If C wins B 80% of the time, and B wins A 80% of the time, how often would C beat A?

Question

I am currently trying to better understand probabilistic skill ranking systems for games, but I find that I have trouble properly understanding the basic concept of how skill as a pairwise comparison can be generalized.

For instance, if all you know is that player C wins player B 80% of the time, while that same player B wins player A 80% of the time, would this be enough data to determine how often C would win against A? How would those calculations work?

Of course it might even be possible for a game to have different styles of play where A might win specifically against C, which would completely confuse the issue, but I am talking about general ranking systems such as ELO or Trueskill that only take winning into account.

Answer 1

this information is not enough. Let's look more precisely what I mean by the lack of information. An event $CA$ means that $C$ wins against $A$ and event $\overline{CA}$ for when $C$ looses against $A$. Then we have:

\begin{equation} p(CB) = p(CB|CA)p(CA) + p(CB|\overline{CA})p(\overline{CA}) \end{equation}

and following that we have,

\begin{equation} p(CB|CA) = p(CB|CA,BA)P(BA) + p(CB|CA,\overline{BA})P(\overline{BA}) \end{equation}

however you don't have information on the conditional probabilities, having the prior probabilities such as $p(CA)$ and $p(BA)$ is not logically sufficient to infer $p(CB)$.

Answer 2

In the Elo model, C would be over one class interval (+240) stronger. B the average player (=0) and A over one class interval weaker -(240). The Elo model predicts a win for C against A with 95% probability (rating diff = 480).

See here for the rating probabilities: https://www.fide.com/docs/regulations/FIDE%20Rating%20Regulations%202022.pdf.

In a real world example, A, B and C could be equally strong. Where C is the "fear opponent" of B and B is the "fear opponent" of A.