Friday, 4 January 2013

Predicting Countdown's 30th Anniversary Tournament (Part 2)

This post summarizes the findings of my Countdown 30th anniversary tournament predictions, the background to which can be found here. Below are the top 10 favourites, complete with their online ratings and their rank (out of 41) of all those competing.

Name Rating Rank Win %
Conor Travers 2099 2 20%
Innis Carson 2100 1 15%
Jack Worsley 1992 5 12%
Chris Davies 1998 4 10%
Jack Hurst 2015 3 9%
Steven Briers 1960 8 8%
Kirk Bevins 1988 6 6%
Mark Deeks 1964 7 5%
Jon O'Neill 1847 13 3%
Edward McCullagh 1873 10 2%

Conor Travers and Innis Carson are the top two ranked players in the field, and the top two favourites for the title. Perhaps surprisingly, Travers is 5% more likely to win than Carson, despite their near-identical ratings and Travers having to play an extra game (having been drawn into a preliminary match). This reflects the challenge facing Carson during the first two rounds, where he faces the number 8 and (probably) number 6 seeds before meeting number 2 seed Travers in the quarter-final.

Full results can be found here where, should you feel so inclined, you can play around with the ratings and see how things change.

Predicting Countdown's 30th Anniversary Tournament (Part 1)

To celebrate its 30th anniversary Countdown is hosting a very special tournament. 41 past champions have been invited back for a battle royale of letters and numbers. With such a high quality field, picking a likely winner is no easy task, and so I thought I'd take a statistical approach to predicting who might come out on top. If you just want to see the results, I've put them in a separate post, but if you're interested in some of the stats behind them read on...

Before I could start I needed some measure of how good each player is. Fortunately, a lot of them play a very similar game at apterous.org, which calculates a rating based on their online play (you may be familiar with similar systems in, for example, the chess world). From here we get ratings for 28 of the 41 players, which is good, but still leaves us with some work to do.


One of the competitors - series 65 champion Graeme Cole - has produced some statistics for all those taking part. (Along with a handy diagram of the tournament structure.) One of these stats will serve as a proxy measure of ability: 'max percentage'. This is the proportion of the total possible points available players actually scored during their televised appearances. Obviously there are lots of reasons why these data aren't ideal, but they're the best we've got, and for those players who do play online they correlate pretty well with their ratings. Fitting a simple linear regression model of ranking against max percentage (which worked surprisingly well) we can then estimate ratings for those players who don't play online. The figure (click for a bigger version) summarizes the relationship: it's fairly noisy but good enough for a bit of fun.

Now that we have an estimate of every player's skill, the next challenge is estimating the probability of, say, a player with a rating of 1600 beating a  player with a rating of 1500. Once again, apterous data lend a hand, with tens of thousands of recorded games giving an insight into just this sort of question. This time, a logistic regression model does the trick, allowing us to estimate the relationship between the difference in ratings between two players and the probability that each player will win. For example, a player who is rated 100 points higher than their opponent would be expected to win about 63% of the time, shooting up to 93% for a 500 point advantage.

From here, it's relatively straightforward to calculate the probability that each player will be crowned champion. To find out what they are just go to part 2.

Sunday, 30 December 2012

From the archives: goalfest in the Premier League

Yesterday evening saw one of the highest scoring English Premier League matches of all time. Arsenal's 7-3 drubbing of Newcastle joins just three other 10-goal games, although Portsmouth's 7-4 victory over Reading in 2007 remains the outright claimant for this particular accolade.

This was just one part, however, of a Saturday chock-full of goals, with a total of 35 scored across the eight matches played. This put me in mind of an article I wrote for Significance early in 2011 about an even more extraordinary day of football. Back then, Arsenal and Newcastle were at it again, with the latter's stunning four-goal comeback contributing to a whopping 41 scored across that day's eight Premier League games. In the article I used this as an excuse to show off the Poisson distribution, demonstrating how goals scored in football matches can be modeled surprisingly well by what is ultimately a (fairly simple) mathematical formula.

