Coin tossing posterior density calculation I know that my prior distribution is Beta(3,3) and that after tossing 12 coins, the number of 'heads' is less than 4 but I don't know the exact number.
How do I calculate the posterior density?
What I've tried to do is:
If $X=\#$ of heads in $n=12$ tosses then $X\sim Bin(12,\theta)$
$$\pi(\theta|x)=\frac{p(x|\theta)\pi(\theta)}{f(x)}$$
Where $f(x)$ is the marginal density of the Likelihood.
I tried using 4 cases for $X={0,1,2,3}$
And calculate 4 different values of the posterior density. But is that correct?
 A: What use is four posterior densities? I would have thought you wanted one.  It would be a weighted average of them, but perhaps difficult to find the weights.
If $p$ is the probability of a head then, if my calculations are correct,

*

*The prior density is proportional to $p^2(1-p)^2$ from your Beta distribution;


*The likelihood given $3$ or fewer heads from $12$ attempts is proportional to ${12 \choose 0}(1-p)^{12}+{12 \choose 1}p(1-p)^{11}+{12 \choose 2}p^2(1-p)^{10}+{12 \choose 3}p^3(1-p)^9$; and


*the posterior density is proportional to the product of these, but needs to integrate to $1$ over $[0,1]$
which suggests to me a posterior density of $$\dfrac{185640}{1271} p^2(1-p)^{11}\left(1+9p+45p^2+165p^3\right)$$
which is in a sense a weighted average of $\operatorname{Beta}(3,15)$, $\operatorname{Beta}(4,14)$, $\operatorname{Beta}(5,13)$ and $\operatorname{Beta}(6,12)$ densities, but I do not see a quicker way which involves calculating the weights required
In the chart below, the information from the observation has shifted the red prior density to the blue posterior density

