Probability of having rolled the unfair die given the result of a roll. Let's say I have a box with two dice. One of them is a regular die and the other has 3 on every face. Without looking, I randomly pick one of the dice, roll it, and see that I got a 3. I still don't know whether I rolled the fair or unfair die.
Does the fact that I rolled a 3 on my first roll change the likelihood that the die I rolled was the fair or unfair one, and if so, by how much?
 A: Find $P\left(\text{result is }3\right)$, $P\left(\text{result is }3\mid\text{fair die is rolled}\right)$
and $P\left(\text{fair die is rolled}\right)$.
This enables you to find:
$$P\left(\text{fair die is rolled}\mid\text{result is }3\right)$$ on base of equality:$$P\left(\text{fair die is rolled}\mid\text{result is }3\right)P\left(\text{result is }3\right)=$$$$P\left(\text{result is }3\mid\text{fair die is rolled}\right)P\left(\text{fair die is rolled}\right)$$
A: The probability of choosing either die is 0.5, and the probability of seeing a three is approximately 0.583 (easy exercise).
Similarly the probability of picking the fair die and seeing a three is approximately 0.083, whereas the probability of picking the biased die and seeing a three is simply 0.5.
Use Bayes’ rule to calculate the probability of choosing either die, given that a three was observed. 
A: Suppose your prior (belief before the event) is that you picked the normal die with probability $p$. Under this prior, there is a $p/6$ chance of picking the normal die and rolling a $3$ but a $1-p$ chance of picking the same-sided die and rolling a $3$. Therefore the posterior (belief after the event) is that you picked the normal die with probability
$$\frac{p/6}{p/6+1-p}=\frac p{6-5p}$$
Since $6-5p\ge1$, you are more confident that you picked the same-sided die after observing a $3$. In the case of $p=\frac12$ this works out to a revised probability of $\frac67$ that you picked the same-sided die.
