assuming, though in real life that will often not be the case). So on average the second person's announced rank will be 1.5. Thus the average of both ranks, 1 and 1.5, is 1.25.
Similarly, a third competitor will be ranked 1, 2 or 3. On average it'll be 2. The average of all three ranks would be (1 + 1.5 + 2) / 3 = 1.5.
In general, then, the average of the ranks for n competitors would be (n+3)/4, not (n+2)/3.
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Posted by tomarken
on 2022-02-16 19:13:16 |
re: solution
