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?
For the individual case, the total of the places for n=1, 2, 3, ... are
n total
1 1 (1)
2 4 (1 + (1+2))
3 10 (1 + (1+2) + (1+2+3))
4 20 etc.
5 35
6 56
7 84
8 120
9 165
10 220
11 286
12 364
13 455
14 560
15 680
16 816
17 969
18 1140
19 1330
20 1540
21 1771
22 2024
23 2300
24 2600
25 2925
The OEIS identifies this as Tetrahedral numbers C(n+2,3) = n*(n+1)*(n+2)/6.
To get the average, they need to be divided by the nth triangular number, n(n+1)/2. That counts each announced rank as just one rank.
That division results in (n+2)/3.
n tot average
1 1 1.0000000000
2 4 1.3333333333
3 10 1.6666666667
4 20 2.0000000000
5 35 2.3333333333
6 56 2.6666666667
7 84 3.0000000000
8 120 3.3333333333
9 165 3.6666666667
10 220 4.0000000000
11 286 4.3333333333
12 364 4.6666666667
13 455 5.0000000000
14 560 5.3333333333
15 680 5.6666666667
16 816 6.0000000000
17 969 6.3333333333
18 1140 6.6666666667
19 1330 7.0000000000
20 1540 7.3333333333
21 1771 7.6666666667
22 2024 8.0000000000
23 2300 8.3333333333
24 2600 8.6666666667
25 2925 9.0000000000
In the 2-at-a-time version, the place totals and the numbers of them go:
n tot number of ranks
2 3 2
4 13 6
6 34 12
8 70 20
10 125 30
12 203 42
14 308 56
16 444 72
18 615 90
20 825 110
22 1078 132
24 1378 156
26 1729 182
28 2135 210
30 2600 240
32 3128 272
34 3723 306
36 4389 342
38 5130 380
40 5950 420
The OEIS identifies the totals as n*(n+1)*(4*n+5)/6, but the n here is really our n/2, while the number of the ranks we can see is 2 times the n/2 triangular number, or 2*(n/2)*(n/2 + 1)/2 = (n/2)*(n/2 + 1).
The former is divided by the latter to get the average:
n tot number average
2 3 2 1.5000000000
4 13 6 2.1666666667
6 34 12 2.8333333333
8 70 20 3.5000000000
10 125 30 4.1666666667
12 203 42 4.8333333333
14 308 56 5.5000000000
16 444 72 6.1666666667
18 615 90 6.8333333333
20 825 110 7.5000000000
22 1078 132 8.1666666667
24 1378 156 8.8333333333
26 1729 182 9.5000000000
28 2135 210 10.1666666667
30 2600 240 10.8333333333
32 3128 272 11.5000000000
34 3723 306 12.1666666667
36 4389 342 12.8333333333
38 5130 380 13.5000000000
40 5950 420 14.1666666667
By inspection, the average seems to be 5/6 + n/3, which is verified if we consider
(n*(n+1)*(4*n+5)/6)/((n)*(n + 1))
with the understanding we really want the result for n/2. The simplification
5/6 + (2*n)/3
when n/2 is substituted for n, agrees with the observation.
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Posted by Charlie
on 2022-02-16 11:20:12 |
solution
