After $n - 2$ unchosen doors are opened, how does the probability of the $n - 2$ unchosen doors "shift" or "transfer" to the lone unchosen door?
The second quotation below uses the verb "shift" to describe how Monty Hall's opening the 98 unchosen doors (revealing a goat each) ""shifts" [boldening mine] that 99/100 chance to door #100"? The first quotation below uses "transfer". But how can probabilities "shift" or "transfer"?
This analogy sounds featherbrained, but I hope it explains my bafflement. Imagine a doorman beside each door. Each time after the host opens an unchosen door (revealing a goat), each doorman walks over and congregates at another random unchosen door. After the host opens $n - 2$ unpicked doors, then $n - 2$ doormen will have congregated at the lone unchosen door. But probabilities aren't like doormen! From the $n - 2$ unchosen doors, they can't then "relocate" or "shift" to the lone unchosen door!
- Judea Pearl, The Book of Why (2018). Page 181 of 402.
As you can see from Figure 6.1, Door Opened is a collider. Once we obtain information on this variable, all our probabilities become conditional on this information. But when we condition on a collider, we create a spurious dependence between its parents. The dependence is borne out in the probabilities: if you chose Door 1, the car location is twice as likely to be behind Door 2 as Door 1; if you chose Door 2, the car location is twice as likely to be behind Door 1.
It is a bizarre dependence for sure, one of a type that most of us are unaccustomed to. It is a dependence that has no cause. It does not involve physical communication between the producers and us. It does not involve mental telepathy. It is purely an artifact of Bayesian conditioning: a magical transfer of information without causality. Our minds rebel at this possibility because from earliest infancy, we have learned to associate correlation with causation. If a car behind us takes all the same turns that we do, we first think it is following us (causation!). We next think that we are going to the same place (i.e., there is a common cause behind each of our turns). But causeless correlation violates our common sense. Thus, the Monty Hall paradox is just like an optical illusion or a magic trick: it uses our own cognitive machinery to deceive us.
Why do I say that Monty Hall’s opening of Door 3 was a “transfer of information”? It didn’t, after all, provide any evidence about whether your initial choice of Door 1 was correct. You knew in advance that he was going to open a door that hid a goat, and so he did. No one should ask you to change your beliefs if you witness the inevitable. So how come your belief in Door 2 has gone up from one-third to two-thirds?
Here’s another way to visualize this. Imagine that instead of three doors, Monty Hall presents you with 100 doors; behind 99 of them are goats, and behind one of them is the car. You select door #1, and your initial odds of winning the car are now 1/100:
Then, let’s suppose that Monty Hall opens 98 of the other doors, revealing a goat behind each one. Now you’re left with two choices: keep door #1, or switch to door #100:
When you select door #1, there is a 99/100 chance that the car is behind one of the other doors. The fact that Monty Hall reveals 98 goats does not change these initial odds -- it merely "shifts" [boldening mine] that 99/100 chance to door #100. You can either stick with your original 1/100 odds pick, or switch to door #100, with a much higher probability of winning the car.
At this point the switch shouldn’t seem attractive, there’s no benefit to changing. That all changes when Monty opens door C. This action means that there is then still a 2/3 chance that the car isn’t behind door A. This means there is a 2/3 chance the car is behind B or C. The opening of door C hasn’t affected the probabilities associated with door A but it has affected those associated with door B. It’s almost as if door B must carry the full load [boldening mine] of probabilities that was previously held by both B and C. This is because C isn’t chosen randomly. There is a 2/3 chance the car is behind the B&C pairing, that burden is left to B. Let’s look at that diagrammatically representing B & C as a pair with joint probabilities.
Now let’s have Monty open door C again and see how that affects the probabilities.
The Hundred Doors Solution to The Monty Hall Problem.
In this hypothetical version of the problem the contestant chooses not from 3 doors but from a hundred, for convenience we’ll label these doors 1–100 and say our contestant selects door 1. She has a 1/100 of selecting the door hiding the car and a 99/100 chance of not selecting a car. Let’s represent that with a stripped-down diagram. Again, black denotes the chance of selecting the car. Red the chance of not selecting the car.
Now alternative universe Monty works his way through doors 2–100 careful revealing bearded goats until the bored contestant is left with two doors. Door 1 which she initially chose and door 2 the last remaining unchosen door. Here’s how the odds stack up at this point.