The remaining two matches of that particular weekend of football only produced two more goals, bringing the total for a complete 10 match 'round' of Premier League fixtures to 43. Based on the Poisson distribution (and assuming an average of 2.6 goals per game) I estimated there was a roughly 1 in 720 chance of seeing at least that many goals in a set of 10 games. This weekend's football was almost - but not quite - as remarkable, with six goals today bringing us to a total of 41 across 10 matches. Based on the same theory, this works out to a 1 in 250 occurrence.

Thursday, 20 December 2012

Christmas games theory

Through a coincidence almost as implausible as the Champions League draw, today also sees a new article of mine go up on Significance. The idea was simple: find interesting stats/maths things to say about popular Christmas-time board games. Not making the cut was a discussion of the inevitable unfairness of dice (and an excuse to talk about Awesome Dice Blog's 20,000 roll experiment to compare two manufacturers), and how to use the Markov chain nature of a game of Snakes and Ladders to estimate how long it will take you to finish (40 rolls in the MB version, apparently).

What are the chances: Champions League draw exact repeat of rehearsal

The draw for the UEFA Champions League knockout stage took place earlier today, but it wasn't just the prospect of some truly mouthwatering ties making the headlines.

Observant eyes spotted a seemingly spectacular coincidence: today's draw was a near-identical repeat of that produced during Wednesday's rehearsal. While the order the ties came out was different, every single match featured the same pair of teams. So what were the chances of that?

It seems almost unbelievably unlikely, and an 'ESPN statistician' apparently gets an answer of roughly 1 in 2 million. On face value this makes sense: there are about 2 million possible ways of drawing 8 pairs of matches between 16 teams (the first team has 15 opponents to 'choose' from, then once that match is decided the next team has 13 opponents to choose from, and so on, giving 15 x 13 x 11 x 9 x 7 x 5 x 3 = 2 million (ish) possibilities). Unfortunately, this overlooks a large number of factors that drastically reduce the number of possible matches.

First of all, those teams who qualified as group winners in the previous stage of the competition can only be drawn against teams who qualified as group runners-up. With 8 teams in each 'half' of the draw (so to speak), this immediately drops us down to 8 factorial, or 40,320 possible matches. On top of this, however, no team can be drawn against the other team who qualified from their group, or even a team from the same football association. With 2 Spanish, 1 English, and 1 Italian team on each half of the draw, the number of 'valid' draws becomes even smaller.

The upshot of all of this (and an admittedly lazy brute-force approach) is that there were just 5,463 possible draws that could have been made that satisfied all of these rules, giving chances of two identical draws in a row of about 0.02%. That's still pretty staggering, but nowhere near the 1 in 2 million we started off with.

Update: It has been pointed out to me that while there are 5,463 possible draws, due to the mechanics of the draw process itself (the specifics of which I was not aware of at the time of writing) not all draws were equally likely. However, the different draws do still have very similar probabilities, and there is nothing obviously special about this particular combination of fixtures. More on this to come.

Monday, 3 December 2012

Tuesday, 6 November 2012

Winning the White House (the easy way)

In a bid to create a more easily accessible list of my articles for the Significance website, I'm going to start updating this thing with links to them (maybe with some bonus content if there is anything I couldn't fit in).

It's (US) election day, so here's an article about how to win the Presidency with as small a share of the popular vote as possible. In the process of doing this, I tried a few other things that I didn't think quite made the cut.

After dealing with some trivial(ish) cases, I base my calculations in the article on turnout at the 2008 election. However, I thought I'd see what happens if I assume every member of the electorate voted. Unsurprisingly, not much changes: the minimum popular vote share required increases only a little to 22.3%.

I also looked at the analogous problem for becoming Prime Minister (assuming I had somehow persuaded a major political party to appoint me their leader). The maths is a lot more straightforward for the UK, as we just have one parliamentary seat per constituency (so none of this electoral college business). With 650 seats available, I'd simply have to win the 326 constituencies with the lowest turnout to form a (very) minority government. Admittedly, our politics are rather more complex than the two-party system stateside, but if we continue to assume 50% of the vote is required to win a constituency, then at the last general election a party could have won with just 22.1% of the national vote (the Liberal Democrats, by comparison, received 23%).