# 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.

## 2 answers

Maybe it's a good idea to start from a situation that's intuitive, and then gradually chance it to the Monty Hall version.

So let's start with the following situation:

Monty lets you choose a door. Then, after you've chosen it, he offers you to either select the chosen door, or open *all* the other doors. Well, it should be clear to everyone that staying at the one door gives you only a chance of $1/n$ to get the car, while opening all the other doors gives you a chance of getting the car. Note that if the car is behind those $n-1$ doors, opening all those doors gives you an additional bit of information, but one that is irrelevant to you: Behind *which* of those doors the car sits.

Now let's assume that in order to save time, Monty does not open all those doors, but only the one with the car if it is among them, and a random door with a goat otherwise. Obviously that doesn't affect the probability of winning the car.

Now let's, on the contrary, assume Monty wants to increase the tension by first opening all those of the doors neither you nor he did choose. In other words, he first reveals $n-2$ goats. Since you know in advance that those $n-2$ doors will show goats, this gives you no information about whether you've won the car. It *does* give you some information about in which of those doors the car is in case it is behind one of them, but as noted above, that information is irrelevant. Therefore your probability of having won the car still stays the same, as you didn't get any new information about whether you've won it.

But if opening those doors doesn't give you any relevant information, we can move it to before you made the choice without affecting the probabilities. That is, Monty will first open the $n-2$ goat doors, and then present you with the choice whether you want to stay with the door you chose, or select the other $n-1$ doors. But since $n-2$ doors are already open, choosing the $n-1$ doors is equivalent to you choosing the one of them which hasn't been opened.

But now we've arrived at the original Monty Hall situation. And since at no point during the transformation the probabilities changed, you still have the probability of $(n-1)/n$ if switching versus $1/n$ if staying.

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This is the way of explaining it that works the best for me.

Forget about the probabilities for a moment - after the host opens 98/100 doors, there are 2 doors left. Therefore, you either have selected a door with a goat or a door with a car. Now, if you have selected a goat, then switching will get you the car. If you've selected a car, then switching will unfortunately get you the goat.

Now think back to the beginning. There were 100 doors and only 1 car. Therefore at the beginning you had a 99/100 chance to pick a goat. And we know that if you picked a goat, switching gets you the car, so you have a 99/100 chance to receive the car if you switch.

The probabilities are congregating on the last door because of how the strategy of switching works.

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