The last post discussed the two alternatives for the Consolation bracket needed for an upcoming tournament. But in addition to a Consolation, my friend also needs a Last Chance bracket. Third brackets are almost always rather difficult brackets to build, and they depend on the second brackets they’re attached to in the same way that second brackets relate to first brackets.
In the last post, I started showing how to build a bracket for the consolation in my friend’s tourney, which he expects to draw up to 48 entrants. In showing the method, I digressed to show how it generated the more familiar 64 brackets. Now let’s return to the problem of building the 48.
Along the way, I’ll introduce a new half maxim of tourney design: be squirrelly early.
A friend asked me for a 48 bracket for a tournament he’ll be running next month. The tourney will have a main flight, a consolation, and a last chance. He wants to pay two places in the main, and two more in the consolation.
Rather than just giving you the bracket I came up with, let me show you the process by which the bracket was built. The will take a few posts. For the first one, I’ll begin the process, and then digress to show how second brackets are built for a 64 main.
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.
A friend asked me for a specialized bracket for a backgammon tourney he’s running soon. He wants a 128 bracket that’s a full double elimination with a progressive consolation as a third bracket. There’s an unexpectedly nice one available.
An interesting design problem from a reader.
The tourney is for “Cold War”, a two-handed version of the famous board game Diplomacy. As many as 64 players are expected. The chief limitation is time – apparently even in its two-handed form, the game takes a long time to play. So minimizing the number of rounds is crucial.
A single elimination 64 bracket takes six rounds to play. A full double elimination can take as many as 13 rounds, but since time is at a premium, that can be pared down to 10 rounds. One round is saved by not having a recharge, and two more by shifting the lower bracket.
But an unusual feature of this particular game offers a third possible bracket structure. Each player can easily play two games at the same time! So perhaps we can give players a second chance by simply playing two separate 64 brackets at the same time, with a playoff between the two bracket winners as a seventh round if the two brackets are not won by the same person.
This is marvelously efficient. Each player gets to play until they lose twice, but you save at least three rounds, and maybe four if you would otherwise insist on a recharge, because a recharge is never needed.
How should such a bracket be drawn, and how does it compare with conventional single and double elimination brackets in the item of fairness (C)?
Here is another draft from my slowly-growing manuscript, Tourneygeek’s Guide to Tournaments. TGT Fairness
Only the first five pages, discussing fairness in general and fairness (A), are new, but because this now completes a first full chapter I’m also including the previously posted parts of the chapter than discuss fairness (B) and fairness (C).
As before, comments and criticisms are welcome.