For $X,Y,Z\in\{A,B,C\}$, let $P_{XY}$ be the probability that $X$ wins against $Y$, and let $P_{XYZ}$ be the probability that the race ends in the order $X$ first, $Y$ second, $Z$ third.
Then $$P_{XY} = P_{XYZ} + P_{XZY} + P_{ZXY}.$$
Writing down those equations explicitely for the $6$ possibilities to distribute $A,B,C$ to the letters $X,Y,Z$, we get the linear equation system
$$
\begin{pmatrix}
1 & 0 & 1 & 0 & 1 & 0 \\
1 & 0 & 1 & 0 & 0 & 1 \\
0 & 1 & 1 & 0 & 1 & 0 \\
0 & 1 & 0 & 1 & 1 & 0 \\
1 & 0 & 0 & 1 & 0 & 1 \\
0 & 1 & 0 & 1 & 0 & 1
\end{pmatrix}\cdot
\begin{pmatrix}P_{ABC} \\ P_{ACB} \\ P_{BAC} \\ P_{BCA} \\ P_{CAB} \\ P_{CBA}\end{pmatrix} = \begin{pmatrix}P_{AB} \\ P_{BA} \\ P_{AC} \\ P_{CA} \\ P_{BC} \\ P_{CB}\end{pmatrix}
$$
If the equation system matrix $M$ was invertible, we could reconstruct the values $P_{XYZ}$ (and thus the probability that $A$ wins) from the values $P_{XY}$.
But it turns out that $M$ is not invertible. Instead, it has the nontrivial kernel spanned by $(1,0,-1,0,-1,1)$ and $(0,1,-1,1,-1,0)$.
This knowledge can be used for the construction of a counterexample, showing that in general, it is not possible to predict the winning probability of $A$.
Three equal horses give $P_{ABC} = P_{ACB} = \ldots = 1/6$ and $P_{AB} = P_{BA} = \ldots = 1/2$. Horse $A$ wins with probability $1/3$.
The vector $v = (1/12,1/12,-1/6,1/12,-1/6,1/12)$ is in the kernel of $M$. Modifying the above example for the three equal horses by this vector, we get the probabilities $P_{ABC} = 1/6 + 1/12 = 1/4$, $P_{ACB} = 1/4$, $P_{BAC} = 0$, $P_{BCA} = 1/4$, $P_{CAB} = 0$, $P_{CBA} = 1/4$. Horse $A$ wins with probability $1/2$.
Since the vector $v$ is in the kernel, the values $P_{AB}, P_{AC}, \ldots$ are the same as before.
This shows that for $P_{AB} = P_{AC} = \ldots = 1/2$, it is possible that $A$ wins with probability $1/3$, as well as that $A$ wins with probability $1/2$.
(Note that the constructed counterexample is the same as in the post of Micah.)
