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Technical
The Best Player is not Guaranteed to Win the Knockout

On 19/04/2012 15:09, Jonathan Kirby wrote:

One is the player who happens to play best in the event. Hopefully we'll know at the end of the event who that is because they are world champion.

David Maugham argues:

I think that there is a degree of confusion amongst almost everyone involved in this discussion about one aspect of it. I think it deserves to be spelt out very clearly.

The knock-out format does not guarantee that the person who wins was the best* player.

Let me repeat that:

The winner of a knock-out is not guaranteed to be the best player.

Right, now, having got that out of the way. What does it imply?

Basically everyone playing in a knock-out (KO) has a chance of winning EVEN IF everyone plays to their level. If we want a format where the best player wins all (or more) of the time, then a KO is not a particularly good choice. But we accept it for a number of reasons, so given that we have to live with it, can we decide beforehand what the right percentages are for each player and ensure that the "luck of the draw" does not change these by a significant margin.

Let's take a simple 4 player example. But before we do, let's clear up another misunderstanding that seems to be prevalent. The ranking system is a just guess at a player's ability and, as such, a player has TWO grades. He has a ranking system grade, which we know and is recalculated after every game, but he also has a "real" grade, i.e. the level to which he is actually playing at. We do not (probably cannot) ever really know this (not least because it is almost infinitely variable). As a consequence we can use a number to represent a players "true" playing ability when doing simulations, but it should not be confused with his ranking system number.

Before I get to the example, there is a small point which probably deserves attention too. Every simulation like this relies on a couple of assumptions. The first is that a player plays at his level continuously throughout the tournament. The other (which applies to the ranking system too) is that ability is essentially transitive and follows a simple probability distribution (e.g. if A will beat B 66% of the time and B will beat C 66% of the time, then A must beat C 80% of the time)1.

So let's take 4 players A, B, C and D with real ability levels 2700, 2550, 2400 and 2250

Now there are 3 possible 4 player draws AvB/CvD, AvC/BvD, AvD/BvC

For each of these we can work out the probability of any of the players winning (because, if you remember, the best player (A in this case) does not win every time). In the first one, the calculations are:

A beats B then beats C after C beats D = 66.6% * (79.9% * 66.6%) = 35.5%

A beats B then beats D after D beats C = 66.6% * (88.8% * 33.4%) = 19.8%

So A's overall probability of winning is 35.5% + 19.8% = 55.2%

Similarly B, C and D's chances are:

B beats A then beats C after C beats D = 33.4% * (66.6% * 66.6%) = 14.8%

B beats A then beats D after D beats C = 33.4% * (79.9% * 33.4%) = 8.9%

So B's overall probability of winning is 14.8% + 8.9% = 23.7%

C beats D then beats A after A beats B = 66.6% * (20.1% * 66.6%) = 8.9%

C beats D then beats B after B beats A = 66.6% * (33.4% * 33.4%) = 7.4%

So C's overall probability of winning is 8.9% + 7.4% = 16.3%

D beats C then beats A after A beats B = 33.4% * (11.2% * 66.6%) = 2.5%

D beats C then beats B after B beats A = 33.4% * (20.1% * 33.4%) = 2.2%

So D's overall probability of winning is 2.5% + 2.2% = 4.7%

And just for confirmation, 55.2%+23.7%+16.3%+4.7% = 100% (with some rounding)

It's worth pointing out again, that D will win this particular draw with these players nearly 5% of the time despite being (and playing as) the worst player in the event.

Similarly we can work out the percentages for the other 2 draws:

AvC/BvD  A=56.8%, B=32.0%, C=8.0%, D=3.1%

AvD/BvC  A=63.1%, B=27.7%, C=8.4%, D=2.7%

As a sideline, this demonstrates at least part of the justification for seeding -  A's chance of winning varies between 55.2% and 63.1% purely based on "luck of the draw". Absolutely nothing to do with playing ability - an 8% swing that is just random. The larger the draw the bigger this random swing will be.

Anyhow, if we take these three draws, and average them, we get these four percentages:

A=58.4%, B=27.2%, C=10.9%, D=3.54%

It is my contention that these percentages are "fair" and we should be aiming, as far as possible, to use a draw such that the percentages in that draw are as that are as similar to them as we can.

The article at http://oxfordcroquet.com/tech/knockout/index.asp gives a better indication of why I think that the "traditional" seeded draw is, if not completely equivalent, at least within a reasonable tolerance of this "fair" draw (because it deals with the probabilities of an 8 player draw).

Now I fully accept that once we decide on a "fair" draw, we then have to decided what the best way of determining what the "real" playing ability is in order to put people in their fairest place in the draw. (This is where the blocks vs rankings debate comes in.) But it is, I think, worth separating the two different (albeit related) factors involved in performing seeding.

Dave

* Yes, I know that one of us here thinks that "win" automatically means "better" but I hope everyone else knows better.

----

See also discussions in Is a Knock-out Fair? and How the Draw Influences Outcome in Knock-Out Events

1. If t is the probability of A beating B and s is the probability of B beating C, then the probability, x, of A beating C is given by:

x = ts /(ts + (1 - t)(1 - s), e.g. 0.66*0.66/(0.66*0.66 + 0.33*0.33) = 0.8

 

Author: David Maugham
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Updated 28.i.16
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