Intuition behind independence result The following problem is from Wasserman's $\textit{All Of Statistic}s$. I have worked through the algebra to arrive at the result, but it still seems very strange to me, so I would appreciate any intuition that could be offered.
Let $N$~Poisson($\lambda$). Suppose we toss a coin $N$ times, with $p$ being the probability of heads, and let $X$ be the number of heads, and $Y$ be the number of tails. Show that $X$ and $Y$ are independent.
Now, as I said, I have shown that $X$ and $Y$ are independent, but it $\textit{feels}$ like they shouldn't be. At first glance it seems like a very large value of $X$ should make $P(Y|X)$ smaller for large $Y$. What's going on here?
 A: Maybe it helps to look at this 'the other way around.'
Suppose you have two beakers of radioactive liquid. One of
then emits $N_1 \sim Pois(100)$ particles a minute into your Geiger counter, and independently, the other emits $N_2 \sim Pois(100)$
particles a minute. Knowing that in a particular minute you got
$N_1 = 107,$ would that help you figure out the actual count from
the other beaker in that period of time? No, because of independence.
Now, suppose you pour the contents of both beakers into one
larger beaker. You would now have an average of 200 particles
a minute coming from the combined liquid: $N \sim Pois(200).$
You could view $N$ as $N = N_1 + N_2.$ You could even pretend
that a toss of a fair coin dictated whether each particle came
from a molecule that used to be in the first (Success) or in the second (Failure)
beaker. (But of course that coin would have to be flipped by a quantum-mechanical 'god' that does play dice with the Universe.)
A: You have coin flips being made at Poisson intervals, rate $\lambda$, and are tallying the counts of heads and tails separately.
If we have $x$ heads and $y$ tails, then we have made exactly $x+y$ flips.  We can calculate the probability of making so many flips. $$\mathsf P(N=x+y)= \dfrac{\lambda^{x+y}e^{-\lambda}}{(x+y)!}$$
The conditional probability of having so many heads(and tails) given that many flips is: $$\mathsf P(X=x, Y=y\mid N=x+y) = \dbinom{x+y}{x}p^x (1-p)^y$$  
Thus the probability that the count of heads equals $x$ and the count of tails equals $y$ will be:
$\begin{align}
\mathsf P(X=x, Y=y) & = \mathsf P(N=x+y)\mathsf P(X=x, Y=y\mid N=x+y)
\\ & = \dfrac{\lambda^{x+y}e^{-\lambda}}{(x+y)!}\cdot\dfrac{(x+y)!}{x!y!}p^x(1-p)^y
\\ & = \dfrac{(p\lambda)^xe^{-p\lambda}}{x!}\dfrac{((1-p)\lambda)^ye^{-(1-p)\lambda}}{y!} & \because \lambda = p\lambda+(1-p)\lambda
\end{align}$
That looks like the result of two Poisson processes, with the rate of heads occurring at $p\lambda$, and the rate of tails occurring independently at rate $(1-p)\lambda$.
To confirm we can look at the marginal distributions:
$\begin{align}
\mathsf P(X=x) & = \sum_{n=x}^\infty \mathsf P(N=n)\cdot\mathsf P(X=x\mid N=n)
\\ & = \sum_{n=x}^\infty \dfrac{\lambda^{n}e^{-\lambda}}{(n)!}\cdot\dfrac{(n)!}{x!(n-x)!}p^x(1-p)^{n-x}
\\ & = \dfrac{(p^\lambda)^xe^{-p\lambda}}{x!}\cdot\sum_{n=x}^\infty \dfrac{((1-p)\lambda)^{n-x}e^{-(1-p)\lambda}}{(n-x)!}
\\ & = \dfrac{(p^\lambda)^xe^{-p\lambda}}{x!}\cdot\sum_{m=0}^\infty \dfrac{((1-p)\lambda)^{m}e^{-(1-p)\lambda}}{m!}
\\ & = \dfrac{(p^\lambda)^xe^{-p\lambda}}{x!}\cdot 1
\end{align}$
And indeed, heads do occur in a Poisson distribution of rate $p\lambda$.  By a symmetrical argument we find that tails occur at rate $(1-p)\lambda$.
And we can quite see clearly that we have independence, since: $\mathsf P(X=x\mid Y=y) = \mathsf P(X=x)$.

tl;dr
Pennys from heaven are dropped in a Poisson process at rate $\lambda$, and we find that each one lands heads up with probability $p$.   This means heads will show up at rate $p\lambda$ and tails at rate $(1-p)\lambda$.   This distribution is the same if the angels are letting the coins tumble as they fall, or if they are perfectly dropping at those two rates to land in a predetermined way.
Thus the tally of heads tells us nothing about the tally of tails, as Poisson processes are memoryless.
A: Suppose you're talking about a fair coin, and you're told that 50 tosses (from an unknown number of tosses) are heads. Intuitively, to infer $Y=N-X$ given $X=50$, say, you would need to first infer $N$, which in turn is inferred based on the probability of heads relative to tails (the next best thing you have as you don't know X), and this probability has absolutely nothing to do with your observed values.
Essentially, the detail is that you have no idea about what the entire sample size is, so if you are only given that a million people in city X are male, the best guess you have for the number of females is also a million, i.e. $\mathbb P [\text{individual is female}]=0.5$ regardless of the number of males in the city when you don't know the full population.
