A die is thrown five times, what is the probability that you get 20 as the sum of the values

Question

This is supposed to be a Inclusion-Exclusion problem.

We have $6^5=7776$ different results.

Now, with the Inclusion-Exclusion principle i resolve the number of solutions for the equation:

$d_1+d_2+d_3+d_4+d_5=20 , 1 \leq d_i \leq 6 ,\forall 1 \leq i \leq 5$

That, i think, is equivalent to resolve:

$d_1+d_2+d_3+d_4+d_5=15 , 0 \leq d_i \leq 5 ,\forall 1 \leq i \leq 5$

This is:

$\binom{19}{15}-\binom{5}{1} \binom{13}{9}+\binom{5}{2} \binom{7}{3}=651$

So mi answer is: $\frac{651}{7776}$.

However, the "correct answer" is: $\frac{116}{7776}$. Whats the problem with mi reasoning?

EDIT I found the solutions for my problem in the book, the result is: $\frac{651}{7776}$.

I was right, but the solution from mi teacher was wrong. Thats why the confusion.

Answer 1

You can calculate this by finding the coefficient on $x^{20}$ in the polynomial $$ \left(\frac{1}{6}x+\frac{1}{6}x^2+\frac{1}{6}x^3+\frac{1}{6}x^4+\frac{1}{6}x^5+\frac{1}{6}x^6 \right)^5 $$ Using PARI/GP, I get that this polynomial is $$\frac{1}{7776} x^{30} + \frac{5}{7776} x^{29} + \frac{5}{2592} x^{28} + \frac{35}{7776} x^{27} + \frac{35}{3888} x^{26} + \frac{7}{432} x^{25} + \frac{205}{7776} x^{24} + \frac{305}{7776} x^{23} + \frac{35}{648} x^{22} + \frac{5}{72} x^{21} + \frac{217}{2592} x^{20} + \frac{245}{2592} x^{19} + \frac{65}{648} x^{18} + \frac{65}{648} x^{17} + \frac{245}{2592} x^{16} + \frac{217}{2592} x^{15} + \frac{5}{72} x^{14} + \frac{35}{648} x^{13} + \frac{305}{7776} x^{12} + \frac{205}{7776} x^{11} + \frac{7}{432} x^{10} + \frac{35}{3888} x^9 + \frac{35}{7776} x^8 + \frac{5}{2592} x^7 + \frac{5}{7776} x^6 + \frac{1}{7776} x^5 $$ This shows that your value, $\frac{651}{7776}=\frac{217}{2592}$ is correct.

Answer 2

Using generating functions, we get a generating function of $$f(x) = 1+x+\ldots+x^5 = \frac{1-x^6}{1-x}$$ for each variable so we are now looking for the coefficient of $x^{15}$ in $f(x)^5$: $$ \begin{split} \left[x^{15}\right]\frac{\left(1-x^6\right)^5}{(1-x)^5} &= \left[x^{15}\right]\frac{1-5x^6+10x^{12}}{(1-x)^5} \\ &= \left[ \left[x^{15}\right] -5\left[x^{9}\right] + 10 \left[x^{3}\right]\right] \frac{1}{(1-x)^5}\\ &= 3876 - 5 \cdot 715 + 10 \cdot 35 \\ &= 651. \end{split} $$