Calculating odds of Minesweeper is this correct?

This is slightly unnecessarily complicated.

There are two partly identified sections on the board, and in the right-hand one (with the $8$ red squares) the total of mines is known to be $1$. Thus, each red square has probability $\frac18=0.125$ of containing a mine, and we can deduct this one mine from the total of $25$ mines and calculate the left-hand section using $t=124-8=116$ and $m=25-1=24$ in my answer to the other question (with $s=15$ and $n=3,4,5$ as before). Note the correction to that answer (which you didn't have a chance to take into account yet in this calculation).

Some errors in the post:

There are some percent signs where there shouldn't be any; all the numbers seem to be probabilities out of $1$, not out of $100$.

It's not generally the case that the player should click on the square with the lowest marginal probability of containing a mine. The optimal strategy also depends on future opportunities to gain information. For instance, in the extreme case, there's no use in clicking on a square, no matter how low its marginal mine probability, if you already know that you won't gain any information by doing so.

This is also not correct:

For example, having 4 mines in total is more likely than having 5 mines in total. We know this without knowing the total mines because in Minesweeper there are always a greater number of unmined cells than mined cells.

If you have a total of $t=100$ unidentified squares with $m=20$ mines in them, even though there are a lot more unmined cells than mined cells, if you assign mines to $s=30$ squares you'd expect about $6$ mines to be in those $30$ squares, and solutions with $5$ mines will be more probable than solutions with $4$ mines.

I would like to give a (somewhat long-winded) discussion of an excerpt from Joriki's answer.

$\underline{\text{Excerpt}}$

This is also not correct:

For example, having 4 mines in total is more likely than having 5 mines in total. We know this without knowing the total mines because in Minesweeper there are always a greater number of unmined cells than mined cells.

If you have a total of t=100 unidentified squares with m=20 mines in them, even though there are a lot more unmined cells than mined cells, if you assign mines to s=30 squares you'd expect about 6 mines to be in those 30 squares, and solutions with 5 mines will be more probable than solutions with 4 mines.

$\underline{\text{Discussion}}$

I agree with Joriki's analysis (excerpted above). However...

I regard the intuition around assuming that a region is more likely to have 4 mines than 5 as reasonable. The reason that it is in error in Joriki's example is that the region in his example is a relatively large percentage of the number of remaining unidentified squares.

Joriki's rebuttal was based on $\;1/5 \times 30 = 6\;$ and 6 is closer to 5 than 4. Based on this approach, one would guesstimate that if the region is 22 instead of 30, since $\;1/5 \times 22 < 4.5\;$, the chances of the region (of 22 unknown squares) having 4 mines would be slightly larger than the chances of this region having 5 mines.

Therefore, for a significantly smaller region, (e.g. a region significantly smaller than 22 unknown squares), 4 mines is more likely than 5 mines. So the assertion (i.e. 4 mines is more likely than 5 mines), will generally hold for smaller regions of unknown squares.