If there are
$n$ horses,
$h_i, i=1, ..., n$,
and you know
$p_{ij}$,
the probability that
horse $i$ beats horse $j$,
I would argue like this:
First of all,
I assume that there is
zero probability of
a tie,
so $p_{ij}+p_{ji} = 1$.
For any permutation
$\pi$
of
$[1, ..., n]$,
I would define the probability
of that permutation occurring
is
$\prod_{i=1}^{n-1} \prod_{j=i+1}^n p_{\pi(i)\pi(j)}
=\prod_{j=2}^{n} \prod_{i=1}^{j-1} p_{\pi(i)\pi(j)}
=\prod_{1\le i < j \le n} p_{\pi(i)\pi(j)}
$.
This is because,
for that permutation to happen,
horse $\pi(i)$
has to beat
horse $\pi(j)$
for all combinations of $j > i
$.
I think it would be possible
to prove
(by induction)
that the sum of this probability
of all permutations
is $1$.
(Oops.
Further on
I show that this does
not hold
for $n=3$.)
For $n=2$,
the permutations are
$[1, 2]$ and $[2, 1]$.
The probability of
$[1, 2]$
is
$p_{12}
$
and
the probability of
$[2, 1]$
is
$p_{21}
$
and the sum of these is $1$.
For $n=3$,
the permutations are
$[123], [132], [213], [231], [312], [321]$.
The probabilities are
$\begin{array}\\
123& p_{12}p_{13}p_{23}\\
132& p_{13}p_{12}p_{32}\\
213& p_{21}p_{23}p_{13}\\
231& p_{23}p_{21}p_{31}\\
312& p_{31}p_{32}p_{12}\\
321& p_{32}p_{31}p_{21}\\
\end{array}
$
and the sum of these is
$\begin{array}\\
\sum_{\pi([123])} \prod_{1\le i < j \le 3} p_{\pi(i)\pi(j)}
&=p_{12}p_{13}p_{23}+p_{13}p_{12}p_{32}
+ p_{21}p_{23}p_{13}+ p_{23}p_{21}p_{31}
+ p_{31}p_{32}p_{12}+ p_{32}p_{31}p_{21}\\
&=p_{12}p_{13}p_{23}+p_{13}p_{12}(1-p_{23})
+ p_{21}p_{23}p_{13}+ p_{23}p_{21}(1-p_{13})
+ p_{31}p_{32}p_{12}+ p_{32}p_{31}(1-p_{12})\\
&=p_{12}p_{13}
+ p_{21}p_{23}
+ p_{31}p_{32}\\
&=p_{12}p_{13}
+ (1-p_{12})p_{23}
+ (1-p_{13})(1-p_{23})\\
&=p_{12}p_{13}
+ p_{23}-p_{23}p_{12}
+ 1-p_{13}-p_{23}+p_{13}p_{23}\\
&=1+p_{12}p_{13}
-p_{23}p_{12}
+ p_{13}+p_{13}p_{23}\\
\end{array}
$
Interesting.
Unless I made a mistake
(which is not unlikely),
the sum is not $1$
unless
$p_{12}p_{13}
-p_{23}p_{12}
+ p_{13}+p_{13}p_{23}
= 0
$
or
$p_{23}p_{12}
=p_{12}p_{13}
+ p_{13}+p_{13}p_{23}
=p_{13}(1+p_{12}+p_{23})
$.
I don't know whether
this is a
genuine consistency condition
or a result of an error of mine.
Oh well.
We can always divide each probability
by their sum,
making it into
a genuine probability.
