Why $E[X|X=Y]\neq E[Y]$?

Question

The full question was:

Let $X$ be the number of tosses until a coin with probability of $1\over 3$ to land on Heads does and $Y$ the number of tosses until a fair coin lands on Heads. What is $E[X|X=Y]$?

The correct answer is apperently $3\over 2$, which is lower than both $E[X]$ and $E[Y]$. What I'm confused about is if I know that $X=Y$, and I know that $Y=2$, then also $X=2$, therefore

$$E[X|X=Y]=\sum_k k\cdot P(X=k|X=Y) = \sum_k k \cdot P(Y=k) = E[Y]$$

which seems to be wrong, my guess would be because knowing that $X=Y$ also gives us information on $Y$, but I'm still not able to understand how or why it works.

Answer 1

I do not exclude that there is a shortcut, but I think this must work:

$\Pr\left(X=k\mid X=Y\right)\Pr\left(X=Y\right)=\Pr\left(X=k\wedge X=Y\right)=\Pr\left(X=k=Y\right)$

So:

$$\mathbb{E}\left(X\mid X=Y\right)\Pr\left(X=Y\right)=\sum_{k=1}^{\infty}k\Pr\left(X=k\mid X=Y\right)\Pr\left(X=Y\right)=\sum_{k=1}^{\infty}k\Pr\left(X=k=Y\right)$$

So that:

$$\mathbb{E}\left(X\mid X=Y\right)=\frac{\sum_{k=1}^{\infty}k\Pr\left(X=k=Y\right)}{\Pr\left(X=Y\right)}=\frac{\sum_{k=1}^{\infty}k\Pr\left(X=k=Y\right)}{\sum_{k=1}^{\infty}\Pr\left(X=k=Y\right)}$$

$X$ and $Y$ are independent so $\Pr\left(X=k=Y\right)=\Pr\left(X=k\right)\Pr\left(Y=k\right)$.