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I know we can try and count the number of ways that they can add to 20, however this is much easier when there is only 2 or 3 dice. I believe there is a way to do this using combinations but I am not sure. I thought it was a stars and bars, so I wrote out 20 like -> 1 + 1 + 1 + 1 |+ 1 + 1 + 1 + 1| + 1 + 1 + 1 + 1 + 1 + 1 |+ 1 + 1 + 1 |+ 1 + 1 + 1

Which makes a 20 choose 4, however, this is over counting as each block of 1's can't be greater than 6 or less than 1.

Thanks for your help

Sam Cole
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    If you don't mind using computer assistance... it would be the coefficient of $x^{20}$ in the expansion of $\frac{1}{6^5}(x+x^2+x^3+x^4+x^5+x^6)^5$ – JMoravitz Jan 30 '20 at 16:22
  • @JMoravitz That works, Can you explain a little why that is correct? I am shocked that got the same answer. Thanks! – Sam Cole Jan 30 '20 at 16:42
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    Let us use some colors and make the example smaller... imagine we roll a three sided die instead three times... $(x+x^2+x^3)^3 = \color{red}{(x+x^2+x^3)}\color{blue}{(x+x^2+x^3)}\color{green}{(x+x^2+x^3)}$ after foiling and without simplifying will look like $\color{red}{x}\color{blue}{x}\color{green}{x}+\color{red}{x^2}\color{blue}{x}\color{green}{x}+\color{red}{x}\color{blue}{x^2}\color{green}{x}+\dots+\color{red}{x^3}\color{blue}{x^3}\color{green}{x^3}$. Each term in the expansion before simplification refers to a specific sequence of die rolls. – JMoravitz Jan 30 '20 at 16:48
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    Each of these outcomes in the pre-simplified pre-aggregated expansion are equally likely to have occurred and there are $3^3=27$ of them in that example (or more generally $k^n$ of them for a $k$-sided die being rolled $n$ times). After having simplified and aggregated the terms, the coefficients there are the number of ways of having rolled a total of that amount having "forgotten" which numbers actually appeared and in what order. Dividing by $k^n$ then gives the probability of having achieved that sum. – JMoravitz Jan 30 '20 at 16:50
  • have a look at this [related post][http://math.stackexchange.com/questions/992125/rolling-dice-problem/1680420#1680420] – G Cab Jan 30 '20 at 19:06

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It can be done without a computer. As explained by others, and in the linked question, the generating function of your problem is given by $$\eqalign{ g(x)&=(x+x^2+x^3+x^4+x^5+x^6)^5 \cr &=x^5\left({1-x^6\over 1-x}\right)^5=x^5(1-x^6)^5(1-x)^{-5}\cr &=x^5\ (1-5x^6+10x^{12}+{\rm higher\ terms})\ \sum_{j=0}^\infty{5+j-1\choose j}x^j\ .\cr}$$ Now you have to find the coefficient of $x^{20}$ on the RHS. This coefficient is $${19\choose15}-5{13\choose9}+10{7\choose3}=651\ .$$ This is the number of lucky outcomes among the $6^5$ possible, and equally probable, histories.

Christian Blatter
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  • There is a "-?" there, what is it supposed to be? – HellSaint Jan 30 '20 at 17:14
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    @HellSaint: It collects the terms in $(1-x^6)^5$ that have no effect in our problem, since their degrees are too large. – Christian Blatter Jan 30 '20 at 18:47
  • I am confused on how you got the coefficient for the x^5 term. I guess I am mostly confused on how you got j and also the first step when you get 1-x^6)/(1-x), could you explain this? – Sam Cole Jan 30 '20 at 18:57
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    @SamCole: $1+x+x^2+\ldots+x^5$ is a finite geometric series with a summation formula. $(1-x)^{-5}$ is computed with the binomial series. – Christian Blatter Jan 30 '20 at 19:06