Let there be types $\omega\in\{0,1\}^n$ drawn according to some probability distribution. Suppose that these types are relayed through some imperfect message service. Specifically, any type $\omega$'s message, $m$, is always the product of some garbling so that $m\in\{0,1\}^n$, but $\sum_{i=1}^n m_i = k$. So there are $\binom{n}{k}$ different messages. A type $\omega$ will send message $m$ with probability $p_m^\omega$. I'd like to construct these distributions $p^\omega$ for each $\omega$ so that the message is "believed" according to a cost-weighted error. If $\omega_i=1$ and $m_i=0$, then the cost is 1. If $\omega_i=0$ and $m_i=1$, then the cost is $c$. Insofar as we want to minimize error, I think it's obvious that for any type $\omega$ it should send a message where $m_i=\omega_i$ when $\omega_i=1$, but for types where $\sum_{i=1}^n\omega_i>k$, then some ones must be turned off so to speak. Similarly, we have to turn on some false ones for types $\sum_{i=1}^n\omega_i<k$. Then, I think the problem is finding a distribution $p^\omega$ across messages for each $\omega$ so that for any $m$,
$$\sum_{\omega\in \{\omega:\sum_i^n\omega_i<k\}} c(k-\sum_i\omega_i )\Pr(\omega)p_m^\omega + \sum_{\omega\in \{\omega:\sum_i^n\omega_i\geq k\}} (\sum_i\omega_i-k)\Pr(\omega)p_m^\omega$$ $$\leq \sum_{\Omega}\Pr(\omega)p_m^\omega(\sum_i\omega_i) .$$
This inequality says that the cost of errors made from believing $\omega_i=m_i$ when $\omega_i=0$ and $m_i=1$ is less than the number of errors made from disregarding the message and believing every $\omega_i=0$. I think we can always find these $p^\omega$s whenever $\Pr(\sum_i^n\omega_i < k)\leq \frac{1}{c+1}$, but I don't believe this is necessary and I don't see how to actually construct the distributions to guarantee the inequality above.
Any help would be appreciated, and let me know if anything is unclear.
Example: Let $n=3$, $c=1$, and $\Pr(\omega_i=1)=.4$, i.i.d. Then for $k=2$, we can have $p^{(0,1,1)}_{(0,1,1)}=1$,$p^{(1,0,1)}_{(1,0,1)}=1$, and $p^{(1,1,0)}_{(1,1,0)}=1$. Then for the low types, set $p^{(1,0,0)}_{(1,1,0)}=p^{(1,0,0)}_{(1,0,1)}=1/2$. Similarly, set $p^{(0,1,0)}_{(1,1,0)}=p^{(0,1,0)}_{(0,1,1)}=1/2$ and set $p^{(0,0,1)}_{(0,1,1)}=p^{(0,0,1)}_{(1,0,1)}=1/2$. Finally, let the remaining types, $\omega=(1,1,1)$ and $(0,0,0)$ have $p^\omega_{(0,1,1)}=1$. I think all of the coniditions are met, yet $\Pr(\sum_{i=1}^n\omega<2)=0.648$ is not less than $1/2=\frac{1}{c+1}$.