A fair dice is been tossed until "6" is achieved. Let Y be the random variable stating the number of tosses required. Let X be the random variable stating the number of tosses where "1" was achieved until "6" is achieved. What is the joint distribution of X and Y?
What I have tried to do
First I know that $$ X\sim G(\frac{1}{6}) $$
$$ P_{X|Y}(x|y) = \frac{P_{X,Y}(x,y)}{P_{Y}(y)} \Leftrightarrow {P_{Y}(y)}*P_{X|Y}(x|y) = P_{X,Y}(x,y) $$
So from here I tried to find the probability function of $$P_{X|Y}(x|y)$$
I created a table $$ \begin{array} {|r|r|}\hline k & P_{X|Y}(x|y=k) \\ \hline 1 & \frac{1}{6} \\ \hline 2 & \frac{1}{6^2}+\frac{4}{6^2} \\ \hline 3 & \frac{1}{6} *((\frac{4}{6})^2 + 2*\frac{4}{6}*\frac{1}{6} + (\frac{1}{6})^2)) \\ \hline \end{array} $$
and so on until we get $$ P_{X|Y}(x|y=k) = \frac{1}{6}*(\frac{4}{6}+\frac{1}{6})^{k-1} = \frac{1}{6}*(\frac{5}{6})^{k-1} $$ if k=1 thus first throw is 6 if k=2 thus first throw is either 1 or 2-5 but second throw is 6 and so on. If I sum all the values it helps me to make sure it is indeed a probability function (equals 1)
but then, returning to what we derived from results in
$$ P_{X,Y}(x,y) = \frac{1}{6^2}*(\frac{5}{6})^{2k-2} $$
where $$k=1,2,...$$
The sum on all k values isn't 1, where did i go wrong or maybe i was wrong from the begining?
Sorry for the title, couldn't come up with something short and precise for that
The joint distribution of $X$ and $Y$ is the function $p(x,y)$ that gives the probability that $X = x$ and $Y = y$. That is, $p(x,y) = P(X = x\cap Y= y)$. We know from conditional probabilities that $P(X = x\cap Y= y) = P(X = x | Y= y)P(Y = y)$ and $P(X = x\cap Y= y) = P(Y = y|X=x)P(X=x)$. Since $P(Y= y)$ is much easier to find than $P(X=x)$, we'll use the first form.
$Y$ clearly follows a geometric distribution with success probability $1/6$. This gives $P(Y = y) = 5^{y-1}/6^y$
For a fixed value $Y=y$, each of the $y-1$ rolls prior to it has an equal chance to be any of the numbers $1$ through $5$, which means a $1/5$ chance to be $1$. Therefore, given $Y=y$, $X$ follows a binomial distribution with $y-1$ trials and success probability $1/5$. This allows us to find $P(X=x | Y=y) = \binom{y-1}{x}4^{y-1-x}/5^{y-1}$.
We now use these together to find the joint distribution $p(x,y)$: $$ p(x,y) = P(X=x|Y=y)P(Y=y) = \left[\binom{y-1}{x}\frac{4^{y-1-x}}{5^{y-1}}\right]\left[\frac{5^{y-1}}{6^y}\right] =\binom{y-1}{x}\frac{4^{y-1-x}}{6^y}. $$ As a sanity check, WolframAlpha tells me that these probabilities do sum to $1$.