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Using $5$ $6$-faced dice, what's the probability of hitting a cumulative sum $S$, on a particular roll (each roll incorporates all $5$ dice), if $15 \leq S\leq 20$?

I reckon this particular situation could be considered equivalent to a similar set of circumstances in which we'd roll the same die $5$ consecutive times and take the total sum.

I thought the way to a possible solution would be to consider the following generating function :

$$f(x)=\frac{1}{6^{5}}(x+x^2+x^3+x^4+x^5+x^6)^5$$

Next up, I believe I should compute the sum of polynomial coefficients going from degree $15$ up to $20$, but I'm not sure this is the way, so please share your thoughts.

But if I'm right, how can I calculate this tedious sum?

ViHdzP
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1 Answers1

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Answering your actual question, how to calculate the tedious sum, you can use a computer algebra software such as Wolfram alpha to get the exact solution, $361/648$.

Another option is to use dynamic programming directly. Let $P(n,S)$ be the probability that the sum of $n$ cubes equals $S$. Then $P(0,0) = 1$, $P(0,S) = 0$ for $S \neq 0$, and for $n \geq 1$, $$ P(n,S) = \frac{1}{6} \sum_{s=1}^6 P(n-1,S-s). $$ You can implement this recurrence using dynamic programming efficiently.

Yuval Filmus
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