[rspeele]

> It doesn't make a difference what causes Monty to reveal a goat.

Oh, but it does! See the "Monty Fall" version of the problem, in which Monty accidentally trips and opens a door, which just happens to reveal a goat. In this variant there is no advantage gained by switching, because no more information was revealed about the remaining unopened door.

The information gain only happens in the original game because we know that Monty was forced to avoid the winning door in the (66% likely) case where we didn't already pick it.

[bemusedthrow75]

Nope.

But I am going to leave this to someone else to explain. I'm tired out now.

[jncfhnb]

Suppose Car is C.

The only possible options are (equally likely)

You choose A and get shown B You choose B and get shown A You choose C and get shown A You choose C and get shown B

4 options. You lose 50% of the time.

The 2 options where you would normally get the 2/3 odds are explicitly ignored when you are told Monty chosen randomly and randomly got a goat.

You choose A and get shown C or you choose B and get shown C are forbidden.

50/50 probability on the dot.

If Monty chooses with intention then the options where you choose goat and are shown goat double in probability so you get back to 2/3

[rspeele]

They are two different problems. Map out all the scenarios exhaustively and you'll find the difference.

In both cases we were originally 1/3 chance of being right. That is not in dispute.

In the original (fully defined) "Monty Hall", Monty was going to show us a goat no matter what. It's part of the rules, he has to show a goat. So the fact that we see a goat behind the revealed door is no surprise, and no new information. But which of the two unchosen doors was the goat, is valuable information because in 2/3s of the scenarios Monty's hand was tied and he HAD to show that door to avoid revealing the remaining car.

In the "Monty Fall" problem, the fact that we see a goat at all is interesting information. This becomes more likely when we picked the car in the first place, because if we had initially picked the car, and a random other door is opened, it's 100% going to be a goat, whereas if we had picked the goat in the first place, we are only 50% likely to see a goat when a random other door is opened. Let's call the goats Alice and Bob to illustrate this point. We know we DID see a goat, but we don't know which of these equally probable scenarios led to that:

1. We picked the car and saw Alice

2. We picked the car and saw Bob

3. We picked Alice and saw Bob

4. We picked Bob and saw Alice

Notice how "we picked the car" originally had 1/3 odds but represents half the scenarios that remain possible, because there are two ways to see a goat with that start, while only one way to see a goat with the others.

This kind of brings the problem back around to similar territory as Bertrand's Box[0] where the fact that you drew a gold coin is already hinting to you that you're more likely on the "both gold" box than on the "half gold" box.

[0]https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox

[rspeele]

Since this bugged me all day, and I suspect you are the kind of person where it bugged you all day, too, here is a better description of the "fall"/"hall" distinction.

I think we can agree that these are the six possible, equally likely, configurations of the problem starting from me having chosen door 1. G1 here is "goat 1" and G2 is "goat 2". For each possible prize behind my chosen door, there are two possible configurations of the remaining prizes.

    My Door | Door 2 | Door 3
    Car     | G1     | G2
    Car     | G2     | G1
    G1      | Car    | G2
    G1      | G2     | Car
    G2      | Car    | G1
    G2      | G1     | Car

With the "Monty Hall" problem, Monty uses his knowledge to always open a goat door. Thus we see the following revealed options and resulting 2/3s probability of switching succeeding. This is the classic version of the problem.

    My Door | Door 2 | Door 3 | Monty Reveals | Switch Result
    Car     | G1     | G2     | Either        | Lose
    Car     | G2     | G1     | Either        | Lose
    G1      | Car    | G2     | G2            | Win
    G1      | G2     | Car    | G2            | Win
    G2      | Car    | G1     | G1            | Win
    G2      | G1     | Car    | G1            | Win

With "Monty Fall", the first thing that happens is a randomly chosen door, that isn't our own, reveals a goat. This is interesting. In the classic problem we were always going to see a goat next, because those are the rules Monty plays by. But in this case, the fact that we randomly found one wasn't guaranteed.

Essentially, you are blindfolded and throw a dart at the 2x6 grid of cells under the headers "door 2" and "door 3", and I tell you that the cell you've hit is a goat. What do you know about the row you hit being a switch-or-stay row? Well, half the possible goats you might've hit are in the first 2 scenarios where you should stay, and half the possible goats are in the last 4 scenarios where you should switch. So you're at 50/50. You don't have any new information to switch on.

    My Door | Door 2 | Door 3
    Car     | G1(a)  | G2(b)
    Car     | G2(c)  | G1(d)
    G1      | Car    | G2(e)
    G1      | G2(f)  | Car
    G2      | Car    | G1(g)
    G2      | G1(h)  | Car

You are just as likely to be looking at (a), (b), (c), or (d) (so you should stay) as you are to be looking at (e), (f), (g), or (h) (so you should switch). It is 50/50 in this version of the problem.[footnote]

This may make it confusing going back to the original. I seem to have shown that both ways make sense but still, how is it different? Imagine it like Monty is doing a random dice roll for which door to open, and he simply juices the outcome by correcting it to the goat door when a car door is selected, since he can't reveal a car and spoil the game. Now we have these equally possible scenarios (a) through (l) for his fair dice roll...

    My Door | Door 2 | Door 3
    Car     | G1(a)  | G2(b)
    Car     | G2(c)  | G1(d)
    G1      | Car(e) | G2(f)
    G1      | G2(g)  | Car(h)
    G2      | Car(i) | G1(j)
    G2      | G1(k)  | Car(l)

Which he corrects, avoiding cars, to:

    My Door | Door 2 | Door 3
    Car     | G1(a)  | G2(b)
    Car     | G2(c)  | G1(d)
    G1      | Car    | G2(f,e)
    G1      | G2(g,h)| Car
    G2      | Car    | G1(i,j)
    G2      | G1(k,l)| Car

Now we are back to the original game scenario where we see a goat no matter what. And we can see that 8 of the possible ways we might have arrived at seeing this goat come from "switch" rows while 4 come from "stay" rows.

[rspeele]

[footnote] If this part of the explanation is bugging you, consider these related problems:

Problem 1. There are two opaque, externally identical bags, each containing 2 marbles. One bag contains 2 black marbles. The other bag contains 1 black marble and 1 white marble.

You choose a bag and draw a marble from it, without looking inside. The marble is black. What should you conclude are the odds that the remaining marble in the chosen bag is black?

Answer: .elbram kcalb dnoces a dnif ot ylekil sdriht-owt era ew oS .gab etihw-dna-kcalb eht morf si eno ylno dna ,gab kcalb-lla eht morf era elbram kcalb a werd ew hcihw ni soiranecs elbissop eht fo owT

Problem 2. There are three opaque, externally identical bags. One bag contains 2 black marbles. The other two bags each contain 1 black marble and 1 white marble.

Again you choose a bag and draw a marble from it, without looking inside. The marble is black. What should you conclude are the odds that the remaining marble in your chosen bag is black?

Answer: .tnecrep ytfif era elbram kcalb dnoces a gniward fo sddo ruO .gab kcalb-lla eht nesohc gnivah fo sddo ytfif-ytfif ta won era ew oS .sgab etihw-dna-kcalb tnereffid owt eht morf era owt dna ,gab kcalb-lla eht morf era elbram kcalb a werd ew hcihw ni soiranecs elbissop eht fo owT

You can connect Problem 2 to our random door opening and a goat being revealed, in the Monty Fall (with an F) problem.