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)
Lets start with bottom left
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
Since R2C6 dictates that R2C5 has to be a mine, and the rest follows.
& for the bottom right
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
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:
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%
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%
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%
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 :
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.
Adding for bonus:
As mentioned in my answer, there is no guesswork required to reach the additional step provided by TS. Probability remains the same (:
The scenario arrived by TS currently is Scenario 3, which means that it is 100% possible to solve it from here onwards.