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

Question

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.

Answer 1

Quoting from this answer:

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 $140$.

Answer 2

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

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