Geometric distribution, tossing a die 
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geometric distribution throwing a die 

Yesterday I posted a question which was answered but I disagree with the answer so I'd like to ask again so we can discuss it together :)
The problem says as follows: We throw a die repeatedly. X  and Y  denote, respectively, the number of rolls until we reach a 5  and 6 .
The aim is to compute $E[X|Y=5]$.
First question: If Y=5 (this means no 6 have come up in the first 4 rolls) then the probability of getting 5 in those rolls should be 1/5 instead of 1/6?
I have done some simulations to guess the expected number of rolls and I get always something around 5.8 which seems reasonable but I can't arrive to that answer algebraically or analitically.
Thanks a lot for you help! :)
 A: Yes, if the only information we have about a roll of a die is that we did not get a $6$, then the probability we got a $5$ is $\frac{1}{5}$. For given only that we did not get a $6$, the numbers $1$ through $5$ are equally likely. 
One sloppy but intuitive argument goes as follows. Suppose we toss the die $6000$ times. Then we will get roughly $1000$ of each number. (More precisely, the proportion of each is very likely to be close to $\frac{1}{6}$.)
Now concentrate attention on the about $5000$ times we did not get a $6$. About $1000$ of these times, we got a $5$, so the probability we got a $5$ given we did not get a $6$ is $\frac{1}{5}$. 
A: You can proceed in a straightforward manner:
$$\tag{1}\Bbb E(X|Y=5)=\sum\limits_{i=1}^5 i\cdot P[X=i|Y=5]+E(X|Y=5, X> 5)P[X>5|Y=5].$$
We have 
$$\eqalign{
\sum\limits_{i=1}^5 i\cdot P[X=i|Y=5]
&=(1/5)\cdot 1+ (4/5)(1/5)\cdot 2+(4/5)^2(1/5)\cdot 3+(4/5)^3(1/5)\cdot 4+0\cr  
&=\textstyle{821\over625}.}
$$
and, noting that if both $Y=5$ and $X>5$, it's as if we "started over"
$$
E(X|Y=5, X> 5)= (\Bbb E(X)+5)=6+5= 11.
$$
Computing the required sum $(1)$, using $P[X>5|Y=5]=(4/5)^4$, we obtain
$$\Bbb E(X|Y=5)
=\textstyle{821\over625}+(4/5)^4\cdot11
\approx 5.8192.$$
A: If you know that $Y=5$, then you know that you did not roll a $6$ on any of the first four rolls. Thus, $X=1$ with probability $\frac15$, $X=2$ with probability $\frac45\cdot\frac15$, $X=3$ with probability $\left(\frac45\right)^2\cdot\frac15$, and $X=4$ with probability $\left(\frac45\right)^3\frac15$. Clearly the probability that $X=5$ is $0$. For $n>5$ the probability that $X=n$ is $\left(\frac45\right)^4\left(\frac56\right)^{n-6}\cdot\frac16$. Thus,
$$\begin{align*}
\Bbb E[X\mid Y=5]&=\sum_{k=1}^4k\left(\frac45\right)^{k-1}\frac15+\left(\frac45\right)^4\sum_{k\ge 6}k\left(\frac56\right)^{k-6}\frac16\\\\
&=\frac15\sum_{k=1}^4k\left(\frac45\right)^{k-1}+\frac16\left(\frac45\right)^4\sum_{k\ge 0}(k+6)\left(\frac56\right)^k\\\\
&=\frac15\sum_{k=1}^4k\left(\frac45\right)^{k-1}+\left(\frac45\right)^4\sum_{k\ge 0}\left(\frac56\right)^k+\frac16\left(\frac45\right)^4\sum_{k\ge 1}k\left(\frac56\right)^k\\\\
&=1.3136+\frac{0.4096}{1-5/6}+\frac16\left(\frac45\right)^4\left(\frac56\right)\sum_{k\ge 1}k\left(\frac56\right)^{k-1}\\\\
&=3.7712+\frac{256}{36\cdot125}\frac1{\left(1-\frac56\right)^2}\\\\
&=3.7712+\frac{256}{125}\\\\
&=5.8192\;.
\end{align*}$$
