Probability estimation of minesweeper undecidable choices

Question

Someone asked what was his probability to win in this situation?

This is an interesting problem and I'll be curious to know an efficient solution.

So far, I wanted also to generalize, with a simplification, we consider only:

• pairs of blocks such as the one on top right, and model them with a binary variable which value is the presence of the mine
• twin pairs, such as the one on bottom left, which is a couple of pairs linked together, the knowledge of one value of a pair gives the value of the other pair and vice versa

So we omit groups of more than 2 pairs linked together, for simplicity.

Now in this simple example, we have 2 single pairs and 2 twin pairs, let's denote twins with binary variables: $A_1, A_2$, and the other one with $B_1, B_2$ and the other single pairs with $C$ and $D$

So we could think the probability is $(\frac 1 2)^4$ but I'm not sure at all

The goal is to generalize with m twins pairs and n single pairs

Answer 1

You are correct with your answer of $\left(\frac{1}{2}\right)^4$. You have arrived at the unfortunate situation of having less than a 7% chance of winning!

Single and double pairs both have probability $\frac{1}{2}$ of being correctly solved assuming no mistakes are made after the first click. For single pairs this is obvious, for double pairs the first click will always either reveal a mine or a number, both of which provide enough information to complete the rest safely. This logic can be extended to any pattern of this sort, and is also true for triple pairs etc.

Each single/double pair is independent as long as they are not next to each other. As there is a fixed number of mines in each pair, the configuration has no effect on another pair, as long as they don't share any revealed numbers.

So, your answer can in fact be generalised to:
For a total of $n$ groups of pairs left at the end of a game, which may be single, double, triple etc., the probability of successfully completing the game assuming no mistakes are made is $\left(\frac{1}{2}\right)^n$.

Note
We do not actually need to limit ourselves to these configurations alone, although they are the simplest. It isn't too hard to calculate some of the slightly more complicated ones.

Answer 2

I'm not a specialist in statistics but i think the twin pairs isn't different from single pairs, because your chance for first correct select in 4 choose is 2/4 and after select that if you select a non-bomb cell you will know state of 3 other cell. so you have 4 single pairs and your answer will be 1/2 power 4. please correct me if i wrong. thanks.