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Not sure how to do this question: Four six-sided dice are thrown. Find the probability that the total sum of the faces is 6. Help is greatly appreciated

James
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2 Answers2

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Hint
We can use multinomial theorum to find the number of ways in which the faces add up to $6$
That will be equal to the coefficient of $x^6$ in $(x+x^2+x^3\cdots+x^6)(x+x^2+x^3\cdots+x^6)(x+x^2+x^3\cdots+x^6)(x+x^2+x^3\cdots+x^6)=(x+x^2+x^3\cdots+x^6)^4=x^4\frac{(1-x^6)^4}{ (1-x)^4}$

which is same as the coefficient of $x^2$ in $(1-x)^{-4}$

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You get total sum of 6 with arrangements of (a) {1,1,2,2} and (b) {1,1,1,3}.

Number of arrangements of a) $\frac{4!}{2!2!}$ and b) $\frac{4!}{3!1!}$

Total number of arrangements (c) $= 6^4$

So probability of sum 6 $ = \frac {a + b} {c} = \frac{5}{648}$

Math Lover
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  • Btw this is a small number so we could easily find arrangements adding to 6. If you have to generalize it, you should use $mHn$ which is m numbers adding to $n = (m+n-1)Cn$. But this will include zeroes so you can subtract m from n and place 1 each in m places. Then do the above for (n-m). In our case, m = 4, n = 6. Take out 4 from 6 as no dice will have zero. Now, $4H2 = 5C2 = 10$. So, answer will be as above $\frac{10}{6^4} = \frac{5}{648}$. – Math Lover Jul 08 '20 at 13:55
  • To understand this easily, please see the accepted answer to another question on MSE. https://math.stackexchange.com/questions/1462099/number-of-possible-combinations-of-x-numbers-that-sum-to-y. – Math Lover Jul 08 '20 at 14:22