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Announced rank average (Posted on 2022-02-16) Difficulty: 3 of 5
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?

No Solution Yet Submitted by Jer    
Rating: 4.0000 (1 votes)

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Solution solution | Comment 1 of 3
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.

  Posted by Charlie on 2022-02-16 11:20:12
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