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.