Asking for an "intuitive" explanation of a probability problem The problem is as follows:

We pick a real number p in $(0,1)$ randomly and uniformly, then construct a coin such that when tossed, $P(H) =p$ and $P(T) = 1- p$.
Now fix a positive integer $n$, if we were to toss the coin $n$ times (independently), what is the probability that exactly $k$ ($0 \leq k \leq n$) heads occur?

I have done this by evaluating the integral $$\int_0^1{{n \choose k}p^k(1-p)^{n-k}}dp = \frac {1}{n+1}.$$
However, the evaluation was far from easy and I had to use the "snake oil" method mentioned in the book "generatingfunctionology" by Wilf.
Yet this result is so simple and beautiful, in particular, it is very suprising to me that the probability is independent of $k$. I tend to believe there must be a simpler reasoning for this, something I failed to notice that can draw the conclusion "the probability is independdent of $k$". Can someone show me such a way(if there is one)?
 A: The desired random construction is equivalent to the following: we first choose a random real number $x_0 \in [0, 1]$ uniformly. Next, we choose $n$ more random real numbers $x_1, \dots x_n \in [0, 1]$, also uniformly and independently of each other and $x_0$, and we want to know the probability that exactly $k$ of them lie to the left of $x_0$. Here $x_0$ plays the role of the probability $p$ and the $x_i$ simulate tossing the coin.
Rephrasing the construction this way allows us to take advantage of an extra symmetry which is not at all apparent in the original construction: the question is equivalent to choosing $n+1$ i.i.d. real numbers $x_0, \dots x_n \sim U(0, 1)$, then computing the probability that $x_0$ is exactly the $(k+1)^{st}$ number if they are sorted. But the $x_i$ are i.i.d., so by symmetry this probability must be the same for each of the $x_i$! So it must be $\frac{1}{n+1}$.
The integral you need to evaluate to do this problem more directly is called the Beta function integral, and this is actually a probabilistic argument that evaluates it. The same symmetry argument shows more generally that each possible order that the $x_i$ could occur in occurs with the same probability, which must therefore be $\frac{1}{(n+1)!}$.
