Suppose I ask $2N$ people a yes or no question. The answer are independent binomially distributed random variables. Each individual has his own probability of saying yes. When the $i$th responds, he says yes with probability $p_i$. Moreover,
$$\\p_1,p_2,...,p_{2n} \sim Uniform[0,1]$$
and are independently and identically distributed. What is the probability half of the responders say yes?
I know that
$$\int_0^1p dp$$
Is the probability of voting yes for some individual. I'm inclined to express the probability half of the responders say yes as,
$$\\\int_0^1\binom{2N}{N} p^{N}(1-p)^{N}dp.\\$$
But this seems to impose that $p=p_1,p_2,...,p_{2n}$, which doesn't need to be the case.
I am not entirely sure, but here what looks right to me.
Conditioned on $p=(p_1,\dots,p_{2N})$, what you have is a random variable $X_p$ following a Poisson Binomial distribution with parameters $2N$ and $p$. So what you want is the probability that your random variable $X$ equals $N$, which is $$ \mathbb{P}\{X=N\} = \mathbb{E}\mathbf{1}_{\{X=N\}} = \mathbb{E}[\mathbb{E}[\mathbf{1}_{\{X=N\}}\mid p]] = \mathbb{E}[\mathbb{P}[X_p = N]] $$ From the known expression (cf. Wikipedia article) of the pmf of a Poisson Binomial r.v., we have $$ \mathbb{P}[X_p = N] = \sum_{A\in F_N} \prod_{i\in A} p_i \prod_{j\in A^c} (1-p_j) $$ so that, by linearity of expectation and independence of the $p_i$'s, $$\begin{align} \mathbb{P}\{X=N\} &= \sum_{A\in F_N}\mathbb{E}\left[\prod_{i\in A} p_i \prod_{j\in A^c} (1-p_j)\right] = \sum_{A\in F_N} \prod_{i\in A} \mathbb{E}[p_i] \prod_{j\in A^c} \mathbb{E}[(1-p_j)]\\ &= \sum_{A\in F_N} \left(\frac{1}{2}\right)^{\lvert A\rvert}\left(\frac{1}{2}\right)^{\lvert A^c\rvert} = \frac{\lvert F_N\rvert}{2^{2N}}\\ &= \boxed{\frac{1}{2^{2N}}\binom{2N}{N}} \end{align}$$ (I followed the notation of Wikipedia, writing $F_k$ for the set of all subsets of $k$ integers that can be selected from $\{1,2,3,\dots,2N\}$).