# Coin Toss with Prior Distribution for $p$

I am attempting to answer the following question from my actuarial exams:

We are asked to find $$P(p=0.5|X=7)$$.

Using Bayes Rule, I believe this is

$$P(p=0.5|X=7) = \frac{P(X=7|p=0.5)P(p=0.5)}{P(X=7|p=0.5)P(p=0.5) + P(X=7|p \sim Unif[0.5,1])P(p \sim Unif[0.5,1])}$$

Straightforwardly, using $$X \sim Bin(8,p)$$:

$$P(X=7|p=0.5)P(p=0.5) = {8 \choose 7}0.5^70.5 \times 0.8$$

I am less sure of $$P(X=7|p \sim Unif[0.5,1])P(p \sim Unif[0.5,1])$$, if even this is the right expression. Given it's continuous, I thought it might be:

$$P(X=7|p \sim Unif[0.5,1])P(p \sim Unif[0.5,1]) = \int^1_{0.5}{P(x|p)f(p) dp}$$

where $$P(x|p) = {8 \choose 7}p^7(1-p)$$ and $$f(p) = 2$$ (being the pdf of $$Unif[0.5,1]$$). But this doesn't account for the probability of p being uniformly distributed, $$P(p \sim Unif[0.5,1]) = 0.2$$.

Many thanks

Steven

• This question was probably inspired by en.wikipedia.org/wiki/Paul_the_Octopus. Commented Aug 13, 2021 at 16:34
• @fblundun I'm sure it is, and appreciated the reference when I saw the question Commented Aug 14, 2021 at 10:44

The question states the prior distributions on the hypotheses as $$\pi(H_0)=0.8$$ and $$\pi(H_1)=0.2$$. From Bayes' theorem: $$p(H_0\,|\,k)=\frac{P(k\,|\,H_0)\pi(H_0)}{P(k\,|\,H_0)\pi(H_0) + P(k\,|\,H_1)\pi(H_1)}$$ We observe $$k=7$$ correct predictions out of $$n=8$$ total predictions. Using the binomial mass function, we can calculate the posterior as follows: $$P(k\,|\,H_0) = {n\choose k}(0.5)^k(1 - 0.5)^{n-k} = {8\choose 7}(0.5)^7(1 - 0.5)^{1} = \frac{1}{32}$$ and for the alternative hypothesis, we integrate over its range, weighted by the density of the $$U(1/2,1)$$ distribution (which is $$2$$): $$P(k\,|\,H_1) = \int_{0.5}^{1}2{n\choose k}(p)^k(1 - p)^{n-k}\,\mathrm{d}p = 2\int_{0.5}^{1}{8\choose 7}(p)^7(1 - p)^{1}\,\mathrm{d}p=\frac{251}{1152}$$ So finally we have: $$p(H_0\,|\,7)=\frac{\dfrac{1}{32}0.8}{\dfrac{1}{32}0.8 + \dfrac{251}{1152}0.2} = \frac{144}{395} = 0.364557$$
$$P(X=7|p=0.5)\cdot P(p=0.5)$$ $$+\int_{0.5}^1 P(X=7|p= r\cap 0.5.
Let me know if I have made a mistake. The term $$P(p=r|0.5 is the pdf of the uniform distribution over (0.5,1) and the term $$P(0.5 would account for the belief mass of 20%. Here is a related thread asking about how to apply these probability statements to the coin and octopus under investigation.