Given a simple head-to-head game that can't end in a draw, between two players $P_A$ and $P_B$, whose chances of winning are $Pr_A$ $Pr_B$ respectively, and where $0 < Pr_A, Pr_B < 1$, and where $Pr_A + Pr_B$ does not necessarily have to add up to one.
We define the probabilities of $P_A$ and $P_B$ winning as follows:
$$ \begin{align*} Pr(P_A wins) = \frac{Pr_A(1 - Pr_B)}{(Pr_A(1 - Pr_B)) + Pr_B(1 - Pr_A)}\\ Pr(P_B wins) = \frac{Pr_B(1 - Pr_A)}{(Pr_A(1 - Pr_B)) + Pr_B(1 - Pr_A)} \end{align*} $$
Now if we were to have more than two players - instead of a one-on-one match, we have a race between 3 or more contestants, eg: $P_A$, $P_B$ and $P_C$, would using the above definition and taking into account the extra players as is done below, be considered correct?
$$ \begin{align*} Pr(P_A wins) = \frac{Pr_A(1 - Pr_B)(1 - Pr_C)}{D}\\ Pr(P_B wins) = \frac{Pr_B(1 - Pr_A)(1 - Pr_C)}{D}\\ Pr(P_C wins) = \frac{Pr_C(1 - Pr_A)(1 - Pr_B)}{D} \end{align*} $$
Where:
$$D=(Pr_A(1 - Pr_B)(1 - Pr_C)) + (Pr_B(1 - Pr_A)(1 - Pr_C)) + (Pr_C(1 - Pr_A)(1 - Pr_B))$$
This is a modelling question. Your proposal assumes 3-wise independence of the collection of events of the form $$(E_A,E_B,E_C)$$ where for $$X \in \{A,B,C\}$$ $E_X$ is either $\{ \omega: X~wins\}$ or its complement.
The validity of this will depend on the players and the game. It may be approximately valid in many contests.