I have difficulty formulating this question, but I hope I've explained what I'm after. Feel free to downvote it if I haven't succeeded, but please also tell me what I need to clarify.
Another question on this site, which I cannot locate at the moment for some reason, posed a question something like this.
During a boring lecture, a student gazes out the window at a tree, whose leaves fall off according to a Poisson distribution with parameter $\lambda$. Every time a leaf falls, the student tosses a fair coin, and records the outcome. Let $H$ and $T$ be random variables giving the number of heads and tails, respectively, recorded during the lecture. Show that $H$ and $T$ are independent.
We must show that for every pair of nonnegative integers $(h,t)$, $$\Pr(H=h,T=t)= \Pr(H=h)\Pr(T=t)$$
This is easy enough to do; it's just a matter of writing out the definitions and manipulating the series.
I found the result very surprising, however. (In fact, when I saw the problem, my first thought was that it must be false.) Clearly it depends on the Poisson distribution, but does it just happen to work out this way, or is there some general property of the Poisson distribution that would lead one to guess that $H$ and $T$ are independent? Is it related in some way to PASTA ("Poisson arrivals see time averages,") for example? I haven't been able to recast this in terms of the arrival theorem, which doesn't seem to have anything to do with it, except that's it's another somewhat mysterious property of the Poisson distribution.
If you know that the sum of two independent Poisson random variables is another Poisson random variable with rate which is the sum of the two original rates, then I do not think H and T being independent is intuitively difficult to accept.
Suppose instead you had two trees, a Hawthorn and a Teak, each dropping leaves independently each with rates $\frac \lambda 2$. Then leaves are falling as a Poisson process with rate $\lambda$ and each leaf is H or T with equal probability independently of the other leaves. So this system is equivalent to yours.
I would say this is a consequence of the memoryless property of Poisson processes.