# Unexpectedly small likelihood function

My question is premised on this one, which I'm trying to do in a Bayesian style. I'm a Bayes noob.

In a recent poll of 755 randomly selected adults, 587 said X. Test the claim that 75% of adults think X.

I'm starting out just by working out a likelihood function (before testing any claims; I don't want to announce a prior just yet). So I did the following in Mathematica:

l[ev_] := Function[{param},
Probability[var == ev,
Distributed[var, BinomialDistribution[755, param]]]


The idea being that l[evidence] is the likelihood function under that evidence, where param is the underlying probability of the binomial distribution.

I then calculated the integral of l[587] over all param between $0$ and $1$, and was a little surprised to get an integral of $0.00132$. According to my previous understanding, that means no hypothesis in my hypothesis space (i.e. no probability for the binomial distribution) makes the outcome very likely.

If I treat this function l[587] as a Bayesian update from the uniform prior (i.e. I normalise, by dividing through by that integral), I get a function which peaks at the value param == 0.777 with value 26.3756; a lot of the probability mass is now concentrated around that value of the parameter, which is basically what I expected. But I didn't want to treat any priors at this point, and normalising is effectively forcing the likelihood function to become a PMF (which it shouldn't be, because that implies a prior has been picked).

What is the One True Way(TM) of interpreting my unexpectedly-small likelihood function? Why is it so small? Does it matter?

Although I am an ignoramus in Bayesian statistics, I can explain the number $0.00132$. In fact, it is just equal to $1/756$. So this is neither small nor large; this is not unexpected.
Indeed, the probability to observe $k$ (here it is $587$, but this won't matter, as you will see that the result is independent of $k$) is
$$L(k,p) = {n \choose k} p^{k}(1-p)^{n-k},$$ where $n=755$ and $p$ is the success probability. Integrating, we get $$\int_0^1 L(k,p) dp = {n \choose k} \int_0^1 p^{k}(1-p)^{n-k}dp = {n \choose k}\mathrm{B}(k+1,n-k+1),$$ where $\mathrm{B}$ is the beta function. We know that $$\mathrm{B}(k+1,n-k+1) = \frac{\Gamma(k+1)\Gamma(n-k+1)}{\Gamma(n+2)} = \frac{k!(n-k)!}{(n+1)!}.$$ Therefore, $$\int_0^1 L(k,p) dp = \frac{n!}{k!(n-k)!}\frac{k!(n-k)!}{(n+1)!} = \frac{1}{n+1},$$ as claimed.