I'm sorry in advance for rather long questions. This is an example in "Bayesian logical data analysis for physical sciences" by P. C. Gregory and I have some questions about the example.
In a poll of 800 decided voters, 440 voters supported the political party A. Let's denote the poll result as $D$. The quantity of interest is the probability that the party A will achieve a majority of at least 51% in the upcoming election, assuming the poll will be representative of the population at the time of the election.
The book regards the problem as a model selection problem.
$M_1$ : The party A will achieve a majority with a parameter $H$ that has uniform prior in the range $0.51 \le H \le 1$.
$M_2$ : The party A will not achieve a majority with a parameter $H$ that has uniform prior in the range $0 \le H < 0.51$.
If we have no prior reason to prefer $M_1$ over $M_2$, we can write the odds ratio $\begin{aligned} O_{12}&=p(M_1|D,I)/p(M_2|D,I)\\ &=p(D|M_1,I)/p(D|M_2,I)\\ &=\frac{\int_{0.51}^1 p(H|M_1,I)p(D|H,M_1,I) dH }{\int_{0}^{0.51} p(H|M_2,I)p(D|H,M_2,I) dH}\\ &=\frac{\int_{0.51}^1 (1/0.49)p(D|H,M_1,I) dH }{\int_{0}^{0.51} (1/0.51)p(D|H,M_2,I) dH}\\ &=87.68 \end{aligned}$
Here are my questions. The book don't give explicit expressions for $p(D|H,M_1,I)$ and $p(D|H,M_2,I)$. If I use binomial distribution $p(D|H,M_1,I)=p(D|H,M_2,I)=\frac{800! H^{440}(1-H)^{800-440}}{440!(800-440)!}$ I get $87.03$ as a result. It is not same to the value $87.68$ of the Book. What probability distribution should I use for the likelihoods?
I have another question. Why do I have to introduce the models $M_1$ and $M_2$? Is
$ O_{12}=\frac{\int_{0.51}^1 p(H|D,I) dH}{\int_{0}^{0.51} p(H|D,I) dH} $ not an appropriate aproach for the problem? It does not have the factor $(1/0.49)/(1/0.51)$ introduced with the models $M_1$ and $M_2$.