Olympic medal table - epilogue (more maths)

This is my last word (for now) on the Olympic medal table problem. One or two people have rightly pointed out to me that the choice of three medals is somewhat arbitrary, and that really we should be counting all the placings when trying to rank countries. A country whose athletes came fourth in everything would surely merit some recognition, after all!

One problem with increasing the number of "medals" is that differences in the sizes of events begin to become more apparent. Coming tenth out of 100 participants is clearly more of an achievement than coming tenth out of ten! But setting that problem aside, what happens to my medal weighting system, which is based on an average of plausible weighting systems?

With three medals, one way of writing the relative scores that helps us see a pattern is as follows:

Gold   = 1 + 1/2 + 1/3 = 11/6
Silver =     1/2 + 1/3 =  5/6
Bronze =           1/3 =  2/6

and in my original article I multiplied these all by 6 to get my whole-number points system 11, 5 and 2.

So if we had four medals, it would be

Gold   = 1 + 1/2 + 1/3 + 1/4 = 25/12
Silver =     1/2 + 1/3 + 1/4 = 13/12
Bronze =           1/3 + 1/4 =  7/12
Iron   =                 1/4 =  3/12

so 25, 13, 7 and 3 points.

And with ten medals (I'll give up trying to rank metals by value!), leaving out the calculations, we end up with this rather unwieldy set of scores: 7381, 4861, 3601, 2761, 2131, 1627, 1207, 847, 532, 252.

It is tempting to think that the relative scores might tend to some kind of pattern as we keep adding medals.  We could even ask ourselves what would happen if we had an infinite number of medals!  The resulting "universal" scoring system could then be applied to real-world events by just using the top few scores from the list.

Unfortunately we get into problems with an infinite number of medals.  The gold score would be

1 + 1/2 + 1/3 + 1/4 + 1/5 + ...

also written as

Mathematics is sometimes kind to us and allows us to calculate these infinite sums, but sadly this one doesn't work - there is no limit as you add more terms to the sum.  You have to stop somewhere, like I did at 10 in the example above.  Some insight into the finite sum can be gained by replacing the discrete sum by a continuous sum known as the integral, which in this case happens to have an easy solution.  The right hand expression above can be replaced by

So having chosen our value of n, the number of medals, we have an approximation to the gold medal score (I'll call it S(1) meaning score for 1st place) which is

Then if we go back to our patterns to work out the scores for the other medals, we get

or, cutting out the working:

This gives a reasonable approximation given the arbitrary nature of all the original assumptions.  It could probably be improved by tweaking the approximation of the discrete sum by the continuous integral (I just placed each term of the sum in the middle of one unit's worth of the integral).  Oh, so it's probably not the last word after all...