[rsa25519]

> In this: "the host will open the door at random and in this example it happen to have a goat" vs "the host will never open a door with a car behind it".)

If the host opens a door with a goat, then it doesn't matter whether or not it was intentional.

[dllthomas]

Yes it does. Or more precisely, it matters whether the host could be counted on to do so reliably; the mechanism for that doesn't matter.

There's a difference here, that our language obscures, between procedure and hypothetical.

[tpush]

The intention of the host only matters if the contestant would have to choose the subsequent action (switching or not) before the hosts opens a door.

If the host has revealed a goat door, and the contestant then has to decide what to do, the intentions of the host for having chosen the door are irrelevant.

[dllthomas]

(Noting again that the mechanism doesn't matter, what matters is the odds of various behavior by the host, but - I think reasonably - using "intentions of the host" as a proxy for that.)

The intentions of the host do matter.

Imagine the host picks the correct door by the following procedure: 1) picks an available door at random; 2) if that door has a goat, opens it; 3) if that door has the car, opens the other door.

I hope you will agree that this is equivalent to the problem as originally intended - Monty can be relied on to reveal a goat, and exactly why doesn't matter.

Breaking it down into equally likely cases, assuming the contestant picks door 3:

    A) The car is behind door 1, Monty picks door 1, Monty corrects.
    B) The car is behind door 1, Monty picks door 2
    C) The car is behind door 2, Monty picks door 1
    D) The car is behind door 2, Monty picks door 2, Monty corrects
    E) The car is behind door 3, Monty picks door 1
    F) The car is behind door 3, Monty picks door 2

When Monty reveals the goat behind (say) door 2, we know we're in case A, B, or F. All remain equally likely, and switching wins in A and B.

If Monty would not have corrected, then revealing the goat behind door 2 eliminates (the new) A as well, leaving us with only B and F, again equally likely.

If all of this remains unconvincing, I encourage you to write a simple simulation of the problem.

[tpush]

So, thinking about it really hard and reading about it online:

My comment was definitely wrong: If Monty could have opened a car door, but just didn't, then duh the probabilities for the car to be behind the doors are different than if Monty always opens a goat door. So in that way, the intentions of Monty, meaning how he chooses, definitely matter.

But I think your example here doesn't show that? Are you trying to illustrate the Monty Fall variation?

[dllthomas]

Sorry it took me so long to get back to this; real life intervenes sometimes.

I think what I was trying to do was frame the original Monty Hall problem as a variant of Monty Fall, in a way that (I hoped) makes it clear where Monty is doing work to convert some outcomes into other outcomes (and therefore producing different likelihoods).