TITLE: Minesweeper probability
QUESTION [12 upvotes]: I ran into the situation pictured in the minesweeper game below. Note that the picture is only a small section of the entire board.
Note: The bottom right $1$ is the bottom right corner tile of the board and all other tiles have been marked/deemed safe.

What we know:

There are exactly $2$ bombs left on the board
Bombs can't be: (A & B) || (A & C) || (B & D) || (C & D)
Bombs can be: (A & D) || (B & C) 

Can anyone prove if there is a square that is more likely to be safe/unsafe or is every square equally likely to be safe/unsafe?

REPLY [6 votes]: 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 $C$.