Calculating probabilities in horse racing! I've seen a few similar threads to this on different forums but they don't seem to conclude to a satisfactory answer. My question is this:
If you have 3 horses, A, B, and C and you know the winning probabilities of each horse racing against each other as a pair, how do you work out their winning probabilities if all 3 horses race together? Is there a formula for n number of horses?
So you know the probabilities of:
A beating B
A beating C
BA
BC
CA
CB
I initially thought that: p(A winning) = p(A beat B). p(A beat C)
But this is clearly wrong.
Can someone please help!!
Many thanks
Matt
 A: Expanding on Tony K's comment, here's an (extreme) example of why it's impossible.
Let's say that we have three very strange horses:


*

*Horse A runs the race in $1$ second with probability $1/2$, and in $9$ seconds with probability $1/2$.

*Horse B runs the race in $3$ seconds with probability $1/2$, and in $7$ seconds with probability $1/2$.

*Horse C always runs the race in $5$ seconds.


Then it's not hard to check that, in head-to-head races, $P(A \text{ beats } B) = P(A \text{ beats } C) = P(B \text{ beats } C) = 1/2$.
In a three-way race, A will win whenever it runs fast, so $P(A \text{ wins})=1/2$. If A doesn't run fast, B will win if it runs fast, and C will win if B runs slowly. So $P(B \text{ wins})=P(C \text{ wins})=1/4$.
On the other hand, let's say we have three more normal horses; in fact, horses D, E, and F are so ordinary that they're identical to each other in every way. Then it must be true that $P(D \text{ beats } E) = P(D \text{ beats } F) = P(E \text{ beats } F) = 1/2$ again. But in this case, each horse must have a $1/3$ probability of winning the entire race.
That is, we've found two sets of three horses such that:


*

*The head-to-head probabilities are identical between them.

*The probabilities of winning a 3-way race are different.


So there can't possibly be any way of calculating the 3-way probability, given only knowledge of the head-to-head probabilities.
EDIT:
The problem is that the three-way win probabilities depend not just on the average abilities of the horses, but also on how reliable they are. A horse that's very unreliable (sometimes runs really fast and sometimes runs really slow) will generally do better in a 3-way race than in a 2-way race, because if they run really fast against one of their opponents they'll run really fast against both of them! Notice that my three horses all finish the course in the same amount of time on average, but horse A still wins out by being unreliable.
In short, if you want to calculate 3-way win probabilities, then you'll need to either know or assume something about how a horse's performance varies, as well as about its average performance. There's no way to get this data from 2-way win probabilities.
A: 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.)
A: We have the following equations:
$$ P(12)=P(123)+P(132)+P(312),P(13)=P(123)+P(132)+P(213),
P(23)=P(123)+P(213)+P(231),
P(123)+P(132)+P(213)+P(231)+P(312)+P(321)=1. $$
$ P(i,j) $ are known, $ P(k,n,m)$ and then
$$ P(1)= P(123)+P(132), 
P(2)= P(213)+P(231), 
P(3)= P(312)+P(321), $$
should be found from these equations (if we find $ P(n) $ we’ll know $ P(k,n,m))$ . The answer is clear.
Added in connection with the issues
$ P(n)$  – probability, that n-th horse wins among the rest,
$ P(i,j)$  - probability, that the i-th horse beats the j-th,
$ P(k,m,n)$ - probability, that the k-th horse beats the m-th and the m-th horse beats the n-th.
As I noted earlier, if we find $ P(n) (via P(i,j))$, then  probabilities $ P(k,m,n) $ will be known. Therefore, the solution of these equations for $ P(k,n,m) (0 \le P(k,n,m)\le 1)$ provides an accurate and complete answer to the question of the topic.
Depending on the values of $ P(i,j) $ the following cases are possible: 
1) The equations have many solutions for $ P(k,n,m)$. In this case probabilities $ P(k,m,n)$ are not uniquely determined via $ P(i,j)$ (respectively, $ P(n) via P(i,j))$.
2) The equations have a unique solution, and therefore give unique dependence $ P(n) via P(i,j)$. Example: $ P(1,2)=1, P(2,3)=1$. You can find more interesting examples.
3) The equations are unsolvable (initial data are incorrect). Example: $P(1,2)+P(2,3)<P(1,3)$.
