One of my favorite web sites, FiveThirtyEight.com, runs a weekly feature by Oliver Roeder called the Riddler, which each week poses a couple of interesting (and usually difficult) questions in math, logic, and probability.

This week’s “Riddler Classic” question concerns tournament design. Given a competition in which the better player wins two-thirds of the time, and that you only care about maximizing the probability that that best player wins, how should your construct a blind-draw tournament with these rather severe constraints: four entrants and four total games; and five entrants and five total games.

(Note that this success criterion is the same as tourneygeek’s earliest measure of fairness (C). So we can relate this challenge to the frequent discussions of fairness (C) elsewhere on this site.)

Now, I’m unable to imagine any competition in which the better player wins two-thirds of the time regardless of the size of the skill differential between the two players. The best player ought to beat the worst player rather more often than they beat the next best. I can see why the Riddler doesn’t want to put its readers to the trouble of using a more realistic match model. But since I have one ready to use, I’m going to seek a solution using my simulator, in which skill levels are handled rather better. In deference to the question as posed, however, I’ll use a skill parameter of 2.4, which is about what’s required to give the better of any two randomly-chosen players a two-thirds chance to prevail.

My solution is below the fold – before looking at it, you might want to give it a shot yourself.

**I was dead wrong! For the correct answer (or at least a better one), scroll down to the comment of Donald the Potholer. His “Page Ladder” brackets improve on the ones I found. For the 4/4 case he succeeds 49% of the time, and his 5/5 bracket succeeds 44.8%.**

I’ll rewrite this post with new analysis after the official results have been announced.

Continue reading “Designing Tourneys for the Riddler, Corrected!”