# Bayesian update process on example of unfair dice

Let's say we have an unfair dice with the initial (estimated) probability of rolling a specific number = 1/6. How does probability estimation update when getting this specific number on every next experiment? How does probability change when getting other number (on failure)?

Thanks.

-
You should not start with a point estimate but with a prior distribution, which we might call $f(p)$ where $p$ is the unknown probability.
For example, suppose you are trying to roll a 6. You might get $x$ successes and $y$ failures. The likelihood of this is proportional to $p^x (1-p)^y$. So your posterior distribution is
$$f(p|x,y) = \frac{p^x (1-p)^y f(p)}{\int p^x (1-p)^y f(p) \, dp}$$ with the integral taken over the possible range of $p$, which is typically $[0,1]$ for a probability.
If you start with a prior distribution which is $f(p)=\text{Beta}(\alpha,\beta)$ then you will get a posterior distribution of $f(p|x,y)=\text{Beta}(\alpha+x,\beta+y)$: this is called a conjugate prior and assuming it makes the mathematics easier. The expected value of a Beta distribution $\text{Beta}(\alpha,\beta)$ is $\frac{\alpha}{\alpha+\beta}$, so you might decide to start with $\beta = 5 \alpha$, and set $\alpha$ according to how confident you are at the start that $p$ is close to $\frac{1}{6}$. This would make the expectation value of your posterior distribution $\frac{\alpha+x}{\alpha+x+\beta+y}$.