Take the famous Monty Hall problem. There are three doors, one has a prize. If you pick the door with a goat, he reveals the other door with the goat. If you picks the door with the car, he randomizes and reveals one of the goats.
The calculations of the probabilities are what is confusing to me. So say you pick door 1 initially. and I define event $A_i =$ car is behind door $i$, and event $B_i$, Monty Hall reveals door $i$.
I have seen this calculation made:
$\mathbb{P}(A_1 \cap B_2)=$ probability the car is behind door 1 and he opens door 2 $= \frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$.
$\mathbb{P}(A_1 \cap B_3)= \frac{1}{3}\times\frac{1}{2} = \frac{1}{6}$
$\mathbb{P}(A_2 \cap B_3)= \frac{1}{3}\times 1 = \frac{1}{3}$
$\mathbb{P}(A_3 \cap B_2)= \frac{1}{3}\times 1= \frac{1}{3}$
It can then be shown using conditional probabilities that you have a $\frac{2}{3}$ chance of winning if you switch, and $\frac{1}{3}$ if you stay, and this is because Monty's choice is not random, it is conditional on your choice.
I actually do not understand why we can calculate the above probabilities. For instance, take the first one, $\mathbb{P}(A_1 \cap B_2)$. We are calculating above it seems as $\mathbb{P}(A_1)\times \mathbb{P}(B_2)$. Are we assuming $A_1$ and $B_2$ are independent? How can we do that - isn't the probability that he reveals door 2 dependent on whether or not the prize is behind door 1? So shouldn't this itself be a conditional probability, i.e. something like $\mathbb{P}(A_1)\times \mathbb{P}(B_1\mid A_1)$?
It should, and it is. That is why the second factor in each expression have differing values: they depend on the first event.
$\qquad\begin{align}\Pr(A_1\cap B_2)&=\Pr(A_1)\cdot\Pr(B_2\mid A_1)\\&=(1/3)\cdot(1/2)&&\raise{1ex}{\small\text{Monty has equal chance to reveal door 2 or 3}\\\small\text{ when the car is behind door 1}}\\&=1/6\\[2ex]\Pr(A_2\cap B_2) &=\Pr(A_2)\cdot\Pr(B_2\mid A_2)\\ &= (1/3)\cdot (0)&&\raise{1ex}{\small\text{Monty can not reveal door 2 when}\\\small\text{ the car is behind that door}}\\&=0\\[2ex]\Pr(A_3\cap B_2)&=\Pr(A_3)\cdot\Pr(B_2\mid A_3)\\&=(1/3)\cdot(1)&&\raise{1ex}{\small\text{Monty can only reveal door 2}\\\small\text{ the car is behind door 3}}\\&=1/3\end{align}$
Hence:
$\qquad\begin{align}\Pr(A_3\mid B_2)&=\dfrac{\Pr(A_3\cap B_2)}{\Pr(A_1\cap B_2)+\Pr(A_2\cap B_2)+\Pr(A_3\cap B_2)}\\&=\dfrac{1/3}{1/6+0+1/3}\\&=2/3\end{align}$