Dice probability theory questions Given $3$ normal $6$-sided dice, 


*

*what is the probability the sum will be 9?,

*What is the probability that the first die shows 5? and 

*What is the conditional probability that the first die shows $5$, given that the sum of the dice is $9$?
This was the given solution for question 1: $x_1+x_2+x_3=9$, where each $x_i\in\{1,2,3,4,5,6\}$. We can rephrase this as the number of solutions to the problem $x_1+x_2+x_3=6$, where each $x_i\in\{0,1,2,3,4,5\}$ I'm confused why and how they changed $9$ to $6$.
Then solving using combinations with repetition, $\binom{6+4-1}{6} -3 = 25$ because $(6,0,0)$, $(0,6,0)$, $(0,0,6)$ don't meet our constraints subtract $3$. I'm confused where the constraint came, since dice don't have $0$ it can't be the rolls.
 A: The set of integer solutions to the system $\begin{cases} x_1+x_2+x_3=9\\ 1\leq x_1\leq 6\\ 1\leq x_2\leq 6\\ 1\leq x_3\leq 6\end{cases}$ can be seen to be in direct bijection with the set of integer solutions to the system $\begin{cases} y_1+y_2+y_3=6\\ 0\leq y_1\leq 5\\ 0\leq y_2\leq 5\\ 0\leq y_3\leq 5\end{cases}$ via the change of variable $x_i-1=y_i$ for each $i$.
Notice what the change of variable accomplishes.  $y_1+y_2+y_3=(x_1-1)+(x_2-1)+(x_3-1)=(x_1+x_2+x_3)-1-1-1=9-3=6$.  We also have $1\leq x_i\leq 6$ so subtracting one from everything we have $0\leq x_i-1\leq 5$ and noting that $x_i-1=y_i$ this gives us $0\leq y_i\leq 5$
In a similar problem, it might have been worded "how many ways can we give $9$ identical cookies to three distinct children such that each child receives at least one cookie but not more than six cookies."  In that setting we could "give each child a cookie to begin with before deciding how to distribute the rest."  That is all we are doing here for the first part of the problem.  We are "giving a pip to each of the dice before deciding how to distribute the remaining pips."
So... a result of $(y_1,y_2,y_3)=(3,0,3)$ corresponds to a die roll result of $(4,1,4)$.  Notice that since $1\leq x_i\leq 6$ we have $0\leq y_i\leq 5$ and so we do not allow $y_i$ to be equal to six.  As a result we remove the three results $(6,0,0), (0,6,0),(0,0,6)$ because those correspond to the dice rolls $(7,1,1),(1,7,1),(1,1,7)$ and it is impossible to roll a seven on a single six-sided die.

For the second part of the problem, you should know that already.
For the third part of the problem, note that a triple $(x_1,x_2,x_3)$ simultaneously satisfying $x_1=5$ and $\begin{cases} x_1+x_2+x_3=9\\ 1\leq x_1\leq 6\\ 1\leq x_2\leq 6\\ 1\leq x_3\leq 6\end{cases}$ is equivalent to the system $\begin{cases}y_1+y_2=2\\0\leq y_1\leq 5\\ 0\leq y_2\leq 5\end{cases}$

 Give five cookies to the first child, then give one cookie to each of the other two children.  Then decide how to distribute the remaining two cookies among the other two children.

From there, apply your definitions for conditional probability.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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The $\underline{first\ one}$:

\begin{align}
&\sum_{d_{1} = 1}^{6}{1 \over 6}\sum_{d_{2} = 1}^{6}{1 \over 6}
\sum_{d_{3} = 1}^{6}{1 \over 6}\braces{\bracks{z^{9}}z^{d_{1} + d_{2} + d_{3}}} =
{1 \over 6^{3}}\bracks{z^{9}}\pars{\sum_{d = 1}^{6}z^{d}}^{3} =
{1 \over 216}\bracks{z^{9}}\pars{z\,{z^{6} - 1 \over z - 1}}^{3}
\\[5mm] = &\
{1 \over 216}\bracks{z^{6}}{\pars{1 - z^{6}}^{3} \over \pars{1 - z}^{3}} =
{1 \over 216}\bracks{z^{6}}{1 \over \pars{1 - z}^{3}} -
{1 \over 216}\bracks{z^{0}}{3 \over \pars{1 - z}^{3}} =
{1 \over 216}{-3 \choose 6} - {1 \over 72}
\\[5mm] = &\
{1 \over 216}{8 \choose 2} - {1 \over 72} = \bbx{\ds{25 \over 216}} \approx
0.1157
\end{align}
