Context: I am trying to figure out the probability of rolling doubles in sets on a given number of dice. I want to compare this to a standard probability of rolling doubles on 4d6 for a game I am running to see if its reasonable or not for my players (an RPG).
Question: I have the following 10 dice pools - 1d6, 2d6, 3d6, 4d6, 5d6, 7d6, 8d6, 9d6 and 10d6. And I want to figure out the odds of rolling doubles, doubles x2 (two sets of doubles) and doubles x3 (three sets of doubles) per dice pool.
Doubles can be of any number. Don't care about triples or quads etc. Just sets of doubles.
Some are obvious - cant roll doubles on 1d6 so probability is zero. While others have me scratching my head - chance of rolling 3 sets of doubles on 10d6.
Thanks in advance.
As I see it, it's most naturally approached via recursion, and then implemented as a program.
Let $p(n,f,d)$ be the probability of getting $d$ doubles with $n$ dice, where each die has $f$ equally likely faces with values $1,...,f$.
For example $p(10,6,3)$ is the probability of getting $3$ doubles with $10$ standard dice.
Then we have the recursion $$ p(n,f,d)= \left\lbrace \begin{align*} &\text{if}\;n\le 1\;\text{then}\\[4pt] &\;\;\;\;\;\;\text{if}\;d=0\;\text{then}\;1\;\text{else}\;0\\[4pt] &\text{else if}\;n < 2d\;\,\text{then}\\[6pt] &\;\;\;\;\;\;0\\[0pt] &\text{else if}\;d < \left\lfloor\frac{n-(f-1)}{2}\right\rfloor\;\,\text{then}\\[0pt] &\;\;\;\;\;\;0\\[4pt] &\text{else if}\;f=1\;\text{then}\\[4pt] &\;\;\;\;\;\;\text{if}\;d=\left\lfloor\frac{n}{2}\right\rfloor\;\,\text{then}\;1\;\text{else}\;0\\[0pt] &\text{else}\\[4pt] &\;\;\;\;\;\; \sum_{x=0}^{\min(n,2d+1)} \binom{n}{x} \Bigl(\frac{1}{f}\Bigr)^x \Bigl(\frac{f-1}{f}\Bigr)^{n-x} p\Bigl(n-x,f-1,d-\left\lfloor\frac{x}{2}\right\rfloor\Bigr) \\[4pt] \end{align*} \right. $$ Implemented in Maple, for the example case of $10$ standard dice we get $$ \begin{array}{|c|c|} \hline d&p(10,6,d)\\ \hline 0&0\\ \hline 1&0\\ \hline 2&{\large{\frac{245}{8748}}}\approx .028\\ \hline 3&{\large{\frac{11875}{26244}}}\approx .452\;\\ \hline 4&{\large{\frac{38185}{78732}}}\approx .485\;\\ \hline 5&{\large{\frac{2717}{78732}}}\approx .035\;\\ \hline \end{array} $$