A prize is placed behind one of three closed doors. You can pick (but not open) any door. An assistant will open one of the remaining two doors. Suppose that prize is not behind the door opened by the assistant. You are given the choice of staying with your original pick or switching to the other closed door. You win the prize, if it is behind the door you ultimately pick.
There are two variants. In the first (the original "Monty Hall problem"), your assistant is the host of the show, Monty Hall himself. He has complete information about the game and will not open the door hiding the prize. In the second variant, your assistant is completely clueless and picks the door at random among the two possible choices.
What is the best strategy in each case: stick with the original choice or switch to the other closed door?
In order to establish a framework that allows to analyze both cases (Monty and a clueless assistant), it is useful to assume that both Monty and assistant select the door that they open at random. It really is so in the case of the assistant (to be precise, when we say that the assistant selects the door at random, we mean that the probability of the assistant picking one door among the two doors that you did not select is $\frac{1}{2}$). In the case of Monty picking the door, we can assume the following. If your original pick was wrong, then Monty selects the door hiding the prize with probability $0$ and the one without the prize — with probability one. If your original pick was right, Monty selects either door with probability $\frac{1}{2}$. We are using the fact that we can always add to the sample space any fake outcomes with probabilities zero without changing the probability of any event.
We label the doors as A, B and C, with the originally picked door labeled A and the door opened by assistant — as C. We now compute the probability of winning if we switch, i.e. select the door B.
\begin{multline*} P \left( \begin{matrix} \text{winning with}\\ \text{Monty's help} \end{matrix} \right) = P \left( \begin{matrix} \text{the prize} \\ \text{is behind} \\ \text{door B} \end{matrix} {\Huge|} \begin{matrix} \text{Monty}\\ \text{opened} \\ \text{door C} \end{matrix} \right) = \\ \\ \\ = \frac { P \left( \begin{matrix} \text{the prize} \\ \text{is behind} \\ \text{door B} \end{matrix} \right) \cdot P \left( \begin{matrix} \text{Monty}\\ \text{opened} \\ \text{door C} \end{matrix} {\Huge|} \begin{matrix} \text{the prize} \\ \text{is behind} \\ \text{door B} \end{matrix} \right) } { P \left( \begin{matrix} \text{the prize} \\ \text{is behind} \\ \text{door A} \end{matrix} \right) \cdot P \left( \begin{matrix} \text{Monty}\\ \text{opened} \\ \text{door C} \end{matrix} {\Huge|} \begin{matrix} \text{the prize} \\ \text{is behind} \\ \text{door A} \end{matrix} \right) + P \left( \begin{matrix} \text{the prize} \\ \text{is behind} \\ \text{door B} \end{matrix} \right) \cdot P \left( \begin{matrix} \text{Monty}\\ \text{opened} \\ \text{door C} \end{matrix} {\Huge|} \begin{matrix} \text{the prize} \\ \text{is behind} \\ \text{door B} \end{matrix} \right) + P \left( \begin{matrix} \text{the prize} \\ \text{is behind} \\ \text{door C} \end{matrix} \right) \cdot P \left( \begin{matrix} \text{Monty}\\ \text{opened} \\ \text{door C} \end{matrix} {\Huge|} \begin{matrix} \text{the prize} \\ \text{is behind} \\ \text{door C} \end{matrix} \right) } = \\ \\ \\ = \frac { \frac{1}{3} \cdot 1 } { \frac{1}{3} \cdot \frac{1}{2} + \frac{1}{3} \cdot 1 + \frac{1}{3} \cdot 0 } = \frac{2}{3} .\end{multline*}$ P \left( \begin{matrix} \text{assistant}\\ \text{opened} \\ \text{door C} \end{matrix} {\Huge|} \begin{matrix} \text{the prize} \\ \text{is behind} \\ \text{door C} \end{matrix} \right) = 0 $ because this situation would result in a replay of the game.
Now we can also find the probability of winning when following "no switch" strategy. It is obvious that when Monty is assisting you, that probability is $\frac{1}{3}$. Somewhat surprisingly, that probability is $\frac{1}{2}$ when the clueless assistant is helping you. That is because the game is replayed if the assistant opens the door with the prize, which would have resulted in your loss otherwise.
This problem illustrates one very important point: the future depends not only on the present state of the world, but also on the past history. Indeed, observe that the physical world before you make your ultimate choice is the same, regardless whether Monty or the clueless assistant is helping you.