Conditional probabilities with impossible outcomes 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.
 A: I don't think you can determine $P(A>B>C)$ from the individual probabilities unless you know the actual order in which $A,B$ and $C$ finish which can change $P(A>B>C)$ for the same individual probabilities.
Example, in one $10$ race series the finishing order was:
$ABC, ABC, ABC, ABC, ABC, ACB, ACB, BAC, BAC, BCA$ where $P(A>B) = 0.7; P(A>C) = 0.9$ and $P(B>C) = 0.8$.
In another $10$ race series the finishing order was:
$ABC, ABC, ABC, ABC, ABC, ABC, ACB, BAC, BAC, CBA$ with identical $P(A>B) = 0.7; P(A>C) = 0.9$ and $P(B>C) = 0.8$ as the first $10$ race series.
However, $P(A>B>C)$ for the first race series is $0.5$ and $0.6$ for the second.
Then again, for a given number of races, one may just have to consider all the possible outcomes for the given individual probabilities assuming each one is equally likely. But to do that, one has to know the number of races.
A: 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 $p_6$.
A: Along the lines of what the other responder has said, you can condition the probabilities as shown in the table below


The first finish is A>B>C which is the product of $P(A>B)*P(B>C)*P(A>C)$  You have six such finishes as below {ABC, ACB, BAC, BCA, CAB, and CBA}. The way you condition is $\frac{ABC}{\text{sum of all}}$
