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

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.

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By generalizing drunken rock-paper-scissors I'd say C wins 20% of the time over A. –  Marc Claesen Feb 4 '14 at 15:03
Winning is not necessarily transitive. Even in games of chance that is not the case: see Efron's intransitive dice. These ranking systems adopt particular models of winning but they are not appropriate for all forms of competition! (This is particularly true where contests are decided by a voting procedure.) Thus it's possible that your 80%, 80% numbers would be consistent with any winning probability for C against A. –  whuber Feb 4 '14 at 15:14
If there's a single underlying thing (a single overall skill, like 'general ability at chess') that determines win probability and which is transitive, you may be able to get somewhere (it's still not enough to pin it down completely, you need more assumptions). If it's based on multiple separate abilities (or on one thing which isn't transitive), then you really can't say anything from the available information. –  Glen_b Feb 4 '14 at 16:44

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:
$$p(CB) = p(CB|CA)p(CA) + p(CB|\overline{CA})p(\overline{CA})$$
$$p(CB|CA) = p(CB|CA,BA)P(BA) + p(CB|CA,\overline{BA})P(\overline{BA})$$
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 $p(CB)$.