What is the intuition behind P(Y=4) in this example? In my text book I have the following question:
Given two fair dice, one blue and one red.
Each die has two faces of the same number $\lbrace 1,\ 2,\ 3\rbrace$.
We toss the two dice until they both match.
Let $X$ be a RV that counts how many times we threw the dice.
Let $Y$ be a RV that is the sum of the first toss.


*

*What is the distribution of $X$?

*What is $P(Y=4)$?


For 1 I deduced that the distribution is a geometric distribution.
My question is for 2:
Why can't we deduce that: the probability of getting a 4 is 1/9 ?
My reasoning is that for the first toss we need 2 and for the second toss we also need a 2, each with a probability 1/3 of happening.
However, the solution in the textbook relies on the law of total probability and that $Y$ depends on $X$, therefore we sum $$P(X=1,Y=1) + \sum_{k=2}^{\infty} P(X=k,Y=4).$$
Why is my intuition wrong here?
 A: Typo aside, the answer is correct, just needlessly complicated.
Let $T_n$ be the result of the first toss (an ordered pair).
Now $\mathsf P(Y{=}4)$ is the probability that the sum of the first toss is 4 when either that happens on the first toss (when the die match) or later (so the die don't match on the first toss). 
$$\begin{align}\mathsf P(Y{=}4) &= \mathsf P(Y{=}4, X{=}1)+\sum_{k=2}^\infty \mathsf P(Y{=}4,X{=}k)\\[1ex] &={{\mathsf P(T_1{=}{\small(2,2)})} +{\mathsf P(T_1{\in}\{{\small(1,3),(3,1)}\})} {\sum_{k=2}^\infty\mathsf P(T_k{\in}\{{\small (1,1),(2,2),(3,3)}\})} {\prod_{j=2}^{k-1}\mathsf P(T_j{\notin}\{{\small(1,1),(2,2),(3,3)}\})}}\\[1ex] &= \dfrac 19+\dfrac {2}9\cdot \dfrac 13\sum_{k=2}^\infty \left(\dfrac 23\right)^{k-2}\\[1ex]&=\dfrac 19+\dfrac 29\cdot\dfrac 33\\[1ex]&=\dfrac 13\end{align}$$


Why is my intuition wrong here?

You only evaluated the case where both die showed 2; they might also show $(1,3),(3,1)$.
$\mathsf P(Y{=}4) =\mathsf P(T_1\in\{{\small(1,3),(2,2),(3,1)}\})=\tfrac 39$
