# If four dice are thrown, what is the probability that the sum of numbers thrown up will be 15? How about 16?

If four dice are thrown, what is the probability that the sum of numbers thrown up will be 15? How about 16?

What's a good method of finding the answer? I know that overall there are 1296 possibilities.

It is the coefficient of $x^{15}$ in the expansion of the generating function $$(x^6+x^5+x^4+x^3+x^2+x^1)^4=\left(\frac{x(1-x^6)}{1-x}\right)^4$$ which is $$1\cdot{{x}^{24}}+4\cdot {{x}^{23}}+10\cdot {{x}^{22}}+20\cdot {{x}^{21}}+35\cdot {{x}^{20}}+56\cdot {{x}^{19}}+80\cdot {{x}^{18}}\\+104\cdot {{x}^{17}}+125\cdot {{x}^{16}}+140\cdot {{x}^{15}}+146\cdot {{x}^{14}}+140\cdot {{x}^{13}}+125\cdot {{x}^{12}}+104\cdot {{x}^{11}}\\+80\cdot {{x}^{10}}+56\cdot {{x}^{9}}+35\cdot {{x}^{8}}+20\cdot {{x}^{7}}+10\cdot {{x}^{6}}+4\cdot {{x}^{5}}+1\cdot{{x}^{4}}$$

so the answer is $140$.

• Unless my computations are mistaken, this leads to 364/1296. But according to dice probability charts, the answer should be 140/1296 – Eric Sep 18 '17 at 7:15
• Fair enough, I'll see if I've made a mistake. – Gokul Sep 18 '17 at 7:18

There are $11$ different ways to throw 15: $$6621\\6531\\6522\\6441\\6432\\6333\\5541\\5532\\5442\\5433\\4443$$ Now we have 4 types of throws (in brackets there are numbers of occurcences): $$ABCD (2)\\ABCC(7)\\ABBB(2)\\AABB(0)\\AAAA(0)$$ Where $A,B,C,D$ are different numbers. Now let's compute number of possible permutations of the above types (with positive numbers of occurences):

• $P_{ABCD}=4!$
• $P_{ABCC}= 4\cdot 3$
• $P_{ABBB}=4$

Thus there are $$X=2\cdot4! + 7\cdot 4\cdot 3 + 2\cdot 4 = 140$$ events that lead to sum $15$.

Probabiblity is then equal to

$$P(15)=\frac{140}{6^4}=\frac{35}{324}$$