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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.

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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 $140$.

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

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

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