# Questions on Bayesian analysis of an opinion poll (an example in a book)

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 $M_2$.

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$87.03$ looks reasonable to me as the ratio of the integrals.
The reason for the models and the $p(H|M,I)$ terms is to avoid rewarding uncertain hypotheses for their uncertainty. You may disagree, arguing that uncertain hypotheses are more likely to be true.
For example, if one hypothesis was that $H=0.55$ exactly and the other was $0 \le H \lt 0.1$, then your final suggestion would always give $O_{12}=0$ no matter what data was supplied, since the numerator would always be an integration over a zero-length interval, while the denominator would always be positive.