How do I approach a “stuck” game of minesweeper?

I've played many a-game of minesweeper, and I frequently reach positions in which there is no logical, deterministic way to come to a conclusion about how I should solve the puzzle.

How should I best approach solving these puzzles to guarantee the highest statistical victory?

Meta-context: this question is an offshoot of this original question, which feedback indicated was somewhat too broad.

• Don't click the top left corner. Apparently if your first click is a mine, then that is where it is relocated to and your first click is "safe". Aside from that, being lucky is probably your best bet. – Travis J May 22 '14 at 17:40
• Just for your information: Simon Tatham's Portable Puzzle Collection features an implementation of Minesweeper that is guaranteed to be dead-end-free. As its name suggests, it is available for many platforms. chiark.greenend.org.uk/~sgtatham/puzzles – Oliphaunt Feb 23 '16 at 22:40
• with great caution :) – JMP Jun 1 '17 at 10:53
• There's a smiley face button in the top center of the game. Press that. – Ian MacDonald Jun 1 '17 at 14:45

In general, when looking at these situations, you want to click on the squares that are the least likely to contain mines. This will invariably give you more information about your current situation.

Here's an example of a game that can't be beaten without strict testing, which I will use to demonstrate what I mean.

There are two mines remaining. I apologize, but the 3 underneath the square labeled $2$ actually has two adjacent flags, and needs one more flag; one of them was cut off in the screenshot.

Because there is a 1 next to $1$ and $2$, I know that one of these must be a mine. If $1$ is a mine, then $2$ is not, which means one of $4,5,6$ is a mine, and $3$ is not. This gives three cases when $1$ is a mine.

If $1$ is not a mine, then $2$ is a mine. If $2$ is a mine, then none of $4,5,6$ is a mine, which means $3$ must be a mine. This has only one case.

Therefore, the number of cases where $1$ is a mine is $\frac{3}{4}$; the number of cases where it is not is $\frac{1}{4}$. We can proceed assuming that $1$ is a mine - and as it happens to turn out, it is.

Here's another example of a possible setup. The method I use in these cases effectively can be extended across the entire board (which makes it a bit more accurate), but I only execute it a few levels in my head. Here's how my thought process works:

We can see that if $5$ is a mine, then $4$ can't be, which means $3$ is. At this point, either $1$ or $2$ could be a mine, which gives two cases.

If $5$ isn't a mine, then $4$ must be a mine, which means $3$ isn't, which means both $1$ and $2$ are - which gives one case.

Therefore, the situation in which $5$ is a mine covers $\frac{2}{3}$ of the cases, and the situations in which $5$ isn't covers $\frac{1}{3}$. Therefore, we proceed on the assumption that $5$ is a mine.

Of course, there are more than three cases for the entire grid - this is only a rough estimate. I could go into further detail about what the squares above $1$ and $2$ could be, as well as the surrounding squares, but I won't here. It all depends on how long you want to take and how accurate you want to be.

• Fair criticisms. I'll update the answer later tonight. – Aza May 22 '14 at 17:51
• ", said Emrakul a week ago ;-) – John Dvorak May 28 '14 at 14:31
• @Jan .....whoooops. Totally slipped through the cracks there. Thanks for the reminder! – Aza May 28 '14 at 14:32
• The formatting on this answer looks weird. The LaTex is not converting for me... ? – JLee Jun 4 '15 at 15:28
• @Emrakul can i edit it to make the LaTex work or is that in the answer what you intended...? – manshu Feb 23 '16 at 16:21