The Monty Hall Paradox is the answer to this question.
Suppose you're on a game show, and you're given the choice of three
doors: Behind one door is a car; behind the others, goats. You pick a
door, say No. 1, and the host, who knows what's behind the doors,
opens another door, say No. 3, which has a goat. He then says to you,
"Do you want to pick door No. 2?" Is it to your advantage to switch
your choice? (wikipedia)
If you are familiar with the paradox, you know that the answer is to always switch because the odds compress into the other door with 2/3 odds.
But if there are three different things, and the game host is allowed to open either non-picked door, and the revealed item is not the desired item, is it still advantageous to switch? Does the removal of the mechanic where the game host might be limited to only opening one door keep the odds at 1/3 for each door?
My initial instinct is that it stays the same, but this is already a mind puzzle anyway, so I'm not sure.
Best Answer
The Monty Hall problem is basically a shell game. The dealer puts a ball under one of three cups and shuffles them around. You pick one cup, the dealer lifts another, and you have a choice of whether to stay with your original choice, or move.
Basically there are three events. $A$ your first choice is correct, $B$ the dealer lifts an empty cup, and $C$ changing cups is a good idea.
In the classic scenario, the dealer always chooses the empty cup and you are asked to evaluate the probability that changing cups is a good idea given that certain event. $$\begin{align}\mathsf P(C\mid B) ~&=~ \mathsf P(C\mid A,B)~\mathsf P(A\mid B)+\mathsf P(C\mid A^\complement,B)~\mathsf P(A^\complement\mid B) \\ &=~ 0\cdot \tfrac 13+1\cdot \tfrac 23 \\ &=~ \tfrac 23\end{align}$$
Since as it is certain that the dealer always chooses an empty cup, then the event that your first guess is right is independent of the dealer's revelation.
So, how does uncertainty that the dealer could lift an empty cup change this? We go to Bayes' Rule. Assuming the dealer's choice between the two untapped cups is biased to wards favouring the ball with probability $p$ when given your first guess was wrong.
$$\begin{align}\mathsf P(C\mid B) &=~ \dfrac{\mathsf P(C\mid A,B)~\mathsf P(B\mid A)~\mathsf P(A)~+~\mathsf P(C\mid A^\complement,B)~\mathsf P(B\mid A^\complement)~\mathsf P(A^\complement)}{\mathsf P(B\mid A)~\mathsf P(A)+\mathsf P(B\mid A^\complement)~\mathsf P(A^\complement)} \\[1ex] &=~ \dfrac{0\cdot 1\cdot \tfrac 13 ~+~1\cdot p\cdot \tfrac 23}{1\cdot \tfrac 13 ~+~ 0\cdot \tfrac 23}\\[1ex] &=~ \dfrac {2p}{1+2p}\end{align}$$
So when the choice is unbiased, then $\mathsf P(C\mid B)= \tfrac 12$. This is what most peoples intuition about the classic Monty Hall game tells them, and its based in failing to realise the host cheats in favour of the player by never picking the prize.