On 24.6.2011, at 1.56, [email protected] wrote: > In sports round robin tournaments where burial strategy is meaningless, what > is the best measure of > defeat strength? In other words, if you were going to use, say, beatpath to > find a total order of the teams > from best to worst after the conclusion of a tournament, would you use wv, > margins, or something else > to measure defeat strength? > > Let's think about basketball for example. Let W be the winning points and L > be the losing points from a > match between two teams. The margins defeat strength would be W-L. The wv > defeat strength would > be just W. Another possibility would be the point ratio W/L or equivalently > log(W) - log(L). Between > margins and the point ratio would be sqrt(W) - sqrt(L), which seems better > to me from an intuitive point > of view. > > What rational basis could we use to decide between these measures?
We should understand how the results of the winner selection algorithm relate to our real world needs. A good explanation and justification is what we need. The basic explanation behind margins is "number of required extra votes to win everyone else pairwise". For the proportion / ratio approach the corresponding explanation would be "proportion/ratio of required extra votes to win everyone else pairwise". If people think that margins is too much on one side and ratio is too much on the other side, then they might be happy with the sqrt approach. I don't have a good natural explanation available for it. Maybe there is one (a bit more complex than for the other two) (maybe below in your mail). The explanation for winning votes is also a bit more complex and a bit one-sided (since it does not count the defending votes). I think winning votes have some strategic defence flavour and implicit approval flavour in it (but others may have different viewpoints on what is the best explanation for winning votes). > > It seems like we want the defeat strength to reflect the likelihood that if > the match were repeated, the > winner would not change. Yes, that is one good approach. Margins and ratio talk about the size of opposition after some candidate has been elected. They only talk about pairwise oppositions in favour of the candidates that didn't win, one at a time. In real life one could also assume that once elected, there will be some extra support to the elected candidate (maybe loyal civil servants, opportunistic politicians, people who like to agree with the winner, people who support the democratic decision and agreed position of the society), and that would make the winner in practice a Condorcet winner afterwards even if during election that candidate was few votes short of being one. (Well, it is also possible that people would hate the winner and we would have to count the amount of additional problems after the election, but let's be positive this time.) Counting the number of opposition as a whole by measuring the number of voters that would have liked some of the other candidates more, or the number of candidates that were preferred to the winner would be nice sounding natural language explanations too, but can not be used because of the involved problems (both sincere opinion and strategy related). > > Suppose team A beats team B thirty-six to twenty-five, while team C beats > team D forty-nine to thirty-six. > > Which is the stronger defeat? If both teams had rematches, which of the > rematches would be more > likely to turn out the same? > > Note that the C defeat over D has a larger margin, but the A over B defeat > has the larger ratio. > > By the sqrt(W) - sqrt(L) measure, both defeat strengths come out the same. > > Here's a way to resolve it. > > Let N=W+L. Let p=W/N. Let q=L/N. Let sigma = sqrt(N*p*q). > > Measure defeat strength by S=(p-q)*sigma. > > In terms of W and L we have > > S=(W-L)*sqrt(W*L)/(W+L)^1.5 > > I don't have a calculator on me now, but somebody should check this out to > see which of the two defeat > strengths is stronger by this measure. I suspect that they are pretty close. > > What is the heuristic behind S=(p-q)*sigma? > > The Binomial distribution has standard deviation sigma = sqrt(N*p*q). > > From the point of view of the winning team the fraction p is the proportion > of favorable outcomes, while > the fraction q is the proportion of unfavorable outcomes. So (p-q) is the > difference in estimates of the > underlying fractions of favorable and unfavorable outcomes. We multiply this > estimate by the standard > deviation to take into account "sample size." > > If a real statistician, like Jobst were reading this, he (or she) could > refine this estimate, or at least give a > better explanation. > > Thoughts? This gets a bit complex from natural language (and regular politicians and voters) point of view but maybe one can find a solid mathematical (but still real life based) justification for some of the more fine-tuned approaches than margins and ratios. Juho > > Forest > ---- > Election-Methods mailing list - see http://electorama.com/em for list info ---- Election-Methods mailing list - see http://electorama.com/em for list info
