Good and bad numbers of remaining mines

Question

You've been tasked with finishing solving this Minesweeper board:

"How many mines remain?", you ask. "I'm just choosing that now, actually. Tell you what: I was going to consider every possible answer to that equally likely and choose one at random, but I'll let you have a bit of an advantage. Choose one answer to make impossible and another answer to give its probability to. (For example, suppose the possibilities for the number of bombs were either 1, 2, or 3. You could choose to eliminate 2 and give its probability to 3. Now there's a 1/3 chance the board has 1 bomb and a 2/3 chance the board has 3 bombs.)

Once you do that, I'll make the random choice, tell you the answer, and then randomly place that many mines in the remaining 8 squares (consistent with your existing clues, of course)."

What number of mines remaining should you make impossible? What number should you give its probability to? And what's your best strategy for solving the resulting games?

Answer 1

Seems pretty straight forward:

The minimum number of mines is 2.
Label the unopened cells 1-4 left to right on the top row and 5-8 left to right on the bottom row. There are 3 possibilities for 2 mines: 5 and 8; 1 and 6; 1 and 7. If the number of mines is 2 you can open cells 2, 3 and 4 and determine the configuration with no guessing. The probability of winning win this case is 1. This is the best scenario so give it the max probability by adding the eliminated probability to it.

The max number of mines is 5.
This is the 2 required mines plus all 3 in the cells 2, 3 and 4. The best strategy here is to open a cell that has the lowest probability of a mine and that gives you the most information. This is either cell 5 or 6. Both give the same result in the end, but I personally would go with cell 5, because if that cell is empty then so is cell 8. Out of the 3 possibilities for the positions of the 2 required mines, cell 5 is empty 2/3 of the time so I will win with a probability of $\frac{2}{3}$ in this case.

Last possibilities are for there to be either 3 or 4 mines.
For the case of 3 mines the third mine will be in either cell 2, 3 or 4. Using my strategy of opening cells 5 and 8, I would receive information on all 6 possibilities (there are 9 total, but in 3 of them I would die immediately). 3 of the 6 possibilities are unique which will allow me to win the game. The other 3 are the same: 5 and 2 in cells 5 and 8. In this case I would open cell 2 - out of the 3 non-unique probabilities cell 2 is empty 2/3 of the time. This will allow me to figure out the remaining mines. The total probability of me winning in this case is $\frac{2}{3} * (\frac{1}{2} * 1 + \frac{1}{2} * \frac{2}{3}) = \frac{5}{9}$.

For the case of 4 mines they will be distributed in either cells 2,3; 2,4; or 3,4. Again, using my strategy of opening cells 5 and 8, out of the 6 possibilities that I live 1 is unique, 2 are the same (5 and 2 in cells 5 and 8), and 3 are also the same (4 and 3 in cells 5 and 8). If I get the result of 5 and 2, I will open cell number 4. This is a 50% chance. If I get the result of 4 and 3, I will open cell 6. Cell 6 is empty in 2 out of the 3 cases. For a total probability in this case of $\frac{2}{3} * (\frac{1}{6} * 1 + \frac{2}{6} * \frac{1}{2} + \frac{1}{2} * \frac{2}{3}) = \frac{4}{9}$.

The lowest chance of me winning is if there are 4 mines. So I will eliminate the possibility of there being 4 mines and add that probability to there being 2 mines. I will initially open cells 5 and 8 which will determine my next action in either opening cell 2, 4 or 6.

I will win with a probability of $\frac{1}{2}*1 + \frac{1}{4} * \frac{2}{3} + \frac{1}{4}*\frac{5}{9} = \frac{29}{36} \approx 80\%$.

P.S.

I did go through with the strategy of opening cell 6 instead of 5 and 8 and the probabilities came out to be the same, so that's not any better.