# If a fair die is thrown three times, what is the probability that the sum of the faces is 9?

If a fair die is thrown thrice, what is the probability that the sum of the faces is 9?

I did like this.

The total number of cases is $6^3=216$

Now,the number of solutions of the equation $x + y + z = 9$ with each of $x,y,z$ greater than equal to $1$ is ${8 \choose 2}$.

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Please choose more descriptive titles than "probability problem 2". If a problem is short, just put it all in the title, or put an important part of it. Some more advice here. Finally, see math notation guide. –  Yes Aug 13 '14 at 4:20
It depends on how many sides the die has. –  Silver Quettier Aug 13 '14 at 13:53
@SilverQuettier I think, unless otherwise specified, we assume 6-sided dice, as they're the most popular. –  Cruncher Aug 13 '14 at 14:48

You have included the solutions $7+1+1$ and relatives, which cannot occur with dice. So your procedure is fine, except that for the "favourables" we must use $\binom{8}{2}-3$.

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I get that there are only $25$ ways of writing $9$ as a sum of three integers in $[1,6]$, since: $$[x^9](x+x^2+x^3+x^4+x^5+x^6)^3 = 25.$$ Hence the probability is $\frac{5^2}{6^3}$.

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I find you answer interesting but I don't quite understand how you arrived at the formula. can you clarify a bit perhaps? –  Honza Brabec Aug 13 '14 at 8:28
The different ways to obtain x^9 in the polynomial (x+..x^6)^3 represent different ways to obtain 9 by throwing dice three times. –  Kuba Aug 13 '14 at 9:34
Interesting technique, I'd never have thought of doing it like that –  Ben Aaronson Aug 13 '14 at 13:25
@BenAaronson this is called generating functions. look up on google if you are curious –  Varun Iyer Aug 13 '14 at 13:35

I solved it as such.

For each first die roll, I find how many solutions sum to the rest. That is:

For die roll 1: The next 2 dice must sum to 8. There're 5 ways to do this.
For die roll 2: The next 2 dice must sum to 7. There're 6 ways to do this.
For die roll 3: The next 2 dice must sum to 6. There're 5 ways to do this.
For die roll 4: The next 2 dice must sum to 5. There're 4 ways to do this.
For die roll 5: The next 2 dice must sum to 4. There're 3 ways to do this.
For die roll 6: The next 2 dice must sum to 3. There're 2 ways to do this.

You add these up to get $25/216$

Finding the numbers for rolling 2 dice is pretty simple. It's small enough to be enumerated.

This is less elegant than the other solutions here. But I like its simplicity.

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Alternatively: given that $x, y, z \in \{1,..6\}, x+y+z=9$

If $x=1$ then $y\in \{2,..6\}$, else if $x\in\{2,..6\}$ then $y\in\{1,.. 8-x\}$. For each such pairing there is one value of $z$.

\begin{align} \sum_{y=2}^{6} 1 + \sum_{x=2}^6 \sum_{y=1}^{8-x} 1 & = 5 + \sum_{x=2}^{6}(8-x) \\ & = 5+8\times 5 - \frac{6\times 7}{2} + 1 \\ & = 25 \end{align}

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This is the best answer in my opinion, since you actually explained your logic. –  recursive recursion Aug 13 '14 at 19:09
@recursiverecursion The information there that is not explained is trivial enough to omit. –  user26486 Aug 13 '14 at 19:49