Ornaments to the game

It’s surprising to pick up the Age and find out, twice in a week, that talented AFL players are, in fact, wall ornaments. But, that’s what Matthew Lloyd would have us believe about Brett Deledio, and Matt Murnane about Jeff Garlett: they’re barometers.

I’m sure they mean well. Something like, “when Deledio’s playing well, the Tigers are too”, or “when the Blues win, Garlett usually kicks a bag”.

Well, which is it? Do the teams influence the players, or vice-versa? Murnane has a go at answering this:

‘It is an interesting thought for Blues supporters to ponder – do Garlett and Betts play well because Carlton are winning, or do the Blues win because Garlett and Betts are playing well? A group of former Carlton greats posed that question this week said the answer probably was somewhere in the middle. It’s the “chicken and the egg”, so to speak.’

So I guess that’s vice-versa-versa-vice. Well done to the group of Carlton greats for working it out for us. What they may be saying is that there is a feedback loop between the teams and players… which, possibly, makes Lloyd’s and Murnane’s articles redundant (potential headline: Players excels in team, team wins… Player even excels more! Team wins by even more! … continue until you run out of exclamation marks). Continue reading

Kennett’s curse in a neat table

In my last post I talked about the ridiculously complex formula for generating Kennett probabilities. To save you the trouble, here are its outputs for an 11-match streak, within a range of win probabilities and quantities of matches. Requests for more values welcome!

kennett

How often will an 11-match Kennett curse occur for given win probabilities and match quantities? With current curse highlighted in red.

The curse of poor maths skills (oh, all right… Kennett’s curse)

(Update: you can see a table of Kennett curse calculations here.)

When Geelong beat Hawthorn for the eleventh straight time last weekend

When Hawthorn beat Geelong in the 2013 preliminary final, ending a run of eleven straight losses to the Cats, salivating journos had the opportunity to rehash everything they’d already written about Kennett’s curse, on Friday, Thursday and every other day during the week. But of all the things written about Kennett’s curse, the one I’m stunned not to have read is: E(X) = (1 – p^r) / (p^r.(1 – p))

This roughly translates as: if two teams play a whole bunch of football matches, how many would you expect them to play before one team suffers from X: Kennett’s curse?

The answer, with the curse at 11 games, assuming that both teams have an equal chance of winning each match, is 4094 – and Geelong and Hawthorn have only played 152.

The key assumption, though, is “equal chance”. If the Geelong edge is as little as 55-45 – good enough to win 11 games out of every 20 – the calculation for E(Kennett) drops to just 1592. (If Hawthorn has the edge, it blows out to 11864… yikes.)

Surprisingly, providing a general formula for the probability of a string of wins within a longer series of matches isn’t easy, but you can do it using a ridiculously complex formula. Using it for a string of 11 wins within 150 matches, at a fifty-fifty chance of winning for each team: Kennett’s curse will occur 3.3% of the time. (Update: you can see a table of Kennett curse calculations here.)

A quick and handy way to work out the probability of winning streaks within match sequences (K wins in a row within N matches, probability of a win is p, of a loss, q).

A quick and handy way to work out the probability of winning streaks within match sequences (K wins in a row within N matches, probability of a win is p, of a loss, q).

But hasn’t this “experiment” of teams playing sequences of matches against each other been performed many more times than just now, with Geelong and Hawthorn? Continue reading