TITLE: Conditional probabilities with impossible outcomes
QUESTION [6 upvotes]: Suppose we want to predict the outcome of a race between three runners $A$, $B$, $C$. We know the prior probabilities for head-to-head runs: $p_{A>B}$, $p_{A>C}$, $p_{B>C}$, where $A>B$ means that A finishes before B. How do we get the probability $p_{A>B>C}$?
I thought I could just use a tree diagram but I run into the following problem: Some outcomes are impossible. For example, if my first edge is "A finishes before B" and my second edge is "A finishes after C", then "B finishes before C" is no longer possible. I thought about just setting probability of impossible outcomes to 0, but then my outcome probabilities are different depending on which of the three probabilities I start with.
I think I'm making some basic mistake here.
I've seen a lot of similar questions but I don't think they answer this. I've also tried looking at how horse race outcomes are predicted, but couldn't find an answer that I understood.

REPLY [4 votes]: Let $$S_1:\>A>B>C,\qquad S_2: \>A>C>B,\qquad\ldots,\qquad S_6:\>C>B>A$$
be the six possible rankings (in lexicographic order) and $p_i$ $(1\leq i\leq 6)$ their probabilities. Then we have the four equations
$$\eqalign{P[A>B]&=p_1+p_2+p_5\cr
P[A>C]&=p_1+p_2+p_3\cr
P[B>C]&=p_1+p_3+p_4\cr
1&=p_1+p_2+p_3+p_4+p_5+p_6\ .\cr}$$
Here the LHSs are given. These four equations are insufficient to determine the six $p_i$ individually; in particular $p_1$ is not  determined by the given data. E.g., one finds that
$$p_1=P[A>B]+P[B>C]+p_6-1\ ,$$
but we have no information about $p_6$.