Conditional Expectation Dice Roll - 6 after known amount of rolls

Question

Consider the following experiment:

We throw one fair dice till it shows a $6$. $Y$ denotes the total of throws needed to achieve a $6$ and $$ X_i $$ for $$i \in \{1, ..., 5\}$$ the amount of throws with the outcome $i$. Show that $$\mathbb{E}[X_1 | Y] = \frac{1}{5}(Y-1).$$

Currently, I am struggling to find an idea how to solve this excercise, so I would appreciate any help I can get.

Thanks in advance!

Answer 1

Under condition $Y=n$ there are $n-1$ throws that all have a probability of $\frac{1}{5}$ to result in a face $1$.

That gives: $$\frac{1}{5}\left(n-1\right)$$ for the corresponding expectation of the number of times that face $1$ shows up under the mentioned condition.

Proved is now that: $$\mathbb{E}\left[X_{1}\mid Y=n\right]=\frac{1}{5}\left(n-1\right)$$ and on base of that we are allowed to conclude that: $$\mathbb{E}\left[X_{1}\mid Y\right]=\frac{1}{5}\left(Y-1\right)\tag1$$

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Another route:

Evident is that: $$\sum_{i=1}^5\mathbb E[X_i\mid Y]=\mathbb E\left[\sum_{i=1}^5X_i\mid Y\right]=\mathbb E[Y-1\mid Y]=Y-1$$

Then on base of symmetry we can conclude that $(1)$ is true.