What is the probability of when four dice rolled together once,and getting a sum of Thirteen
If we do by just calculating all possible values of sum, then it will take more time; so we can solve the above problem as Multinomial Coefficents of sum, i.e.: $$ x_1+x_2+x_3+x_4 = 13\;\text{ where },1\leq x_i \leq 6\;\;\forall\;1\leq i \leq 4. $$ I got stuck that form the above equation how can we obtain $N_{\text{ways to get $\sum=13$}}$ or is there any easy method to find this kind of problems?
On
I assume that you already know how to calculate the probability for the sum of two dices. One gets the following table. $$ \begin{array}{r|c} k & p(x_1+x_2=k) \\ \hline{} \\ 2 & 1/36 \\ 3 & 2/36 \\ 4 & 3/36 \\ 5 & 4/36 \\ 6 & 5/36 \\ 7 & 6/36 \\ 8 & 5/36 \\ 9 & 4/36 \\ 10 & 3/36 \\ 11 & 2/36 \\ 12 & 1/36 \\ \end{array} $$
Now we can calculate $$ p(x_1+x_2+x_3+x_4=13)=$$ $$\begin{array}{} p( ((x_1+x_2=2) \land (x_3+x_4=11) ) &\lor \\ ((x_1+x_2=3) \land (x_3+x_4=10)) &\lor \\ ((x_1+x_2=4) \land (x_3+x_4=9)) & \lor \\ \cdots &\lor \\ ((x_1+x_2=11) \land (x_3+x_4=2))) &= \end{array} $$
$$ \begin{array}{} p(x_1+x2=2)\cdot p(x_3+x_4=11) &+ \\ p(x_1+x2=3)\cdot p(x_3+x_4=10) &+ \\ p(x_1+x+2=5)\cdot p(x_3+x_4=9)& + \\ \cdots &+ \\ p(x_1+x2=11)\cdot p(x_3+x_4=2)&= \end{array}$$ $$\frac{1}{36} \cdot \frac{2}{36} + \frac{2}{36} \cdot \frac{3}{36} + \frac{3}{36} \cdot \frac{4}{36} + \frac{4}{36} \cdot \frac{5}{36} + \frac{5}{36} \cdot \frac{6}{36} + \\ \frac{6}{36} \cdot \frac{5}{36} + \frac{5}{36} \cdot \frac{4}{36} + \frac{4}{36} \cdot \frac{3}{36} + \frac{3}{36} \cdot \frac{2}{36} + \frac{2}{36} \cdot \frac{1}{36}=$$ $$\frac{(2+6+12+20+30 ) \cdot 2}{36 \cdot 36}=\frac{140}{36 \cdot 36 }=\frac{35}{324}$$
Consider the polynomial $u(x)=(x+x^2+\cdots+x^6)^4$. The number of ways to get $13$ is $N=[x^{13}]u(x)$, that is, the coefficient of $x^{13}$ in $u(x)$. To compute $N$, note that $u(x)=x^4(1-x^6)^4v(x)$ where $v(x)=(1-x)^{-4}$ and that $x^4(1-x^6)^4$ equals the sum of $x^4-4x^{10}$ and of some terms of degree at least $16$.
Hence $N=[x^9]v(x)-4[x^3]v(x)$. Now, $6v(x)$ is the third derivative of the series $(1-x)^{-1}=\sum\limits_{n}x^n$ hence, for every $n$, $[x^n]v(x)={n+3\choose 3}$.
Finally, the probability to get $13$ is $$ \frac1{6^4}N=\frac1{6^4}\left({12\choose 3}-4{6\choose 3}\right)=\frac{35}{324}. $$