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.