Expected number of occurences in Poisson process that can stop under certain conditions 
We have a fisherman who catches fish according to a Poisson distribution with $\lambda = 0.6$ fish per hour. The fisherman always fishes for at least $2$ hours. If during these $2$ hours he catches at least $1$ fish, he goes back home, else, he keeps fishing until he catches his first fish and then immediately leaves (we assume that he cannot catch $2$ fish at once, as per a Poisson process).

Q. What is the expected number of caught fishes?
Let $X_t$ denote the Poisson (counting) process. I made two attempts.

Attempt 1
$$
\begin{align*}
\mathbb{E}\left[\lim_{t \to +\infty} X_t - X_0\right] &= \mathbb{P}(X_2 = 0) \times \mathbb{E}\left[\lim_{t \to +\infty} X_t - X_0\mid X_2 = 0 \right] \\
&+ \mathbb{P}(X_2 > 0) \times \mathbb{E}\left[\lim_{t \to +\infty} X_t - X_0\mid X_2 > 0 \right]
\end{align*}
$$
We know that if $X_2 = 0$, then the fisherman will catch only $1$ fish, and if $X_2 > 0$, then he will catch $X_2$ fish. So, we get
$$
\mathbb{P}(X_2 = 0) \times 1 + \mathbb{P}(X_2 > 0) \times \mathbb{E}\left[X_2 \right] \approx 0.3011942 + 0.8385669 = 1.139761
$$

Attempt 2
$$
\begin{align*}
\mathbb{E}\left[\lim_{t \to +\infty} X_t - X_0\right] &= \mathbb{E}\left[\lim_{t \to +\infty} ((X_t - X_2) + (X_2 - X_0)\right] \\
&= \mathbb{E}\left[X_2 - X_0\right] + \mathbb{E}\left[\lim_{t \to +\infty} X_t - X_2\right]
\end{align*}
$$
Now, $\mathbb{E}\left[\lim_{t \to +\infty} X_t - X_2\right]$ is $1$ if and only if $X_2 = 0$, so $\mathbb{E}\left[\lim_{t \to +\infty} X_t - X_2\right] = \mathbb{P}(X_2 = 0)$. We calculate now
$$
1.2 + \mathbb{P}(X_2 = 0) \approx 1.2 + 0.3011942 = 1.501194
$$

A quick simulation suggests that attempt #2 is correct. However, I do not see why either of these attempts would be incorrect. Could anyone please shed some light?
 A: Let $\lambda=2(0.6)$. The conditional expectation given two hours of frustration is $1$. That contributes $e^{-\lambda}$ to the expectation. 
Now suppose that at least $1$ fish is caught in $2$ hours. Then the conditional probability of $k$ fish is $\frac{1}{1-e^{-\lambda}}e^{-\lambda}\frac{\lambda^k}{k!}$. Thus the conditional expectation is $\frac{1}{1-e^{-\lambda}}\lambda$. This contributes $\lambda$ to the expectation.
Remark: So the expectation of the fisher is $e^{-\lambda}+\lambda$. That is exactly what you got using your (more efficient) Method 2.
If we compare the above with your Method 1, which also used a conditioning argument, we can see that one conditional expectation was computed incorrectly. Your version calculates it as $\lambda$, and it should be $\frac{1}{1-e^{-\lambda}}$, and then the $1-e^{-\lambda}$ at the bottom gets cancelled when we multiply by the probability of at least $1$ fish in $2$ hours.
A: $\newcommand{\E}{\operatorname{E}}$I think bringing in anything like $t\to+\infty$ makes the problem more complicated than it needs to be. You don't need to know anything about how much time it took to catch a fish after the first two hours go by; all you need is that the number of fish caught during any such overtime is always $1$. You're finding the expected value of $Y=\max\{X,1\}$ where $X$ is the number of fish caught in the first two hours and $X\sim\mathrm{Poisson}(1.2)$.
\begin{align}
\E(Y) & = 1\cdot\Pr(Y=1) + \sum_{y=2}^\infty y\Pr(Y=y) \\[10pt]
& = 1\cdot\Pr(X=0) + \underbrace{1\cdot\Pr(X=1)+ \sum_{y=2}^\infty y\Pr(X=y)}_\text{This is $\E(X)$} \\[10pt]
& = 1\cdot e^{-1.2} + \E(X) = e^{-1.2}+1.2.
\end{align}
A: $\lambda=0.6$ is the average number of fishes, that the fisherman is catching during one hour. $\lambda$ is the expected value of the poisson distribution. Thus in two hours he catches on average 1.2 fishes . 
The probability, that he is not catching a fish during the first hour is $P(X=0)=e^{-0.6}\cdot \frac{0.6^0}{0!}=e^{-0.6}$ 
The probabiliy, that he is not catching a fish during the second hour is $P(Y=0)=e^{-0.6}\cdot \frac{0.6^0}{0!}=e^{-0.6}$. 
$P(X=0 \cap Y=0)$=Probability, that the fisherman is not catching a fish during 2 hours.
$P(X=0 \cap Y=0)=P(X=0)\cdot P(Y=0|X=0)$.
$X$ and $Y$ are independent. The number of fishes, which are caught during the second hour does not depend on the number of fishes, which have been caught during the first hour.
Thus $P(X=0 \cap Y=0)=P(X=0)\cdot P(Y=0)=e^{-0.6}\cdot e^{-0.6}=e^{-1.2}$
Therefore he expected number of caught fishes is $1.2+e^{-1.2}\cdot 1=1.501194$
