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.

expectedreturn, and it is easy to ensure that. Set $M$ so that betting on Pass has a nonpositive return for the player, and set $N$ so that betting on Don't Pass has a nonpositive return. That said, I'm voting to close since I don't think the question is on the level of this site. – Douglas Zare Sep 18 '12 at 15:48