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October 13[edit]

Fairest overall result[edit]

Imagine that there are several events where competitors achieve a score. However for each event, the maximum score achievable varies. What is the fairest way of getting an overall or average result?

One method is a percentage calculated as 100 * sum_of_scores / sum_of_maxima.

Another percentage would be calculated as 100 * mean( each_score / its_maximum).

There may be other methods. What is "fairest" - whatever fairest might mean? -- SGBailey (talk) 06:58, 13 October 2019 (UTC)[reply]

Oh yes. Consider each event as important as every other event. -- SGBailey (talk) 07:00, 13 October 2019 (UTC)[reply]

It's difficult to justify an answer without explicit criteria for what "fair" means, but if you want each event to have equal importance then you can eliminate the first option since it weights events according the the maximum points. For example if you multiplied the scores (and maximum) for one event by two then the first option would produce different values, and this seems contrary to what would be expected with equally important events. The second option does not change value under this operation. Some sporting events, such as the decathlon, have to find a balance between different events with widely varying ranges of scores. For the decathlon apparently the formula is hand tuned according to past data, but I gather you're looking for a one-size-fits-all type of solution. --RDBury (talk) 14:30, 13 October 2019 (UTC)[reply]

I would calculate Standard scores for each competitor for each event, then average these z values. Dbfirs 15:48, 13 October 2019 (UTC)[reply]

Thanks. I think I'll go with the second option then. -- SGBailey (talk) 19:44, 13 October 2019 (UTC)[reply]

FWIW: last year in my linear algebra class, I computed homework averages for 12 assignments both ways. Among 18 students, the largest gap between methods was under 2 percentage points in the homework average, far too small to matter when determining final letter grades. --JBL (talk) 20:39, 13 October 2019 (UTC)[reply]