Probability of rolling 'n' dice that are each are greater than or equal to 'x', with a given pool of possible positive modifiers applied on each dice

[SM123]

Hello,
I am currently analyzing a tabletop games' probabilities. Successful rolls in the game are determined by rolling $n$ number of dice and counting the number of dice that are greater than or equal to a certain value (some pre-determined threshold $x$; for instance, values could be between 1 through 6 from a six sided die).

For example, suppose my threshold $x=3$; if I roll 10 six sided die, what is the probability that 4 dice would land on a 3 or greater? For this question, we would use the binomial distribution to solve:

The probability to roll a success is $\frac{4}{6}$. The number of successes need is 4. The number of dice is 10. So let $p=\frac{4}{6}$, and $n=10$. Thus the probability we will get 4 successes is:

$\sum_{k=4}^{n}\binom{n}k\left(p\right)^k\left(1-p\right)^{n-k}=\sum_{k=4}^{10}\binom{10}k\left(\frac46\right)^k\left(1-\frac46\right)^{10-k}\approx98\%\;$​

Additionally, the game allows for a positive pool of modifiers to be applied on the dice after they are rolled, such that it can bring them over the determined threshold $x$ and have it count as a success.

For example, suppose I roll three six sided die. Let $v_i$ represent the value of the die, and let $v_1=2$, $v_2=3$, and $v_3=6$ . Suppose I have a threshold of$x=4$ have a pool of +3 points to apply on any one of those die. I would be able to distribute all those points on both $v_1$ and $v_2$such that I now have three successes instead of only one (where now $v_1=2+2=4$, $v_2=3+1=4$, and $v_3=6$.)

My question is: How are the probabilities of rolling $n$ dice that are each are greater than or equal to $x$ affected by the number of points in the modifier pool (described using a probability distribution where the threshold, modifiers, and number of successes are taken as parameters)?

I have attempted to develop a formula/distribution specifically to answer the last question above, such that I can analyze the distribution of a certain number of dice with different thresholds. I started with attempting to modify a binomial distribution, however, I am uncertain how it is affected by the different combinations of modifiers, thresholds, and successes needed. I've asked this question before on stackexchange where I was kindly suggested a case scenario (at this link), however, attempting to create a distribution from this case example has proven difficult.

Here is a visual I made in excel, showing the previous examples calculation done in excel as well as what is still needed:

To showcase this, the case scenario mentioned was for an example of 10 six-sided dice where you need 4 dice to show at least 5 and you have 3 modifier points that you are able distribute. I have noticed that there is an expanding pattern as I expanded the case scenario; each case was divided by a number of modifier points used, as they are disjoint and can be added together (i.e. having to use no points, 1 point, etc.). Here, I've cleaned up some of the work so that the pattern can be seen more clearly:

The probability that you get 4 dice without having to use points (this is done similiarly as the previous binomial question without the additional modifiers):

$\sum_{k=4}^{10}\binom{10}k\left(\frac26\right)^k\left(\frac46\right)^{10-k}=\frac{8675}{19683}\approx44\%\;.$​

Using 1 point from the pool if you roll 4 once and at least 5 three times:

$\binom{10}{1}\left(\frac16\right)^1*\binom93\left(\frac26\right)^3\left(\frac36\right)^6 = \binom{10}{1}\left(\frac16\right)^1*\binom93\left(\frac26\right)^3\left(1-\color{Red}\frac26-\frac16\right)^6 =\frac{35}{432}\approx8\%\;.$
​

We subtract by $\frac{1}{6}$ from the probability of $\frac{2}{6}$ (in the at least 5 three times probability) as we know that the probability of failure isn't truly $\frac{4}{6}$; we account for the 4's, which we only want to roll one of. So the probaiblity of failure is only $\frac{3}{6}$ in the at least 5 three times probability. This pattern also expands to the rest of the cases (highlighted red above and below).

Using 2 points, if you roll 4 twice and at least 5 twice; or 3 once, at least 5 three times and not 4:

$\binom{10}2\binom82\left(\frac16\right)^2\left(\frac26\right)^2\left(\frac36\right)^6+\binom{10}1\binom93\left(\frac16\right)^1\left(\frac26\right)^3\left(\frac26\right)^6 \\= \binom{10}2\left(\frac16\right)^2*\binom82\left(\frac26\right)^2\left(1-\color{Red}\frac26-\frac16\right)^{8-2}+\binom{10}1\left(\frac16\right)^1*\binom93\left(\frac26\right)^3\left(1-\color{Red}\frac26-\frac26\right)^{9-3} \\= \frac{85505}{1259712}\approx7\%\;.$​

Using 3 points, if you roll 4 three times and at least 5 once; or 3 once, 4 once and at least 5 twice; or 2 once, at least 5 three times and not 3 or 4:

$\binom{10}3\binom71\left(\frac16\right)^3\left(\frac26\right)^1\left(\frac36\right)^6 +\binom{10}1\binom91\binom82\left(\frac16\right)^1\left(\frac16\right)^1\left(\frac26\right)^2\left(\frac26\right)^6 +\binom{10}1\binom93\left(\frac16\right)^1\left(\frac26\right)^3\left(\frac16\right)^6 \\= \binom{10}3\left(\frac16\right)^3*\binom71\left(\frac26\right)^1\left(1-\color{Red}\frac26-\frac16\right)^{7-1} +\binom{10}1\left(\frac16\right)^1*\binom91\left(\frac16\right)^1*\binom82\left(\frac26\right)^2\left(1-\color{Red}\frac26-\frac26\right)^{8-2} +\binom{10}1\left(\frac16\right)^1*\binom93\left(\frac26\right)^3\left(1-\color{Red}\frac26-\frac36\right)^{9-3} \\= \frac{39095}{1259712}\approx3\%\;.$​

Therefore the probability of getting at least 4 die to have a value of at least 5 with 3 points to distribute is:

$\frac{8675}{19683}+\frac{35}{432}+\frac{85505}{1259712}+\frac{39095}{1259712}=\frac{65155}{104976}\approx62\%\;.$
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