Poisson events distributed uniformly in a given time It is given that $4$ Poisson events occur between $12:00$ to $13:00 $ (interval denoted by T).
Intuitively, Why the probability of each event to occur at time $t \in T$ is uniform across $T$?  
I'm well aware of the calculation for that:
Let's calculate the probability of $3$ (out of $ 4$) events to occur between $12:30$ to $13:00$ ($\frac{1}{2}$ an hour).
Let the following random variables:  


*

*$X$ - the $4 $ events from $12:00$ to $13:00$.

*$X_1$ - the single event that has happened between $12:00 $  to $ 12:30$ .

*$X_2$ - the 3 events which happened between $12:30$ to $13:00$.


So their distributions are:
$$
\begin{align*}
& X \sim Poi(\lambda \cdot t) = Poi(4 \cdot 1) = Poi(4) \\
& X_1, X_2 \sim Poi(4 \cdot \frac{1}{2}) = Poi(2)
\end{align*}
$$
Given that a total of 4 events occurred at $T$, then:
$$
\begin{align*} P(X_2 = 3 \,|\, X = 4) &= \\
& = \frac{P(X_2 = 3 \cap X = 4)}{P(X = 4)}   \\
& = \frac{P(X_2 = 3 \cap (X_1 = 1 \cup X_2 = 3))}{P(X = 4)} \\
& = \frac{P(X_2 = 3 \cap X_1 = 1)}{P(X = 4)} \\
& = \frac{P(X_2 = 3) \cdot P(X_1 = 1)}{P(X = 4)}  \\
& = \frac{\frac{e^{-2} \cdot 2^3}{3!} \cdot \frac{e^{-2} \cdot 2^1}{1!}}{\frac{e^{-4} \cdot 4^4}{4!}}  \\
& = \frac{4!}{3! 1!} \cdot \left(\frac{2}{4}\right)^1 \cdot \left(\frac{2}{4}\right)^3\\
& = \frac{2^4 \cdot 4!}{3! \cdot 4^4}\\
& = 4 \cdot \left(\frac{1}{2}\right)^4\\
& = 0.25
\end{align*}
$$
(Note that at the end we get a binomial distribution)
But I just can't understand this intuitively. 

Edit:
The root of my confusion:
Indeed, it is given that in an hour there are precisely 4 events.
Therefore the total 4 events are NOT distributed like Poission.
But it is not given how much time each event takes to occur, so maybe the second, third and fourth events are distributed like Poission?  Perhaps the time that each event takes to occurs is distributed exponentially (which indicates on a Poission distribution)?
 A: It has to do with the fact that it is given that there are exactly 4 events in the hour, whereas the Poisson distribution allows any number of events in a given time period. Both $X_1$ and $X_2$ are both greater than or equal to $1$ (there is at least one event in both half hours), so we need to divide the remaining two events over $X_1$ and $X_2$. This distribution is no longer Poisson. The probability that the final two events both fall in $X_2$ is simply $0.5^2 = 0.25.$
A: A common way to derive the Poisson is as the limit of a binomial as $n\to \infty, p\to 0: np=\lambda>0$. 
So lets see how this helps:


*

*You fix a time interval $T$, hence you fix the mean number of event to $T\lambda$


At this point, all you've done is specify a particular Poisson distribution, as theoretically we are carrying out the limiting process I outlined above.


*

*Now, also fix the number of observed events, $k$.


At this point, we can continue our limiting process by partitioning the interval $T$ into $n$ equally sized bins and asking what is the probability that there is one event in exactly $k$ bins and $0$ events in the remaining $n-k$ bins. This probability will be binomial with parameters $k=4$ $\rm num.trials=n$ and 
$$p=\frac{\exp(-\frac{\lambda T}{n})\left(\frac{\lambda T}{n}\right)}{1!}$$
The overall formula is:
$${n \choose 4} \left(\exp(-\frac{\lambda T}{n})\left(\frac{\lambda T}{n}\right)\right)^4\left(1-\exp(-\frac{\lambda T}{n})\left(\frac{\lambda T}{n}\right)\right)^{n-4}$$
If we take $n\to \infty$ we get (with the help of Wolfram Alpha):
$$\frac{(\lambda T)^4}{24 \exp(\lambda T)}$$
This is the denominator (which is just the usual Poisson probability of 4 events in T, see here for generalization). Your numerator will depend on how you allocate your events to subsets of $T$. 
Now, notice that this process is the result of a limiting process on a bionomial. So how does this relate to your thought experiment?
What you did is you picked a particular partition of $T$ and allocated event counts to each (in this case, two equal sized partitions with 1 and 3 events, respectively). Hence, mathematically, this will result in a binomial distribution because you partitioned your interval into two bins.
What would happen if you defined a three bin partition? Say $t_1,t_2,t_3$? In this case, you would not get a binomial distribution on the probability of seeing a particular allocation of 4 events to these bins. Instead, you would get a multinomial distribution.
So, intuitively, by fixing the number of events and the interval, we arrive at a multinomial distribution over any partition of that interval, with the bin probabilities proportional to the fraction each bin takes up of the whole interval.
This arises because the expected value of a poisson random variable is linear in the interval length, so we get a nice interpretation of equiprobability for an event in the interval. This would not be the case if the expected value were non-linearly related to the interval length. The theoretical connection of the poisson with the Binomial further reinforces this intuition. 
A: The intuition about this that is that in a homogenous poisson process all the waiting times for are independent and identically distributed. So the differnces between times of occurrence should more ore less same. You can prove it like this.
Suppose $N(t)=1$.
let $\tau$ denote the time of that occurrence.
Now $P(\tau > t'|N(t)=1)=P(N(t')=0,N(t-t')=1|N(t)=n)\\= e^{-\lambda t'}e^{-\lambda (t-t')} \lambda (t-t')/(e^{-\lambda t}\lambda t)\\= (t-t')/t \implies P(\tau \leq t|N(t)=1)=t'/t$ 
which is the distribution of $uniform(0,t)$.
This is why the times of occurrences
 follow uniform. You can prove it similarly for $N(t)=n$ but then this times will have the joint distribution of uniform order statistics.
