help computing the beta likelihood when we only observe the number of successes and failures (not the latent probability of success) Imagine we have a biased coin that generates heads with unknown probability $\theta$ where $\theta$ is drawn from a beta distribution with known parameters $(\alpha, \beta)$.
Next imagine that we flip the coin some number of times and we get $m$ heads and $n$ tails. I'd like to compute the quantity $p(m,n|\alpha,\beta)$.
What I have so far is
$$
\begin{align}
p(m,n|\theta) &= \begin{pmatrix}m+n\\n\end{pmatrix} \theta^m (1-\theta)^n \\
&\\
p(\theta|\alpha,\beta) &= \frac{\theta^{\alpha-1}(1-\theta)^{\beta-1}}{B(\alpha,\beta)} \\
&\\
p(m,n|\alpha,\beta) &= \int_0^1  p(m,n|\theta) \, p(\theta|\alpha,\beta)\,d\theta \\
&= \int_0^1 \begin{pmatrix}m+n\\n\end{pmatrix} \theta^m (1-\theta)^n \frac{\theta^{\alpha-1}(1-\theta)^{\beta-1}}{B(\alpha,\beta)} \, d\theta\\
&= \begin{pmatrix}m+n\\n\end{pmatrix}\bigl(B(\alpha,\beta)\bigr)^{-1}
   \int_0^1  \theta^{m+\alpha-1}(1-\theta)^{n+\beta-1}\,d\theta \\
&= \begin{pmatrix}m+n\\n\end{pmatrix} \frac{B(m+\alpha, n+\beta)}{B(\alpha,\beta)}.
\end{align}
$$
The final expression is pretty nice, but I'm wondering if it can be simplified even further. For instance, is it equivalent to a binomial distribution with some parameter $\theta'$?

The beta binomial and binomial distributions are distinct. The binomial distribution has mean $kp$ and variance $kp(1-p)$ where $k$ is the number of trials and $p$ is the probability of success.
The pmf that you have written is a beta-binomial distribution (as you know). The beta binomial distribution has mean
$$\begin{align}
\mu
&=(n+m)\frac{\alpha}{\alpha+\beta} \\
&=q(n+m) 
\end{align}$$
This is suggestive! Perhaps we can use the substitution $q = \frac{\alpha}{\alpha+\beta}$ and show that the beta-binomial distribution is secretly a binomial distribution.
Unfortunately, this plan falls apart when we look at the beta-binomial variance
$$\begin{align}
\sigma^2 &=
(n+m)\frac{\alpha}{\alpha+\beta}\cdot\frac{\beta}{\alpha+\beta}\cdot\frac{\alpha+\beta+n+m}{\alpha+\beta+1} \\
&= q(n+m)(1-q)\frac{\alpha+\beta+n+m}{\alpha+\beta+1}
\end{align}$$
and the first three factors are exactly what we want to find in a binomial distribution... but the fourth factor is not equal to 1 for $n + m > 1$.
