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$p$, thus the probablity of losing is 1-p$1-p$.
Now imagine n1$n_1$ people bet for win, n2$n_2$ people bet for lose, both ante is 1, and the odds for both are 1:M$1:M$ and 1:N$1:N$, respectively.
The banker doesn't want even a penny out of his wallet, so it's reasonable we have , according to Mean Value FomulaFormula: n1 * (M-1) * p + n2 * (N-1) * (1-p) <= n2 * p + n1 * (1-p)$$n_1 (M-1) p + n_2 (N-1) (1-p) \le n_2 p + n_1 (1-p)$$
specially, when n1 == 0$n_1 == 0$:
we have : n2 * (N-1) * (1-p) <= n2 * p $n_2 (N-1) (1-p) \le n_2 p$,
we get : N <= p / (1-p) + 1$N \le p / (1-p) + 1$
when n2 == 0$n_2 == 0$, we can likewise get
M <= (1-p) / p + 1$M \le (1-p) / p + 1$
To make it general, it makes sense to rewrite them like :
N <= (p / (1-p) + 1) * (n2 / (n1+n2))$N \le (p / (1-p) + 1) (n_2 / (n_1+n_2))$,
M <= ((1-p) / p + 1) * (n1 / (n1+n2))$M \le ((1-p) / p + 1) (n_1 / (n_1+n_2))$
Here comes my question: the n1$n_1$ and n2$n_2$ are influenced by the M$M$ and N$N$ and P$P$. However, the M$M$ and N relys$N$ rely on the n1$n_1$ and n2$n_2$. How to figure out what M$M$ and N$N$ should be chosen?
It seems we should have a transcendental value for n1$n_1$ and n2$n_2$. FYI, there is a restritionrestriction : n1 + n2 <= C - 2$n_1 + n_2 \le C - 2$, C $C$ is a constant.
