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

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

For example, suppose my threshold x=3x=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 46\frac{4}{6}. The number of successes need is 4. The number of dice is 10. So let p=46p=\frac{4}{6}, and n=10n=10. Thus the probability we will get 4 successes is:

k=4n(nk)(p)k(1p)nk=k=410(10k)(46)k(146)10k98%  \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 xx and have it count as a success.

For example, suppose I roll three six sided die. Let viv_i represent the value of the die, and let v1=2v_1=2, v2=3v_2=3, and v3=6v_3=6 . Suppose I have a threshold ofx=4 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 v1v_1 and v2v_2such that I now have three successes instead of only one (where now v1=2+2=4v_1=2+2=4, v2=3+1=4v_2=3+1=4, and v3=6 v_3=6.)

My question is: How are the probabilities of rolling nn dice that are each are greater than or equal to xx 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:
fDdkh9K.png


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):

k=410(10k)(26)k(46)10k=86751968344%  .\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:

(101)(16)1(93)(26)3(36)6=(101)(16)1(93)(26)3(12616)6=354328%  . \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 16\frac{1}{6} from the probability of 26\frac{2}{6} (in the at least 5 three times probability) as we know that the probability of failure isn't truly 46\frac{4}{6}; we account for the 4's, which we only want to roll one of. So the probaiblity of failure is only 36\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:

(102)(82)(16)2(26)2(36)6+(101)(93)(16)1(26)3(26)6=(102)(16)2(82)(26)2(12616)82+(101)(16)1(93)(26)3(12626)93=8550512597127%  . \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:


(103)(71)(16)3(26)1(36)6+(101)(91)(82)(16)1(16)1(26)2(26)6+(101)(93)(16)1(26)3(16)6=(103)(16)3(71)(26)1(12616)71+(101)(16)1(91)(16)1(82)(26)2(12626)82+(101)(16)1(93)(26)3(12636)93=3909512597123%  . \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:

867519683+35432+855051259712+390951259712=6515510497662%  .\frac{8675}{19683}+\frac{35}{432}+\frac{85505}{1259712}+\frac{39095}{1259712}=\frac{65155}{104976}\approx62\%\;.
 
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