A Minesweeper Probability Question

Let the positions of these unmarked squares be denoted as $A,B,C,\ldots, I$ as shown in the picture.

I will denote the value of 1 to a square with a flag/mine; and a value of 0 to a square without a flag/mine. For example, if $B = 1$ , then the top rightmost square has a mine.

Our mission here is to find the probability, ${\mathbf{Pr}}(A = 1)$ , given that precisely 4 of $A,B,C,\ldots, I$ has a value of 1, while the others has a value of 0.

We first partition the boxes as shown below and we will try to analyse each of these boxes separately.

For the top left pink box, looking at the left square of "2", we know that there should be precisely another 1 more mine among $A,C,D$ .

For the top right orange box, looking at the right square of "2", we know that there should be precisely another 1 more mine among $B,E$ .

For the bottom dark red box, looking at the top square of "3", we know that there should be precisely another 1 more mine among $F,G,H$ .

Thus, from the 3 paragraphs above, we know that there are precisely 3 mines among $A,B,C,\ldots , H$ . So, one more mine is not used. And so, $I$ must be a mine.

With $I$ as a mine, then looking at the most bottom left square of "3" tells us that $H$ cannot be a mine. So $I = 1, H= 0$ .

Now, looking at the top square in the bottom dark red box tells us that precisely one of $F$ and $G$ is a mine only.

With all the given information above. Let us list out all possible scenarios for the remaining cells.

Case 1: If $F = 0 , G = 1$ , then $D = E = 0$ . This forces $B = 2\Rightarrow A = 0 \Rightarrow C = 0$ , which is a possible scenario.

Case 2: If $F =1 , G = 0$ , then precisely one of $D,E$ is 0, while the other is 1.

Subacase 2.1: If $D = 0$ , then $E = 1\Rightarrow B = 0\Rightarrow A = 1\Rightarrow C = 0$ , which is a possible scenario.

Subacase 2.2: If $D = 1$ , then $A=C=0$ and $E = 0\Rightarrow B =1$ , which is a possible scenario as well.

Hence, all the 3 cases (inclusive of subcases) above are plausible. In all 3 scenarios, only one of the yields the solution of $A = 1$ . Our answer is $\boxed{\dfrac13}$ .