Calculating Minesweeper Global Probability

Question

In an Expert Minesweeper game, there is a 30x16 board with 99 mines. In this situation, what is the global probability of the flagged tile being a mine without opening other tiles?

Given that the local probability is 50%, with the 2 possible situations being the situation I posted earlier and this, what are the global probability of the first situation happening, taking into account the entire board? I've seen various ways of calculating global probability for minesweeper but I'm not sure which one is more appropriate/correct.

We know from simple deduction that the tiles marked green are safe, while the tiles marked red can have either one or two mines as shown in the last 2 images.

Answer 1

There are either 97 or 96 mines to distribute on 30*16-15 tiles. The probability that there is only one mine in the red area is then $\frac{{465\choose 97}}{{465\choose 97}+{465\choose 96}}$.

Answer 2

The position has 2 equally likely local solutions using either 2 or 3 bombs. The red cell will be a bomb if the local solution has 2 bombs. There are 30*16-15=465 other tiles where all the other bombs must be distributed. There are (465 choose 97) global solutions where the red cell is a bomb and (465 choose 96) global solutions where it is not a bomb. Since it must be one or the other, this accounts for all global solutions. So the probability that the red cell is a bomb is equal to
(465 choose 97) / ((465 choose 97) + (465 choose 96)) = 0.79185

Answer 3

I feel that neither of the two existing (correct) answers clarifies what seems to be most in need of clarification here, namely, what to count as equiprobable. Every arrangement of the mines that's consistent with the given information is equiprobable. There are $\binom{465}{97}$ such arrangements in which the flagged tile is a mine, since this implies that the two other unknown tiles aren't mines and the remaining $97$ mines can be distributed in $\binom{465}{97}$ different ways over the remaining $465$ tiles. There are a further $\binom{465}{96}$ arrangements in which the flagged tile isn't a mine, since this implies that the two other unknown tiles are mines, and the remaining $96$ mines can be distributed in $\binom{465}{96}$ different ways over the remaining $465$ tiles. This exhausts the arrangements that are consistent with the given information. Since they're all equiprobable, the probability for the flagged tile to be a mine is

$$ \frac{\binom{465}{97}}{\binom{465}{97}+\binom{465}{96}}=\frac{\binom{465}{97}}{\binom{466}{97}}=\frac{369}{466}\approx79.2\%\;. $$