Number of Rolls of Fair Dice to get '6' and '5' A Fair Dice is Thrown Repeatedly. Let $X$ be number of Throws required to get a '$6$' and $Y$ be number of throws required to get a '$5$'. Find $$E(X|Y=5)$$
 A: In outline, and in most details, the posted solution is correct. 
However, the probabilities from $7$ on are not correct, since "not $6$" should include the probability of getting a $5$. So for example the probability that $X=7$ and $Y=5$ is $(1/6)^2(2/3)^4(5/6)$. For $X=8$ and $Y=5$, we multiply by $5/6$, and so on. 
For the part past $5$, it would be more efficient to use the fact that given we do not have a $6$ in the tosses before the sixth, then the expected waiting time is $5+ \frac{1}{1/6}$.  
A: This is my Approach, Let me know if i am on right track...
$$E(X|Y=5)=\sum_{x\in \mathbb{X}}xP(X=x|Y=5)=\sum_{x \in \mathbb{X}}\frac{xP(X=x,Y=5)}{P(Y=5)}$$
The set $\mathbb{X}$ is $\left\{1,2,3,4,6,7,8,\cdots \infty \right\}$ So
$$E(X|Y=5)=\sum_{k=1,2,3,\cdots,\infty_{k\ne 5}}\frac{k\,P(X=k,Y=5)}{P(Y=5)}$$ We have
$$P(X=1,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^3$$
$$P(X=2,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^3$$
$$P(X=3,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^3$$
$$P(X=4,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^3$$
$$P(X=6,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^4$$
$$P(X=7,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^5$$
$$P(X=8,Y=5)=\left(\frac{1}{6}\right)^2\left(\frac{2}{3}\right)^6$$ and so on $\cdots \cdots$ 
Thus
$$\sum_{k=1,2,3..}k\,P(X=k,Y=5)=\left(\frac{1}{36}\right)\left(\frac{8}{27}\right)\left(10+6\left(\frac{2}{3}\right)+7\left(\frac{2}{3}\right)^2+8\left(\frac{2}{3}\right)^3+9\left(\frac{2}{3}\right)^4+\cdots\right)=\left(\frac{1}{36}\right)\left(\frac{8}{27}\right)\left(5+5+6\left(\frac{2}{3}\right)+7\left(\frac{2}{3}\right)^2+8\left(\frac{2}{3}\right)^3+9\left(\frac{2}{3}\right)^4+\cdots\right)$$ Now for $|x|<1$
$$ \left(1-x\right)^{-2}=1+2x+3x^2+4x^3+5x^4+6x^5+7x^6+\cdots+\infty $$ $\implies$
$$g(x)=5+6x+7x^2+8x^3+\cdots+\infty=\frac{\left(1-x\right)^{-2}}{x^4}-\left(\frac{1+2x+3x^2+4x^3}{x^4}\right)$$ So
$$g\left(\frac{2}{3}\right)=21$$ So
$$\sum_{k=1,2,3..}k\,P(X=k,Y=5)=\left(\frac{1}{36}\right)\left(\frac{8}{27}\right)(26)=\frac{52}{243}$$
Also $$P(Y=5)=\frac{5^4}{6^5}$$ So
$$E(X|Y=5)=\left(\frac{52}{243}\right)\left(\frac{6^5}{5^4}\right)=\frac{1664}{625}$$
A: Since Y=5, we know the first four shakes are not 5, and the fifth is not 6.
The expectation is thus
  (1/5) * 1 + (4/25) * 2 + (16/125) * 3 + (64/625) * 4 + (256/625) * (E(X) + 5)
with E(X) = 6.
This is 3637/625 = 5.8192
A: We cannot get both 5 and 6 on the fifth trial, so $P(X=5 \mid Y=5)=0$
$\mathrm{\large E}(X\mid Y=5) = \sum\limits_{x=1}^{4} x\cdot\mathrm{\large P}(X=x\mid Y=5) + \sum\limits_{x=6}^{\infty} x\cdot\mathrm{\large P}(X=x\mid Y=5)$
To unconditionally get a 5 on the fifth trial, we need four not 5 before a 5 on the fifth.$$\mathrm{\large P}(Y=5) = \left(\frac{5}{6}\right)^4 \frac 16$$ 
To get a 6 before getting a 5 on the fifth trial, we have $(x-1)$ rolls that are neither, a 6 on the $x^{th}$, $(3-x)$ rolls that are not 5, then a 5 on the fifth trial. 
$$\mathrm{\large P}(X=x\mid Y=5)|_{x<5} = \dfrac{\mathrm{\large P}(X=x , Y=5)|_{x<5}}{\mathrm{\large P}(Y=5)} \\ = \dfrac{\left(\frac{4}{6}\right)^{x-1}\frac{1}{6}\left(\frac{5}{6}\right)^{3-x}\frac{1}{6}}{\left(\frac{5}{6}\right)^4\frac{1}{6}} = \dfrac{4^{x-1}5^{3-x}}{6^5}\cdot\dfrac{6^5}{5^4} = \dfrac{1}{20} \cdot\dfrac{4^{x}}{5^{x}}$$
To get a 6 after getting a 5 on the fifth trial, we have $4$ rolls that are neither, a 5 on the fifth trial, $(x-6)$ rolls that are not 6, then a 6 on the $x^{th}$ trial. 
$$\mathrm{\large P}(X=x\mid Y=5)|_{x>5} = \dfrac{\mathrm{\large P}(X=x , Y=5)|_{x>5}}{\mathrm{\large P}(Y=5)} \\ = \dfrac{\left(\frac{4}{5}\right)^{4}\frac{1}{6}\left(\frac{5}{6}\right)^{x-6}\frac{1}{6}}{\left(\frac{5}{6}\right)^4\frac{1}{6}} = \dfrac{4^45^{x-6}}{6^x}\cdot\dfrac{6^5}{5^4} = \dfrac{4^4}{5^5}\cdot\dfrac{5^{x-5}}{6^{x-5}}$$
So $$P(X=x\mid Y=5) = \begin{cases} \dfrac{1}{20}\cdot\dfrac{4^{x}}{5^{x}} & 0< x < 5 \\ \dfrac{4^4}{5^5}\cdot\dfrac{5^{x-5}}{6^{x-5}} & x>5 \\ 0 & \text{elsewhere}\end{cases}$$
Putting it together:
$$\mathrm{\large E}(X\mid Y=5) = \sum\limits_{x=1}^{4} x\cdot\mathrm{\large P}(X=x\mid Y=5|_{x<5} + \sum\limits_{x=6}^{\infty} x\cdot\mathrm{\large P}(X=x\mid Y=5|_{x>5} \\ = \dfrac{1}{20}\sum\limits_{x=1}^{4} x\left(\dfrac{4}{5}\right)^x + \dfrac{4^4}{5^5}\sum\limits_{x=6}^{\infty} x \left(\dfrac{5}{6}\right)^{x-5} \\ = \dfrac{14901}{3125} \approx 4.76832 $$
