Probability in Minesweeper

Question

Suppose I click a random tile during a Minesweeper game. It is a 1. During each time I click an adjacent square, what are the chances of hitting a mine? How would this change if it were a 2 or another number?

Answer 1

In the absence of any other information (which is rarely the case in Minesweeper), clicking a $1$ implies a $\frac{1}{8}$ chance that any of the adjacent $8$ squares is a mine. Now suppose you click an adjacent cell and it reveals a $1$. At this point, you have something like:

$$ \begin{array} & \star & \star & \star & \star \\ \star & 1 & 1 & \star\\ \star & \star & \star & \star \end{array} $$

Suppose that mines are distributed uniformly at random. There are $10$ unknown squares. Suppose further that the number of mines and number of cells are both much larger than $10$. Let's start counting!

First, keep in mind that there are $2^{10}$ possible mine placements in the cells marked with stars (each is either a mine or not a mine).

But there are much fewer mine placements among the $10$ cells that result in $1\ 1$ being displayed on the two middle cells:

• There are $4$ one-mine configurations (corresponding to placing a mine in one of the four centermost cells).
• There are $9$ two-mine configurations (corresponding to placing one mine in the left-most column, and independently placing one mine in the rightmost column).
• We have enumerated all $13$ configurations.

We can tabulate the mine probabilities per-square as follows: $$ \begin{array} & \frac{3}{13} & \frac{1}{13} & \frac{1}{13} & \frac{3}{13} \\ \frac{3}{13} & \star & \star & \frac{3}{13}\\ \frac{3}{13} & \frac{1}{13} & \frac{1}{13} & \frac{3}{13} \end{array} $$

In addition, by summing up these probabilities we can conclude that the expected total number of mines among the $10$ squares is $\frac{22}{13}$.

This is the kind of analysis used in probabilistic Minesweeper solvers. In general, many configurations must be analyzed separately so your original question is a little bit ill-posed.

Answer 2

For any tile in a MineSweeper game, the number revealed when clicked (if not a mine) tells the player the number of mines adjacent to that tile.

For example, assuming your tile is not on the edge of the playing field, it will have 8 adjacent tiles- three above, one on the left, one on the right, and three below.

So, if our tile $T$ reveals some $x \in {1, 2, 3, 4, 5, 6, 7, 8}$ , then the probability a randomly-selected adjacent tile to $T$ is a mine would be $\frac{x}{8}$ ..

The probability calculations for the number of edge cases (on the playing field) is similar.

EDIT

The key strategy of the game then, is to deduce locations of mines from the already-revealed values of neighboring tiles to T. Then you can 'flag' correct tiles as mines.