The probability of a cell in Minesweeper I'm trying to calculate the probability of a cell in Minesweeper when there's constraint intercepting with each other: 

The related cells is marked with ID as shown in the snapshot. Obviously, X1 and X2 could not be mine, since C1 is surrounded by 3 mines already. Therefore, accordingly to C2, The probability of A3 and A4 is 1/2. Likewise, The probability of A1, A2, A3 is 1/3 due to C3. 
Since A3 is constraint by both C2 and C3, so what is A3's probability? And how would A1, A2, A4's probability change after the change of A3's probability on both constraints. Please help me with some thoughts, thanks.
 A: You have to consider all cases for the whole subset of cells $\{A_1,A_2,A_3,A_4\}$. That is, given the constraints, it can be BeeB, eBeB, eeBe (where 'e' means empty and 'B' bomb). But to find out the probability of each case, you have to know the number of bombs left, and assume a distribution (uniform should be fine).
For instance, if you know for sure there is only one bomb left, then you know where it lies, with probability $1$.
If you don't know how many bombs are left, or if there is more than one, you can compute the probability that a uniform sample of bombs ends up with one (resp. two) bombs among $A_1-A_4$, conditioned on the constraint that it's either one or two. But to do this accurately you should also take into account all the other known constraints on the whole game, which is quite cumbersome (but actually they are simpler).
A: You can attach a propositional variable to each cell of the matrix.  A variable is true if and only if its cell holds a bomb.  You can then write a formula $f$ that is true whenever the assignment to the variables is consistent with the given information.
Suppose $f$ has $N$ satisfying assignments and that $x$ is true in $N_x$ of them.   If all satisfying assignments are equally likely, $N_x/N$ gives the probability that $x$ is true in the actual solution.  In particular, if cell $x$ is safe (no bomb is there) $N_x=0$, and if a bomb is present, $N_x=N$.
As mentioned by @Jean-ClaudeArbaut, $f$ depends on whether the total number of bombs is known and, in case it is known, on the exact value.
The computation is better done with a computer.  Here's a couple of matrices. First, the case when no (nontrivial) bound on the total number of bombs is known:
$$\begin{matrix}
1/2 & 1/2 & 1/2 & 1/3 & 1/3 & 1/3 & 1/2 & 1/2 \\
1/2 & 0   & 0   & 0   & 0   & 1   & 1/3 & 1/2 \\
1/2 & 1   & 0   & 1   & 0   & 0   & 1/3 & 1/2 \\ 
1/3 & 1   & 0   & 0   & 0   & 1   & 1/3 & 1/2 \\ 
1/3 & 0   & 0   & 1   & 0   & 0   & 1   & 1/2 \\ 
1/3 & 1/2 & 1/2 & 1   & 0   & 0   & 2/3 & 1/2 \\ 
1/2 & 1/2 & 1/2 & 1/2 & 1/2 & 1/2 & 1/2 & 1 
\end{matrix}$$
Then under the assumption that there are $13$ bombs (the minimum possible in this case):
$$\begin{matrix}
0   & 0   & 0   & 1/3 & 1/3 & 1/3 & 0 & 0 \\
0   & 0   & 0   & 0   & 0   & 1   & 0 & 0 \\
0   & 1   & 0   & 1   & 0   & 0   & 0 & 0 \\ 
1/3 & 1   & 0   & 0   & 0   & 1   & 1 & 0 \\ 
1/3 & 0   & 0   & 1   & 0   & 0   & 1 & 0 \\ 
1/3 & 1/2 & 1/2 & 1   & 0   & 0   & 0 & 0 \\ 
0   & 0   & 0   & 0   & 0   & 0   & 0 & 1 
\end{matrix}$$
