# Minesweeper challenge #4

While solving another Minesweeper puzzle, I ran into yet another unsolvable corner.

Well, actually, this time I have two unsolvable corners in the same puzzle.

Here's the bottom left corner:

...as well as the bottom right:

(Image updated to add more clues from tapping random squares)

Which squares have the best chance of being safe? What is the overall probability I'll be able to solve this puzzle without hitting a mine? There are 7 mines remaining overall in the puzzle.

Bonus:

After opening some further squares based on the clues, can you come up with the solution for this corner? How far can you solve the other corner based on this?

• It's easy just click on thBOOM – LinuxBlanket Oct 13 '18 at 12:08

Mmmmmmm minesweeper

Legend: S - Safe M - Mine

The 7 mines are at

Bottom Left(R2C2, R2C5)
Bottom Right(R2C5, 2 between R3C5/R3C6/R5C5/R5C5, 2 at either R4C3&R5C4 or R4C4&R5C2)

12222M2
??M??4M
?????M2
This suggests a minimum of 2 mines here on second Row R2C1/C2 & R2C4/C5

& Final answer for bottom left (read this after reading Bottom right portion)

Since we have confirmed that there are 5 mines at bottom right, bottom left has to have only have 2 mines. meaning remaining squares in R3 are all safe 12222M2
SMMSM4M
SSSSSM2
Since R2C6 dictates that R2C5 has to be a mine, and the rest follows.

& for the bottom right

1MM4M2
25M???
3MM???
MM??32
M?3???

Around R4C6 there are 2 mines (within R3C5 R3C6, R5C5 R5C6)
& 66% chance that there is a mine at R4/R5C4 and a 33% chance that it is at R3C4

However if it is at R3C4, R4C3 & R5C2 have to be mines to fulfill the 3 mine criteria at R5C3 Since amongst R2C4/C5/C6, there is a 50% chance there is 1 mine (C5) or 2 mines (C4,C6), this brings the total count of mines in this map to be 5/6 But 6 can't be right! Because we have concluded that there are minimum 2 mines in Bottom Left map earlier

Hence it is confident that R2C5 is a mine and R2C4/C6 are safe.

Out of the remaining squares

1MM4M2
25MSMS
3MMS??
MM??32
M?3???
R2C4 will reveal if R3C5 is a M(6) or S(5)
Subsequently, R2C6 will reveal R3C6

Taking a breakdown from R3 onwards,

4 sub-scenarios presents itself for R3C5/C6:
1.
3MMSMM
MM??32
M?3???
R5C5/C6 are both safe and will be 1/0 respectively
Depending whether R3C4 is 2/3/4, there are various subcases.
1.1 R3C4= 2
100% chance mines are at R5C2/C4 1.2 R3C4= 3
50% chance whether it is R4C3&R5C4 or R4C4&R5C2
1.3 R3C4= 4
100% chance mines are at R4C3/C4

Cumulative probability : 83.33%

2.
3MMSMS
MM??32
M?3???
Arguably the worst-case scenario. It is the same guessing effort for the 4 squares as per above, with additional guessing effort for R5C5/C6
2.1 R3C4= 2
Answer for R5C5 will be revealed by R4C4 (100%)
2.2 R3C4 = 3
Strategy will be to click R4C4 or R5C4. If this is not a mine, it will reveal R5C5 (50%)
2.3 R3C4 = 4
Best move will be to click R5C4. If this is not a mine, it will reveal R5C5 (50%)
Cumulative probability: 66.67%

3.
3MMSSM
MM??32
M?3???
R4C4 is revealed by R3C5 (M if it is 2, S if it is 1) and R4C3 is subsequently revealed. The crucial 4 squares as per Scenario 1/2 above could be derived using the same pairing logic (2 mines always at R4C3&R5C4 or R4C4&R5C2). R5C5 could then be derived from R4C4 or R5C4.
Cumulative probability: 100%

4.
3MMSSS
MM??32
M?3???
It becomes clear that R5C5/C6 are mines. R3C4 & R3C5 will determine R4C3/C4 .... and the rest follow the above logic.
Cumulative probability: 100%

Therefore, total probability of solving is :

87.5%

Okay I have to take a break after this lengthy post phew.

Assumptions: I am not entirely sure if Scenario 1 could exist, as if both R5C5/C6 are S, would this be already revealed by clicking R4C5/C6 (which has been done by TS). Would require some more frequent minesweeper players to confirm this.