atoms, the average evidence of a typical atom of $S$. Then $$E(S) = \frac{1}{B} \sum_{a \in Base(S)} E(S, \hat{a}),
\quad \quad ER(mtm) = 1 - E(S),$$ $$ER(rand) = \frac{2}{B} \sum_{a \in Base(S)} E(S, \hat{a})
(1 - E(S, \hat{a})), \quad \quad  ER(worst) = E(S)$$ such that $$\lim_{E(S) \to 1} \frac{ER(rand)}{ER(mtm)} = 2, \quad \quad
\lim_{E(S) \to 1} \frac{ER(worst)}{ER(mtm)} = \infty.$$

Since the most typical model of a given system would be the most trustworthy one, it should be preferred by any rational reasoner. So the existence of a most typical model is a practically important characteristic of any knowledge system. Let $p(mtm)$ denote the probability that a given system $S$ has the most typical model. The probability that a randomly chosen model of $S$ contains all typical atoms is $\prod_{a \in Base(S)} E(S, \hat{a})$. Then $$p(mtm) = 1 - \left( 1 - \prod_{a \in Base(S)} E(S, \hat{a}) \right) ^M$$ where $M = |MOD(S)|$ and $2^{-B} \le \prod_{a \in Base(S)} E(S, \hat{a}) \le (E(S))^B$.\
So $$\label{e-pmtm}
1 - (1 - 2^{-B})^M \: \le \: p(mtm) \: \le \: 1 - (1 - (E(S))^B)^M.$$

Expression (18) provides rather rough bounds for $p(mtm)$. Experimental estimation of $p(mtm)$ is presented in Section 7.

Typical kernel
