An emergency blog post about chess, of which I know nothing. This is not meant as serious analysis; think of it more as “here are some topical maths ideas you can throw at your classes.”

So, I looked up the Elo ratings for the chess world championship players: in rapid chess, champion Magnus Carlsen had, going into the tiebreakers, a world-leading score of 2880; challenger Fabiano Caruana a top-ten score of 2789, 91 points behind. In the even faster blitz version, Carlsen’s lead was larger still: 2939 to 2767, a difference of 172 points. (In classic chess? Carlsen had the edge by just three points, 2835-2832.)

One nice thing about Elo ratings ((I suspect these would be an excellent topic for an A-level class.)) is, they give a (theoretical) prediction of how many points a player would be expected to score in a head-to-head match. Going strictly by a table on Wikipedia, we’d expect Carlsen to pick up somewhere between 0.6 and 0.7 points a game in rapid format, and somewhere between 0.7 and 0.8 per game in blitz.

Let’s simplify things.

As simplifying assumptions, I’ll ignore colour advantage and say any game has a 50-50 change of being a draw. Following the format as I understand it, we’ll play (up to) four matches of rapid — in which Carlsen gets 0.65 points on average; if there’s no winner, we’ll play two further matches of blitz, in which he gets 0.75; these are followed by “Armageddon”, which I won’t try to model; I shall simply split the draw probabilities 50-50 between the players.

So, what are the probabilities for any given match? To win p points on average in a game which is drawn half the time, your win probability would need to be p - 0.25. So, in a rapid game, let’s assume Carlsen wins 40% of the time, draws 50% and Caruana 10%.

… and complicate them again

The simplest way (at least, the simplest that I know of) to model the points distribution for such a setup is to use a probability generating function. If we work out $(0.4x + 0.5\sqrt{x} + 0.1)^4$, the $x^k$ coefficient in the resulting expansion is the probability Carlsen wins that many points over the course of the four games:



















That means, Caruana has about a 9.3% chance of beating Carlsen in Rapid, and 19.2% of taking it to a blitz playoff. Carlsen’s win probability is around 71.5%.

Armageddon wouldn’t be the end of the world

It doesn’t get much better for Caruana: in blitz, he never wins — it’s 50% Carlsen, 50% draw. It’s not hopeless, though: if he were to draw both playoff matches, he’d still have a 25% chance of winning in Armageddon. The chances of that happening (given we’re in the blitz situation) are one-in-sixteen, a little under 6.3% — and Carlsen’s 93.7%.

So, in terms of a playoff, Carlsen’s win probability would be something around 89% altogether — a prohibitive favourite.

So what about Game 12?

His decision in game 12 to offer a draw while in a moderately commanding position attracted a lot of criticism, but in this light, the decision makes a bit more sense. If he assessed his chances from that position as (say) 30% win, 60% draw and 10% loss, it makes perfect sense to offer a draw and go for the playoff.

Why is this? This would give him a 30% chance of winning on the day, plus about 53% of drawing and then winning the playoff. Caruana would have about one chance in six of gaining the title in these circumstances. By offering the draw, Carlsen reduced Caruana’s chances significantly, to one in nine.

Now, this analysis is obviously limited. For a start, I can just about tell a bishop from a horsey; I’m explicitly ignoring first-move advantage, and the assumption that all games are independent and identically distributed is… fanciful. I’m also playing a bit fast-and-loose with the percentages just to get a rough number. And honestly, the zero chance of Caruana winning a blitz match strikes me as an underestimate - but as I say, I’m not an expert.

However, assuming the sums are in the right ballpark, Carlsen was perfectly correct to offer the draw. It was Caruana who made the error in accepting.