Hello,
I am currently analyzing a tabletop games' probabilities. Successful rolls in the game are determined by rolling [imath]n[/imath] number of dice and counting the number of dice that are greater than or equal to a certain value (some pre-determined threshold [imath]x[/imath]; for instance, values could be between 1 through 6 from a six sided die).
For example, suppose my threshold [imath]x=3[/imath]; 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 [imath]\frac{4}{6}[/imath]. The number of successes need is 4. The number of dice is 10. So let [imath]p=\frac{4}{6}[/imath], and [imath]n=10[/imath]. Thus the probability we will get 4 successes is:
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 [imath]x[/imath] and have it count as a success.
For example, suppose I roll three six sided die. Let [imath]v_i[/imath] represent the value of the die, and let [imath]v_1=2[/imath], [imath]v_2=3[/imath], and [imath]v_3=6[/imath] . Suppose I have a threshold of[imath] x=4 [/imath] 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 [imath]v_1[/imath] and [imath]v_2[/imath]such that I now have three successes instead of only one (where now [imath]v_1=2+2=4[/imath], [imath]v_2=3+1=4[/imath], and [imath] v_3=6[/imath].)
My question is: How are the probabilities of rolling [imath]n[/imath] dice that are each are greater than or equal to [imath]x[/imath] 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):
Using 1 point from the pool if you roll 4 once and at least 5 three times:
Using 2 points, if you roll 4 twice and at least 5 twice; or 3 once, at least 5 three times and not 4:
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:
Therefore the probability of getting at least 4 die to have a value of at least 5 with 3 points to distribute is:
I am currently analyzing a tabletop games' probabilities. Successful rolls in the game are determined by rolling [imath]n[/imath] number of dice and counting the number of dice that are greater than or equal to a certain value (some pre-determined threshold [imath]x[/imath]; for instance, values could be between 1 through 6 from a six sided die).
For example, suppose my threshold [imath]x=3[/imath]; 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 [imath]\frac{4}{6}[/imath]. The number of successes need is 4. The number of dice is 10. So let [imath]p=\frac{4}{6}[/imath], and [imath]n=10[/imath]. Thus the probability we will get 4 successes is:
[imath]\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\%\;[/imath]
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 [imath]x[/imath] and have it count as a success.
For example, suppose I roll three six sided die. Let [imath]v_i[/imath] represent the value of the die, and let [imath]v_1=2[/imath], [imath]v_2=3[/imath], and [imath]v_3=6[/imath] . Suppose I have a threshold of[imath] x=4 [/imath] 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 [imath]v_1[/imath] and [imath]v_2[/imath]such that I now have three successes instead of only one (where now [imath]v_1=2+2=4[/imath], [imath]v_2=3+1=4[/imath], and [imath] v_3=6[/imath].)
My question is: How are the probabilities of rolling [imath]n[/imath] dice that are each are greater than or equal to [imath]x[/imath] 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):
[imath]\sum_{k=4}^{10}\binom{10}k\left(\frac26\right)^k\left(\frac46\right)^{10-k}=\frac{8675}{19683}\approx44\%\;.[/imath]
Using 1 point from the pool if you roll 4 once and at least 5 three times:
[imath]
\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\%\;.
[/imath]
We subtract by [imath]\frac{1}{6}[/imath] from the probability of [imath]\frac{2}{6}[/imath] (in the at least 5 three times probability) as we know that the probability of failure isn't truly [imath]\frac{4}{6}[/imath]; we account for the 4's, which we only want to roll one of. So the probaiblity of failure is only [imath]\frac{3}{6}[/imath] 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:
[imath]
\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\%\;.
[/imath]
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:
[imath]
\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\%\;.
[/imath]
Therefore the probability of getting at least 4 die to have a value of at least 5 with 3 points to distribute is:
[imath]\frac{8675}{19683}+\frac{35}{432}+\frac{85505}{1259712}+\frac{39095}{1259712}=\frac{65155}{104976}\approx62\%\;.[/imath]