Exercise in a book should be easy, my solution is too complicated to finish

Question

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 $\mathbb{E}[X_{1}|X_{2}=k]$?

Answer 1

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.

Answer 2

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 $\mathbb E[X_1\mid X_2]=\frac15(n-X_2)$.

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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 ...