Two running teams take part in competitions on the same course, but their competitions differ because of the number of teams entered in each competition.

Results are based solely on position - times are not taken.

Competition 1 has 7 teams of 8 runners Each team enters 2 of its runners in each of 4 races Points are awarded on the basis of 1 for first place down to 14 for fourteenth place The points for all 4 races are added together. The winning team had a total of 40 points.

Competition 2 has 5 teams of 8 runners Each team enters 3 of its runners in each of two races (15 competitors in each) and a further 2 runners in a third race (10 competitors in this one). Points are awarded in each race on a similar basis to competition 1 (2 races award from 1 to 15 and the other, 1 to 10). The winning team has a total of 46 points

Statistically, which is the better team?

I'm not a statistician, but my logic lead me initially to compare each team's performance with what I would (probably incorrectly) call the mean number of points in each competition (average of the most and least number of points the team could score). However this would seem disadvantage the team in competition 2, because their best possible points score is 15 versus 12 for team 1. And the "mean" numbers are in any case similar. I did think about comparing each team's performance against the best case and worst case scenarios in their respective competitions and averaging those scores? The other thought is that I should somehow compare each runner's performance against the best & worst case?

In reality, I'm a bit stuck. Help please!

  • $\begingroup$ You have a contradiction - is a higher score better or lower? $\endgroup$ – Peter Ellis Dec 3 '12 at 19:48

I think you may be making it too complicated. You have two different competitions, with no cross-over of teams. Both systems are basically meritocratic.

The winner of Competition 1 beat six teams, and the winner of Competition 2 beat four teams. So lacking any other prior information the winner of Competition 1 is probably superior to the winner of Competition 2. But this is only a slight lead.

A subtler approach is possible but you'd need more data than you've given (and/or got) - eg the exact breakdown of scores in the second competition so you can scale some of the scores based on 1-10 and some on 1-15. I'm not sure that such an approach would be superior to my simple method above, either. In the end a huge source of randomness is how the pooling was allocated, and with the lack of a direct competition between teams from the two pools you will struggle to benchmark one against the other.

  • $\begingroup$ Sorry, the lower the score the better. Pooling in each competition was allocated using an individual time trial run before the mass start race. The fastest two from each team went in the "A" race, the next two in "B" etc. The race with only 10 contained the slowest 2 competitors from each team in competition 2. I suspect you're probably right that I've made it a but complex. I do have points breakdowns for all teams in the competition and a quick glance shows the winner of competition 1 to be more consistent, with no winners, but no significant poor performances either. Thanks for your help. $\endgroup$ – Will Newton Dec 3 '12 at 20:07
  • $\begingroup$ That pooling information is critical. But its still not clear, is that pooling within each competition, or the decision as to which competition people went to? $\endgroup$ – Peter Ellis Dec 3 '12 at 20:15
  • $\begingroup$ The pooling is within each competition itself. All riders who raced in competition one did an individual time trial which was used to place them in a race in which their potential closest competition from each of the other teams would be racing - obviously this is not perfect, because a team of 8 very fast runners might outclass all but the other runners in the A-race where other teams were of mixed ability. Ideally morning and afternoon competitions would have had identical numbers. While this would still have raised the issues of equivalence, I wouldn't have needed the maths. $\endgroup$ – Will Newton Dec 4 '12 at 9:20

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