Does Monty Fall change the odds in the Monty Hall problem?

[BWV]

Seen it argued that in Monty Fall - the same setup but the host trips and accidentally reveals a door with a goat - does not have the same odds. I disagree.
A) your initial chance of picking the correct door is 1/3
B) the odds are 2/3 one of the other doors has the car

should then not matter if the host deliberately or accidentally opens a door with a goat (other than if the possibility f accidentally revealing a car if the host has no knowledge or opens the door by accident)

another equivalent scenario - the host does not open a door but allows a switch after the initial choice and you see a goat tail under one of the two doors you did not pick. Probability still holds - the odds were 1/3 your initial choice was right. If you then can switch to one of the two remaining doors and know one of those has a goat, then your odds of a win by switching are 2/3

 

[BWV]

However, if I see the goat tail under one of the doors before my first pick, the odds of picking the correct door are 50/50

same as if the host opens a door to reveal a goat before I make my first pick

 

[PeroK]

Seen it argued that in Monty Fall - the same setup but the host trips and accidentally reveals a door with a goat - does not have the same odds. I disagree.
A) your initial chance of picking the correct door is 1/3
B) the odds are 2/3 one of the other doors has the car

should then not matter if the host deliberately or accidentally opens a door with a goat (other than if the possibility f accidentally revealing a car if the host has no knowledge or opens the door by accident)

It does matter. In the Fall scenario there is the possibility that the host opens the door with the car. The three equally likely outcomes are:

The host reveals the car
The host does not reveal the car and stick wins
The host does not reveal the car and switch wins

The host's choice in the original problem is biased towards never revealing the car.

To see the issue more clearly, imagine you lay out all the cards in a pack face down. You have to find the Ace of Spades. You choose your card. Your friend then turns over other cards at random. As each card is turned over, and is not the Ace of Spades, each remaining card (including yours) has more chance of being the ace.

In this scenario, the probability that your card is the Ace does not remain at 1/52.

Usually the Ace is revealing at some stage. But, in only 2 games in 52 you get to the situation where there are only two cards remaining: yours and one other. At this stage both are equally likely to be the Ace.

If, however, your friend knows where the Ace is and avoids revealing it, then the probability that your card is the Ace remains 1/52. In this case, your friend can always take the game down to the last two cards, and your card still has only a 1/52 chance of being the Ace; and the other card 51/52.

If you are in any doubt, you should experiment with these two scenarios. Perhaps using only one suit - the 13 spades perhaps, or even fewer cards - to speed things up. You'll soon see the difference emerge in the data for the two cases.