Hints to calculate Show $E[X_{1}\vert Y]=\frac{1}{5}(Y-1)$

Question

We roll a fair dice until we reach roll a $6$. Let $Y$ be an RV representing the rolls needed to roll a $6$ and $X_{1}$ be the number of throws wherein we threw a $1$.Show $E[X_{1}\vert Y]=\frac{1}{5}(Y-1)$

I do not know how to progress because the partition formula cannot be used

Edit: Let $V_{i}$ be the number thrown at the $i-$th attempt.

Note that $E[X_{1}\vert \operatorname{rolling a 6 for the first time at Y}]=E[X_{1}\vert \operatorname{rolling not a 6 through to Y-1}]=\sum^{Y-1}_{i=1}1P(V_{i}=1)=(Y-1)\frac{1}{5}$

Answer 1

If $X_i$ is the number of $i$'th you get before getting the first $6$, then $\sum_{i=1}^5 X_i = Y-1$, now you know that by symmetry $\mathbb E[X_i|Y]$ are equals and independents of $i$, which means that \begin{align*} \mathbb E[X_1|Y] &= \frac{1}{5}\sum_{i=1}^5 \mathbb E[X_i|Y]\\ &= \frac{1}{5}\mathbb E\left[\sum_{i=1}^5 X_i\middle|Y\right]\\ &= \frac{1}{5}\mathbb E[Y-1|Y]\\ &= \frac{1}{5}(Y-1) \end{align*}