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I would like some help with the following statistical problem.

We have a coin with probability $\theta$ for heads, with prior for $\theta$ being a Beta(a,a) distribution (a is a known parameter). After tossing the coin n times, we get n-1 times heads.

Assuming that we toss the coin again, we would like to find the minimum n for which heads is at least two times more likely than tails.

I tried to address the problem in the following way:

We have that $\pi(\theta)=Beta(a,a)$ and $L(\theta;x)=f(x;\theta)=Bin(n,x)$
Therefore we get: $\pi(\theta|x) \propto Beta(a,a) \times L(\theta;x) \implies \pi(\theta|x) =Beta(a+x,a+n-x)$

Obviously, the posterior of $\theta$ is a function of n and x

But for x=n-1 we get $\pi_n(\theta|x) \propto Beta(a+n-1,a+1)$, that is the posterior of $\theta$ depends only on n.

The n+1 toss is a Bernoulli trial with parameter $\theta$ following the beta distribution described by $\pi_n(\theta|x)$, i.e. $Beta(a+n-1,a+1)$.

In order for heads to be at least two times more likely than tails, we must have $\theta \geq 2/3$.

I am not quite sure, how to continue with the problem after this point.

After skimming some Bayesian Statistics textbooks, I came to the conclusion that the standard Bayesian approach would be to take :$E[\pi_n(\theta)] \geq 2/3$ Still, I am not sure about this.

I have, also, thought of $\frac{\int_{2/3}^{1}\pi_n(\theta)}{\int_{0}^{2/3}\pi_n(\theta)} > 1$

So, summing up, I am not sure how to continue after having calculated the posterior distribution $\pi_n(\theta|x)$ of $\theta$.

Any help would be greatly appreciated.

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    $\begingroup$ The probability of getting heads given a Beta(a, b) distribution is $(a+1)/(a+b+2)$. Should be possible solve for this being at least 2/3. $\endgroup$
    – fblundun
    Dec 29 '20 at 16:40
  • $\begingroup$ I assume that you arrived at this by taking the mean value of Beta, in the manner of the Bayesian approach I described. Could you elaborate on why this is the correct approach? $\endgroup$
    – Filipg
    Dec 29 '20 at 16:48
  • $\begingroup$ Oops, I meant $a/(a + b)$. For example, if you start with a Beta(1, 1) prior and then see 3 heads and 1 tails, you should assign probability 4/6 to getting heads on the next flip. $\endgroup$
    – fblundun
    Dec 29 '20 at 17:19
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After skimming some Bayesian Statistics textbooks, I came to the conclusion that the standard Bayesian approach would be to take :$$𝐸[𝜋_𝑛(𝜃)]≥2/3$$ Still, I am not sure about this.

Correct. We want the probability of the next toss being heads greater than or equal to $2/3$. This means $$P(H|\mathcal D)=\int P(H|\theta)\pi(\theta|\mathcal D)d\theta=\int \theta \pi(\theta|\mathcal D)d\theta = E[\theta|\mathcal D]$$ which is the posterior expectation. $\mathcal D$ represents the data.

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