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 xth?
b) What is the expected average difference for all n competitors?
Note: the order of the competitors is random.
a) The final rank of course will be x. The initial rank depends on how many of those with lower rank precede him. There are x-1 whose ultimate rank are lower out of the n-1 competitors to person x.
For example, if x = 3, there are 2 competitors with lower numbered rank than him. Both can appear before him, one before and one after, or both after. Each has probability 1/3. If both appear befor him, the required difference = 0; one before and one after, the difference is 1; if both after, the difference is 2.
x prob diff
1 1 0
2 1/2 1
1/2 0
3 1/3 0
1/3 1
1/3 2
...
In all instances there are equal probabilities of starting zero, 1, 2, etc up to x-1 positions away and the average is (x-1)/2.
To see this better, remember we can ignore the placement in time of those with higher rank number. Among those with rank x or a lower number, person X has an equal likelihood of being the first, second, ..., xth.
b) Example for n=5:
0
1/2
1
3/2
2
The average is 1 or half of (n-1)/2, i.e. (n-1)/4.
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Posted by Charlie
on 2022-02-23 10:14:42 |
solution
