Let's imagine designing an odds pattern for a game, in which players bet for win or lose.
Suppose the probablity of winning is $p$, thus the probablity of losing is $1-p$.

Now imagine $n_1$ people bet for win, $n_2$ people bet for lose,  both ante is 1,
and the odds for both are $1:M$ and $1:N$, respectively.

The banker doesn't want even a penny out of his wallet, so it's reasonable we have, according to Mean Value Formula:
$$n_1  (M-1)  p + n_2  (N-1)  (1-p)  \le  n_2  p + n_1  (1-p)$$ 

specially, when $n_1 == 0$:  
we have : $n_2  (N-1)  (1-p)  \le  n_2  p$,  
we get  : $N \le p / (1-p) + 1$

when $n_2 == 0$, we can likewise get   
$M \le (1-p) / p + 1$

To make it general, it makes sense to rewrite them like :  
$N \le (p / (1-p) + 1)  (n_2 / (n_1+n_2))$,     
$M \le ((1-p) / p + 1)  (n_1 / (n_1+n_2))$

**Here comes my question:**
the $n_1$ and $n_2$ are influenced by the $M$ and $N$ and $P$. However, the $M$ and $N$ rely on the $n_1$ and $n_2$. How to figure out what $M$ and $N$ should be chosen?

It seems we should have a transcendental value for $n_1$ and $n_2$.
FYI, there is a restriction : $n_1 + n_2 \le C - 2$,  $C$ is a constant.