# Multiple probabilities assigned to the same event

I asked this minesweeper question Two probabilities of the same thing . I think this might be the more general problem:

There are $n$ sets $S_i$, $i=1,2,3.....n$. The set $S_i$ contains $k_i$ elementary events, $i=1,2,3.....n$. All the elements within a set $S_i$ are assigned the same probability $p_i$.

All events within a set are mutually exclusive but multiple events all from different sets can occur simultaneously.

Only one element $C$ is common to all the sets.

A random experiment is performed in which exactly one element from each set is an outcome.

So, if the event $C$ happens, then it is the only outcome. And if $C$ doesn't happen then there are $n$ outcomes, one from each set.

But $n$ different probabilities are assigned to the event $C$, one in each set. Then, what is the actual probability that $C$ happens?

EDIT: I think the probabilities $p_i'^{s}$ can be understood like this: If there was an experiment in which $S_i$ was the sample space, i.e. only an element from $S_i$ could be an outcome, then all the elements of $S_i$ would have the same probability $p_i$. In the experiment being performed here, one element from each set is an outcome, so $p_i'^{s}$ are not the probabilities of these events happening, in this experiment. It explains why the event $C$ does not have $n$ different probabilities in this experiment.

• If you assign $C$ multiple distinct probabilities, then your probability assignment is inconsistent. I'm not sure what the question is here. – Ceph Oct 23 '17 at 15:00
• @Ceph $C$ is assigned different probabilities in different experiments. In an experiment with sample space $S_i$ and only one outcome, $C$ has the probability $p_i$. Now, an experiment is performed in which there is one outcome from each of the sets $S_i$ $i=1,2.....n$. What is the probability that the outcome is $C$ in this experiment? The elements within a set are mutually exclusive. Different elements from different sets are not mutually exclusive. – Dove Oct 23 '17 at 15:52
• @Ceph This is an example version: There are 10 blocks. There is a mine in one of the first 3 blocks.This means the probability of a mine being in the 3th block=$\frac{1}{3}$. Additional information is: there is a mine in one of the last 8 blocks. So, this means the probability of a mine in 3rd block=$\frac{1}{8}$. The 3rd block is common to the set of first 3 blocks and the set of last 8 blocks. Can we use both of these information to get the actual probability of a mine in the 3rd block? – Dove Oct 23 '17 at 16:02