Estimating acceptance probability when offering multiple options simultaneously Suppose we have a set of 5 options, {A, B, C, D, E}, that we want to offer someone. We want to estimate the probability that they accept each option.
We can easily do that if we offer each option separately. E.g., we can offer option A ten times then observe that it was accepted 6 times then we conclude that MLE estimate for option A is 0.6 (here we have a Bernoulli distribution).
However, what if we offer all options {A, B, C, D, E} at once to someone and they can either accept one of the options or not accept any of them. Then this looks to me like a categorical distribution and the corresponding probability vector parameter now corresponds to the probability that each option is selected including the possibility that none are selected rather than the probability that an option is accepted when offered separately.
Is there a way to estimate the acceptance probability of each option by offering all options simultaneously?
 A: I would think not unfortunately due to the Condorcet Paradox.
This paradox shows that the overall ordering of preferences need not be transitive: The group could prefer $A$ over $B$, $B$ over $C$, $C$ over $D$, $D$ over $E$ yet prefer $E$ over $A$!
This means that you'd need to know the average preferences of each letter
Let $X$ represent the event of choosing an option in $S=\{A,B,C,D,E,\emptyset \}$, where I'm using $\emptyset$ to represent "choose none". Then we have the single-choice probabilities:
$$p_x  = P(x|x,\emptyset):=P(X=x|\{X=x,X=\emptyset\})$$
Your experiment will give us the following probabilities:
$$q_x = P(x|S)$$
Let's say that $q_{\emptyset}>0$, then we know that no matter what choice $A$ through $E$ we choose it will lose out to $\emptyset$ at least $q_{\emptyset}$ of the time. Therefore, we can bound $p_x$:
$$q_x \leq p_x \leq 1-q_{\emptyset}$$
Unfortunately, we cannot get much further because we don't know what happens when we remove the other choices. For example, lets focus on getting $p_A$:
What happens when we remove, say, $B$ from the available choices? We know that $q_B$ of the population had $B$ as their top choice, but what if all of them had $A$ as their second choice?
In that case, $p_{A|S\setminus B} = q_A + q_B$
What if none of them had $A$ or $\emptyset$ as their second choice? Then $p_{A|S\setminus B} = q_A$
In general, if $A$ is the second choice for $r$ percent of the people who choose $B$ as a first choice, then:
$$p_{A|S\setminus B} = q_A + r\cdot q_B$$
To make this concrete, let's take $x=A$:
As this example makes clear, you can bound $p_x$ but you cannot directly estimate it without knowing how $x$ stacks up against $S\setminus \{\emptyset, x\}$
