Exercise in a book should be easy, my solution is too complicated to finish Consider the following exercise

Should it be somehow easy arrive at the formula
$$
\mathbb{E}[X_{1}|X_{2}]=\frac{n-X_{2}}{5}\ 
$$ as no explanation regarding the solution was given?
I tried to derive it form first principles and I got stuck. First
I tried to compute 
$$
\mathbb{E}[X_{1}|X_{2}=k]
$$
in hope that that would allow me to see a nice formula for $\mathbb{E}[X_{1}|X_{2}]$
. But computing this formula I obtained
$$
\mathbb{E}[X_{1}|X_{2}=k]=\sum_{i=0}^{n-k}i\cdot P(X_{1}=i|X_{2}=k)=\sum_{i=0}^{n-k}i\cdot\frac{P(X_{1}=i\land X_{2}=k)}{P(X_{2}=k)}= \\ =\sum_{i=0}^{n-k}i\cdot\frac{\binom{n}{n-k-i}\binom{k+i}{k}6^{n-k-i}\frac{1}{6^{n}}}{6^{n-k}\binom{n}{k}\frac{1}{6^{n}}},
$$
at which point I stopped, because I didn't knew how to proceed further.
(For your information, I arrived at numerator in the following way: $\binom{n}{n-k-i}$
counts the number of ways to distribute places in a sequence with
$n$ elements that are neither 1s nor 2s; having these places fixed,
the place where 1s or 2s have to be are alos fixed and there $\binom{k+i}{k}$
ways to distribute the 1s, which also fixed the places of the 2s;
the $n-k-i$places where there aren't 1s or 2s can be filled $6^{n-k-i}$
ways; $\frac{1}{6^{n}}$} is the probability
each of these $\binom{n}{n-k-i}\binom{k+i}{k}6^{n-k-i}$ many choices
has. Similarly one can derive the denominator.)
Question 1 (which is basically my question from above): Is there any easier way to do this?
Question 2: How to carry out my analysis to the end, by computing first $\mathbb{E}[X_{1}|X_{2}=k]$?
 A: There is an easier way. Notice that $X_1+\cdots+X_6=n$. This implies that $X_1=n-X_2-\cdots-X_6$. Since the die has an equal chance of landing on each number, convince yourself that $E[X_1|X_2]=E[X_3|X_2]=\cdots=E[X_6|X_2]$. Thus letting $y:=E[X_1|X_2]$, take conditional expectations of the above with $E[\cdot|X_2]$ and use linearity:
$$y=n-E[X_2|X_2]-5y=n-4y-X_2.$$
Or,
$$y=\frac{n-X_2}{5}.$$
For $E[X_1|X_2,X_3]$, the argument is very similar.
A: A direct route.
At $n-X_2$ of the $n$ throws a number of set $\{1,3,4,5,6\}$ will appear. 
This set contains $5$ numbers and at each of the $n-X_2$ throws these $5$ numbers have equal probability to appear. 
So we expect number $1$ to appear $\frac15(n-X_2)$ times. 
In mathematical notation: $$\mathbb E[X_1\mid X_2]=\frac15(n-X_2)$$
You can do it also a bit less direct by calculation of $\mathbb E[X_1\mid X_2=k]$ starting  with:
At $n-k$ of the $n$ throws a number of set $\{1,3,4,5,6\}$ will appear....
That will lead likewise to $\mathbb E[X_1\mid X_2=k]=\frac15(n-k)$ and from this you can conclude again that $\mathbb E[X_1\mid X_2]=\frac15(n-X_2)$.

Same reasoning for $\mathbb E[X_1\mid X_2,X_3]$, this time starting with:
At $n-X_2-X_3$ of the $n$ throws a number of set $\{1,4,5,6\}$ will appear. 
This set contains $4$ numbers and ...
