Can the Monty Hall paradox be explained by Berkson's paradox?

Question

I just learned about Berkson's paradox, which says that if $A$ and $B$ are independent, then $P(A\mid B,A\cup B) < P(A\mid A\cup B)$ (knowing that $A$ or $B$ occur creates a negative dependence on $A$ and $B$.

The explanation on Wikipedia reminded me of the Monty Hall paradox, and I'm wondering if Berkson's paradox can be used to explain Monty Hall.

Answer 1

Not so much.   They are somewhat different scenarios.

Berkson's paradox is that when $A$ and $B$ are independent, but we know that either one or the other happened, the conditional probability that $A$ happened is greater than the unconditional probability.

$$\mathsf P(A\mid A\cup B)~\geqslant~\mathsf P(A\mid B\cap(A\cup B)) ~=~ \mathsf P(A\mid B)~=~\mathsf P(A)$$

In the Monty Hall scenario $A$ is the event that the contestant didn't choose the prize door, and $B$ is the event that the host didn't choose the prize door.   We are not given that one or the other did not choose the prize: we're given much the stronger condition that the host will not do so.   $\mathsf P(B)=1$

Answer 2

Here is a potential way to look at it. This is a partial answer. I haven't fully been able to work out if this can truly say something about the classic Monty Hall paradox. If anyone has any ideas, let me know.

Berkson's paradox says that if you have two independent binary random variables, $A$ and $B$, a selection bias in the form of rejecting one of the four combinations creates a negative dependence on the conditioned variables (WLOG we reject $\neg A$ and $\neg B$ so that the conditioning is on $A\cup B$).

Now consider Monty Hall. The question is, can we interpret Monty Hall as a rejection of one of the four combinations of two independent variables? I believe we can. Monty Hall picks a door, at random, conditioned that

1. The door does not have a car
2. The door is not the door you chose

Let $d$ be one of the three doors. $A(d)$ means $d$ has a car behind it and $\neg A(d)$ means it has a goat. $B(d)$ means $d$ is the door chosen by you (the contestant), and $\neg B(d)$ means it is not the door chosen by you. We have $P(A(d)) = P(B(d)) = \frac{1}{3}$.

Now Monty Hall will always choose a door with $\neg A(d)$ and $\neg B(d)$, so the remaining doors are $A \cup B$.

Berkson's paradox tells you that conditioned on $A\cup B$, that is conditioned on doors that Monty Hall would never choose, $A$ and $B$ have a negative dependence. In other words, the door you chose is less likely to have a car.

Let's look at just those doors that Monty Hall would never chose ($A \cup B$), that is, your door, and any door that has a car. For the sake of simplicity, suppose you picked door 1. There are three possibilities, all equally likely:

1

--------
|      |
| car  |
|      |
--------
 door 1

2

--------      --------  
|      |      |      |    
| goat |      | car  |
|      |      |      |    
--------      --------  
 door 1        door 2

3

--------      --------  
|      |      |      |    
| goat |      | car  |
|      |      |      |    
--------      --------  
 door 1        door 3

Looking at this limited information (i.e., conditioning on $A \cup B$), we see a negative dependence on the door you chose (door 1), and a door having a car. If you end up going with your initial choice (door 1), it is unlikely to have a car, and if you ended up winning a car, it is unlikely to be from door 1.

Now the question is, does this biased look at the game give you information about the car? In other words, does this say anything about the original Monty Hall paradox that switching doors increases your chances of winning the car.

I've been thinking about this and I really don't know. It seems like it might, because $A\cup B$ is so closely tied to the information. We're saying the door you chose has a negative dependency on the door having a car conditioned on the doors Monty Hall would not choose, but does this really tell you about a conditional dependency in general? After much thought I can't find a proper link to this biased look and the desired "probability of winning after switching or not switching doors".