Here’s a really absurd question: “What’s the expected number of games I’d have to play against Magnus Carlsen before I win one game?”. In the spirit of What if?, let’s try to give a serious scientific answer to this absurd hypothetical.

My Elo is 1420. Magnus’ is 2835. That’s a pretty big gap.

According to the Elo formula1 my expected score against Magnus is 0.0003 (ha!). Let’s call this \(p\). In case it’s not clear, your expected score is the probability you win + 1/2 the probability you draw. So, if there were no draws in chess, then E[#games until I win] = 1/p ~= 3450 games. However, there are draws, which is obviously terrible news for my expected #games. Let’s say that if Magnus doesn’t win outright, he’s 500x more likely to draw against me than to lose to me.

Given that:

\[\begin{align*} \mathbb{P}[win] + \frac{1}{2} \mathbb{P}[draw] &= p = 0.0003 \\ \mathbb{P}[win] + \frac{1}{2} 500 \mathbb{P}[win] &= p \\ 256 \mathbb{P}[win] &= p \\ \mathbb{P}[win] &= \frac{p}{256} \\ \mathbb{E}[\text{#games until win}] = \frac{1}{\mathbb{P}[win]} &= \frac{1}{\frac{p}{256}} \approx \text{850,000} \end{align*}\]

I’m sure Elo breaks down when the rating difference is this large, but I’m not actually sure which way it breaks down2, so I’m just going to stick w/this guess - let’s call it an even million.

Now, another thing to consider is the fact that after playing Magnus in 100,000 games I might actually improve quite a bit! But I’ll ignore that for now…

  1. If Player A has a rating of \(R_{A}\) and Player B a rating of \(R_{B}\), the exact formula for the expected score of Player A is \(\frac{1}{1 + 10^{(R_B - R_A)/400}}\) 

  2. This question motivated a later post. Turns out, Elo overpredicts the ability of the better player to win when the rating difference is extremely large.