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.

This is a (somewhat long-winded) reaction to part of the OP's revised post. My analysis only covers the query up to AND INCLUDING the following excerpt. I ask the OP to consider my reaction, and consider making another edit to his post. Then, I can (iteratively) examine the rest of his query.

$\underline{\text{Excerpt}}$

To normalize the weights:

3 * 2577824781465941808570 = 7.7334743e+21
2 * 668324943343021950370 = 1.3366499e+21
1 * 163006083742200475700 = 163006083742200475700 

7.7334743e+21 + 1.3366499e+21 + 163006083742200475700  = 9.2331303e+21

2577824781465941808570 / 9.2331303e+21 = 0.279%
668324943343021950370 / 9.2331303e+21 = 0.072%
163006083742200475700 / 9.2331303e+21 = 0.018%

So, for each 'scenario' for section1, we can assign the weights:

4 mine scenarios weight = 0.279%
5 mine scenarios weight = 0.072%
6 mine scenarios weight = 0.018%

$\underline{\text{My Reactions}}$

1) As Joriki already indicated, equations like

2577824781465941808570 / 9.2331303e+21 = 0.279%

are confusing, because of the "%" sign. I would eliminate all percent signs from your query, altering equations as needed.

2) Instead of defining $104 \times 8 = 832$ cases, you stuck with 104 cases, capitalizing on the fact that the RSTUVXYZ cells always collectively have 1 mine. This approach is doable but somewhat dangerous, since your intuition is evolving. The rest of my reaction (below) accepts your use of 104 cases, instead of 832 cases.

3) The math you are using is somewhat messy. I would advise labeling your weights as $\binom{101}{21}, \binom{101}{20}, \;\text{and}\; \binom{101}{19}.$ Eventually you will convert to relative weights, so you will avoid messy math. Unfortunately, it is premature to discuss relative weights, because of the issue in my next point (below). After you have edited your query, I will return to the issue of relative weights.

4) If I am not misinterpreting the Excerpt from your query, you are confused about how to assign the weights. You have delineated 104 cases: one of the weights [i.e. $\;\binom{101}{21}, \binom{101}{20}, \;\text{or}\; \binom{101}{19}]\;$ should be assigned to each of the 104 cases. In each of the 104 cases, the weight assigned will depend on how many mines are assigned to Section1 in that case.

Assume that case 1 is assigned W1 (i.e. the weight for case 1), case 2 is assigned W2 (the weight for case 2), ..., case 104 is assigned W104. Here, each of W1, W2, ..., W104 is one of the 3 values : $\;\binom{101}{21}, \binom{101}{20}, \;\text{or}\; \binom{101}{19}]\;$.

Let SUM_OF_WEIGHTS = W1 + W2 + ... + W104.
Let x be any integer in {1, 2, ..., 104}.

Then the probability of Case x is

{Wx : the weight assigned to case x} / {SUM_OF_WEIGHTS}.

Note that under this definition, the
(probability of case 1) + (probability of case 2) + ... + (probability of case 104) = 1.

5) Deferred issues until your posting is edited:

5a) Converting to relative weights to avoid messy math.

5b) Taking the next step in the analysis: once you assign a probability to each of the 104 cases, you then have to decide how to use this information to decide which minesweeper cell to click on. Analysis for taking the next step will probably be so complex as to defy using mathematics. You will probably have to merely consider what factors are relevant and then make an intuition-based judgement call.