Why can't we combine events in a poisson distribution? Let's say that babies in a hospital are delivered according to a Poisson distribution, where on average 1 baby is delivered every hour.
Question 1: What is the probability that at least 1 baby is delivered per hour?
It is 1 - P(X = 0) for $\lambda = 1$, i.e. $1 - e^{-1} = 63\%$.
Question 2: What is the probability that at least 2 babies are delivered per 2 hours?
It is 1 - P(X = 0) - P(X = 1) for $\lambda = 2$, i.e. $1 - e^{-2} - 2e^{-2} = 59\%$.
To simplify the calculation of the question 2, let's combine 2 babies into one superbaby. We could then ask a question which seems (to me) to be identical to question 2:
Question 3:  What is the probability that at least 1 superbaby is delivered per 2 hours?
It is 1 - P(X = 0) for $\lambda = 1$, i.e. $1 - e^{-1} = 63\%$.
Here's what I don't understand: if a superbaby is simply 2 babies, 
Why is the probability that 1 superbaby is delivered per 2 hours not equal to the probability that 2 babies are delivered per 2 hours?
My intuition tells me it has something to do with the assumption that events which follow a Poisson distribution are independent, but that's just a guess.
 A: Intuitively, it is because in the 2 case you can have a half-superbaby but it 2a you cannot.  Let's extend this and ask what is the chance that less than 1000 babies are born in 1000 seconds?  As $\lambda$ gets large, the Poisson distribution becomes a normal distribution with mean $\lambda$ and standard deviation $\sqrt \lambda$.  It is a nice exercise to prove this.  In our 1000 case, you are asking whether there are more or less than the expected number born, and should expect the answer to be that the chances are $1/2$ each way.  
At the other extreme, $\lambda$ is not limited to integers.  If $\lambda=0.001$, the chance that the number of events is less than $\lambda$ is (slightly greater than) $0.999$
A: Your confusion comes from the expectation that the event "at least 1 superbaby" is equivalent to "at least two births". 
Let's construct a random variable that is indeed equivalent to what you stated:
First, I'll recapituate your basic model of the number of normal births in 2 hours:
$$X \sim Poi(2)$$
Now, if we are simply re-casting this in terms of superbabies $S$, then the following transformation of $X$ will capture this new definition:
$$ S\sim \lfloor \frac{X}{2} \rfloor$$
We can still calculate $P(S\geq 1)$:
$$P(S\geq 1) = 1 - P(S = 0) = 1- P(X = 0 \cup X = 1) = 59\% $$
As we would expect. So, what model would apply to your second calculation, lets try the following
$$P(S=n)=P(X=2n)$$
Now, we have a problem, because:
$$\sum_{i=0}^{\infty} P(S=i) = \sum_{i=0}^{\infty} P(X=2i) = \frac{\cosh(2)}{e^2}=\frac{1}{2}+\frac{1}{2e^4}\approx 51\% < 1$$
See here for the Wolfram Alpha calculation.
Therefore, our naive approach of requiring exactly two babies per superbaby results in an invalid probability distribution. However, if we let $c=\frac{1}{2}+\frac{1}{2e^4}$, then:
$$\frac{1}{c}\sum_{i=0}^{\infty} P(X=2i) = 1$$
So, lets define a new, proper distribution:
$$P_S(n):= \frac{P(X=2n)}{c}$$
For your particular case:
$$P_S(S\geq 1) = 1 - P_S(0) = 1-\frac{e^{-2}}{\frac{1}{2}+\frac{1}{2e^4}} = 1- \frac{1}{\cosh(2)} \approx 73\%$$
This is still not what you have. Therefore, to get your result, we cannot think of a superbaby as simple equivalent to two regular babies. It would appear that your definition of a superbaby is not well defined (as pointed out in the comments) and does not match your calculation.
We need to think about what the birth rate, $\lambda$ really means.


*

*In your original problem, its the number of births in 2 hours.

*In your revised problem, it is the number of superbabies in 2 hours. 


We can arrive at a consistent interpretation if we consider the birth of a superbaby a new type of event, and not as equivalent to the statement "two babies were born" (i.e., a new "unit").
Under this model, you only have superbabies being born, and they are born at half the rate of regular babies. This actually matches the model you developed for this case:
$$\lambda_S=\frac{\lambda_X }{2} \implies P(S\geq 1)=1-e^{-1}$$
As expected.
So, what happened here is that you thought you were simply calculating the probability of regular births under a simple change of units but you were actually creating an entirely new event, albeit closely related to the first.
