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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 ${8 \choose 2}$.

But am not sure about my answer. Please help.

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    $\begingroup$ 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. $\endgroup$ – user147263 Aug 13 '14 at 4:20
  • $\begingroup$ It depends on how many sides the die has. $\endgroup$ – Silver Quettier Aug 13 '14 at 13:53
  • $\begingroup$ @SilverQuettier I think, unless otherwise specified, we assume 6-sided dice, as they're the most popular. $\endgroup$ – Cruncher Aug 13 '14 at 14:48
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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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    $\begingroup$ I find you answer interesting but I don't quite understand how you arrived at the formula. can you clarify a bit perhaps? $\endgroup$ – Honza Brabec Aug 13 '14 at 8:28
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    $\begingroup$ 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. $\endgroup$ – Kuba Aug 13 '14 at 9:34
  • $\begingroup$ Interesting technique, I'd never have thought of doing it like that $\endgroup$ – Ben Aaronson Aug 13 '14 at 13:25
  • $\begingroup$ @BenAaronson this is called generating functions. look up on google if you are curious $\endgroup$ – Varun Iyer Aug 13 '14 at 13:35
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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 are 5 ways to do this.
For die roll 2: The next 2 dice must sum to 7. There are 6 ways to do this.
For die roll 3: The next 2 dice must sum to 6. There are 5 ways to do this.
For die roll 4: The next 2 dice must sum to 5. There are 4 ways to do this.
For die roll 5: The next 2 dice must sum to 4. There are 3 ways to do this.
For die roll 6: The next 2 dice must sum to 3. There are 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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  • $\begingroup$ This is the best answer in my opinion, since you actually explained your logic. $\endgroup$ – recursive recursion Aug 13 '14 at 19:09
  • $\begingroup$ @recursiverecursion The information there that is not explained is trivial enough to omit. $\endgroup$ – user26486 Aug 13 '14 at 19:49

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