"Probabilistic ultrafilters?" A naive question.
Let $S$ be a set and let $[0,1]^S$ the set of functions from $S$ to the closed interval $[0,1]$.
Suppose given some function $P \colon [0,1]^S \to [0,1]$ satisfying the following three conditions:


*

*If $f \geq g$ everywhere on $S$, then $P(f) \geq P(g)$;

*$P(\min(f,g)) \geq \min(P(f),P(g))$;

*$P(1-f) = 1 - P(f)$.


This is supposed to model a situation each point in $S$ has a "degree of belief" in some proposition, which yields a function $f$ in $[0,1]^S$; then $P$ is a process which takes all these degrees of belief and aggregates them into a "consensus" degree of belief $P(f)$.
Of course, this is meant to mimic the definition of an ultrafilter, which I think is given by the above definition with [0,1] replaced by {0,1}.
Certainly you have "principal" $P$, which just evaluate $f$ at some point $s$ of $S$.  I suppose you could get other $P$ by sending $f$ to its limit with respect to some non-principal ultrafilter.  
Is that it?
Added:  Actually, the second condition above is perhaps too strong.  I don't see an option for "hide question until I've thought about a bit more about what the best version of the question is" so I will just append this remark.
Added: Thanks, guys, for all the great answers.  I now think the formulation of (2) was misguided (at least if the definition is meant to model consensus about degrees of belief) and I don't know what the "right" formulation is.  One might well, for instance, want P to behave well when f and g refer to independent propositions; that would ask that P(fg) = P(f)P(g), which in the case of {0,1}-valued functions again agrees with the ultrafilter definition.  This rules out averages but leaves in evaluation at ultrafilters.
 A: A natural class of functions $P: [0,1]^S \to [0,1]$ which generalizes ultrafilters in a probabilistic sense is coming from states $\phi \colon \ell^{\infty}(S,\mathbb R) \to \mathbb R$, i.e. linear and monotone functionals with $\phi(1_S)=1$. Each state clearly yields a probabilistic ultrafilter by restriction.


*

*Ultrafilters correspond to special such functionals which in addition satisfy $\phi(fg)=\phi(f)\phi(g)$ since $\ell^{\infty}(S)=C(\beta S)$.

*Any such $\phi$ corresponds to a Radon probability measure on the compact Hausdorff space $\beta S$ by the Riesz representation theorem.

*Many examples have been studied: Let $S=\mathbb N$, $\omega \in \beta \mathbb N \setminus \mathbb N$ and consider
$$\phi(f)=P(f) := \lim_{n \to \omega} \frac1n \sum_{i=1}^n f(i).$$


This example does not come from taking a limit of the function $f$ with respect to some non-principal ultrafilter. In this sense the answer to your question is no.
A: Assume that $P$ satisfies the three conditions in the original question plus Noah Stein's additional requirement (declared " very wise" by the OP) that $P(s\mapsto c)=c$ for all constants $c$.  I claim that then $P$ is the operation of limit along a certain ultrafilter $U$ on $S$.  
As a first step, consider any function $f$ on $S$ that takes only the values 0 and 1; so $f$ is the characteristic function of a subset $A$ of $S$, while $1-f$ is the characteristic function of $S-A$.  Conditions 1 and 2 in the question imply that $P$ commutes with min, i.e., that equality holds in condition 2. In particular, since $\min\{f,1-f\}=0$, we have that one of $P(f)$ and $P(1-f)$ is 0, and then, by condition 3, the other one is 1.  In particular, the characteristic function of any $A\subseteq S$ is sent by $P$ to 0 or to 1.  Then the conditions in the question immediately imply that $U=\{A\subseteq S:P(A)=1\}$ is an ultrafilter on $S$.  Clearly, on functions that take only the values 0 and 1, $P$ agrees with limit along $U$; it remains to prove the same for arbitrary functions $f:S\to[0,1]$.
Let $f$ be such a function, and let $l$ be its limit along $U$.  Consider an arbitrary $c<l$ and note that, by definition of limit, the set $A=\{s\in S:f(s)>c\}$ is in $U$.  Let $g$ be the characteristic function of $A$, and notice that $f\geq\min\{s\mapsto c, g\}$.  Since $P$ sends $s\mapsto c$ to $c$ and sends $g$ to 1, it must send $f$ to a value $\geq c$.  Similar reasoning (using condition 3 to "turn the picture upside down") shows that $P(f)\leq d$ for all $d>l$.  Therefore $P(f)=l$, as required.
