I am trying to create a pdf for the sum of 6 six-sided fair dice. The sums range from 6 to 36 (either have all 1s or all 6s).
My attempt at the solution:
$x$ = sum of the 6 dice
$P(x=6) = C(6,0) (1/6)^6$ <— all 1s
$P(x=7) = C(6,1) (1/6)^5 (1/6)$ <— five 1s and one 2
$P(x=8) = C(6,1) (1/6)^5 (1/6) + C(6,2) (1/6)^4 (1/6)^2$ <—– five 1s and one 3 or four 1s and two 2s
$P(x=9) = C(6,1) (1/6)^5 (1/6) + 2\cdot C(6,2) (1/6)^4 (1/6) (1/6)$ <—– five 1s and one 4 or four 1s, one 3 and one 2
$P(x=10) = C(6,1) (1/6)^5 (1/6) + 2\cdot C(6,2) (1/6)^4 (1/6) (1/6) + C(6,2) (1/6)^4 (1/6)^2$ <—– five 1s and one 6 or four 1s, one 4 and one 2 or four 1s and two 3s
As you can see this process is extremely tedious to count all the possible combinations as the sums get larger, so I was wondering if anyone has a more concise approach/solution to this particular problem.
Thanks!
Best Answer
You are looking for the coefficients of the expansion of $$\left(\dfrac{x+{x}^{2}+{x}^{3}+{x}^{4}+{x}^{5}+{x}^{6}}{6}\right)^6=\left(\dfrac{x(1-x^6)}{6(1-x)}\right)^6.$$ This will give $$\frac{{x}^{6}}{46656}+\frac{{x}^{7}}{7776}+\frac{7\,{x}^{8}}{15552}+\frac{7\,{x}^{9}}{5832}+\frac{7\,{x}^{10}}{2592}+\frac{7\,{x}^{11}}{1296}+\frac{19\,{x}^{12}}{1944}+\frac{7\,{x}^{13}}{432}+\frac{43\,{x}^{14}}{1728}+\frac{833\,{x}^{15}}{23328}+\frac{749\,{x}^{16}}{15552}+\frac{119\,{x}^{17}}{1944}+\frac{3431\,{x}^{18}}{46656}+\frac{217\,{x}^{19}}{2592}+\frac{469\,{x}^{20}}{5184}+\frac{361\,{x}^{21}}{3888}+\frac{469\,{x}^{22}}{5184}+\frac{217\,{x}^{23}}{2592}+\frac{3431\,{x}^{24}}{46656}+\frac{119\,{x}^{25}}{1944}+\frac{749\,{x}^{26}}{15552}+\frac{833\,{x}^{27}}{23328}+\frac{43\,{x}^{28}}{1728}+\frac{7\,{x}^{29}}{432}+\frac{19\,{x}^{30}}{1944}+\frac{7\,{x}^{31}}{1296}+\frac{7\,{x}^{32}}{2592}+\frac{7\,{x}^{33}}{5832}+\frac{7\,{x}^{34}}{15552}+\frac{{x}^{35}}{7776}+\frac{{x}^{36}}{46656}.$$