Minesweeper probability

You've correctly inferred that there are only two possibilities:

1. The two bombs are at $A$ and $D$.
2. The two bombs are at $B$ and $C$.

Common sense now dictates that these two options are equally likely. The only way to prove that the two options are equally likely requires knowledge of how the minesweeper board was generated. It would be in the spirit of minesweeper if information about the positions of the bombs could only be inferred from the board itself, and not from the underlying code generating the boards. So if the minesweeper game was 'properly' coded then yes, both option are equally likely. But unless you can give us the code of the specific minesweeper game you're playing, there is no way to prove such a claim.

You know

• exactly one of A and C is a bomb
• exactly one of B and D is a bomb
• exactly one of A and B is a bomb

So, as you say, there are two possibilities

• A and D are bombs while B and C are not
• B and C are bombs while A and D are not

As far as I can tell, these have the same likelihood

Here's a mathematically rigorous proof of the statement:

Let's assume the board consists of $n$ fields of which $k$ are mines and the field is created by randomly choosing $k$ numbers from $1$ to $n$ without repetitions. This is a reasonable assumption, since the Fisher-Yates-Shuffles would yield an effective and easy way to create a random field and every field would be possible and equally likely.

We introduce the random variables A, B, C, D that are $1$ if the corresponding field has a mine and $0$ else. We can infer the following information from the positions of flags and numbers: $$A + B = A + C = B + D = 1$$

What you now want to calculate is a conditional probability: $$\begin{align*}&P(A = 1 | A + B = A + C = B + D = 1) = \frac{P(A = 1, A + B = A + C = B + D = 1)}{P(A + B = A + C = B + D = 1)} \\ &= \frac{P(A = D = 1, C = B = 0)}{P(A + B = A + C = B + D = 1)} \\ &= \frac{P(A = D = 1, C = B = 0)}{P((A = D = 1, C = B = 0) \vee (A = D = 0, C = B = 1))}\end{align*}$$

Now the two events $\{A = D = 1, C = B = 0\}$ and $\{A = D = 0, C = B = 1\}$ are disjoint, so the probability of their union is the sum of their probabilities. It is easy to see that they both have the same probability, namely ${n - 4\choose k - 2}\cdot {n \choose k}^{-1}$. Since the Event in the numerator also holds with the same probability, this yields $$P(A = 1 | A + B = A + C = B + D = 1) = \frac{1}{2}.$$

So it's a 50:50 change whether the bombs are on $A$ and $D$ or on $B$ and $C$.