"and the host, who knows what’s behind the doors, opens another door [..] which has a goat."
The question is clear: The host (1) knows what is behind each door and (2) always shows a goat. It's clearly a determinate problem.
You're right. The problem is stated multiple times in the article. As is usual, most of the statements do not address knowledge. But Marilyn's own statement did. However she did not address motivation.
Monty himself claims that his motivation varied depending on his mood. The actual game had more complications. And his statement about the real game was, "My only advice is, if you can get me to offer you $5,000 not to open the door, take the money and go home."
You can find the article I got that quote from at https://www.nytimes.com/1991/07/21/us/behind-monty-hall-s-do....
Well you do have that guarantee -- as it is stated in the problem. It's clear that no matter what door you open, you will be shown a goat.
https://www.jstor.org/stable/2683689
On the last line of the problem he opens an empty box.
This paper adds that in as an assumption after the prompt, which I’m pretty sure is not the original prompt
Also, do you have any evidence that this isn’t the original?
Sort of. If you’re saying the next box is chosen at random then there are 2 of 6 possible end games in which the key is chosen; 2 of 6 in which you pick an empty box and can switch for the keys; and 2 of 6 that both of the remaining are empty.
Since the prompt says that you did in fact open an empty box, that removes the 2 where you open the keys. So it’s 50/50.
When you know for a fact that the keys will never be chosen, the probability of picking an empty box when you chose an empty box goes from 50% to 100%. Meaning it now occupies twice the probability space. That’s now it’s 2/3 chance of winning.
You truly learn nothing if Monte randomly opens one of the doors and it is not the keys.
> Also, do you have any evidence that this isn’t the original?
The quote in the article?
The probability of A given B is the probability of A and B divided by the probability of observing B. And the probability of observing B depends on counterfactuals of various sorts. "What would happen if...?" And that's where intention comes in.
In this case B is "Monty opens an empty box". The probability of the event B depends on Monty's knowledge and intent. If Monty knows where the prize is, and always avoids it, then Monty always opens an empty box. Probability 1. If Monty is clueless, then Monty opens an empty box with probability 2/3. And if Monty is knowledgeable and malicious, then Monty opens an empty box with probability 1/3.
Event A is that you have found the prize and Monty found an empty box. We're assuming that this is the probability that you initially found the prize, and so has probability 1/3. And so we get that Savant's Monty leaves you with odds (1/3)/1 = 1/3 of having the prize, ignorant Monty leaves you with odds (1/3)/(2/3) = 1/2 of having the prize, and malicious Monty leaves you with odds (1/3)/(1/3) = 1 of having the prize.
I find it absurd that I've never looked at it this way and recognized the fourth possibility. HELPFUL Monty knows the answer, and is giving you every chance. So if you had the prize, helpful Monty would show you that you're a winner, otherwise helpful Monty will give you another chance. What helpful Monty changes is the probability of A and B. If you had the prize, you would have been shown it. Therefore the probability of A and B is 0, and you really, really want to take Monty's hint and switch.
It states you opened a door and then Monty opened a door.
It does not state that Monty would have opened a door if you had picked a different one.
If this is not communicated you don’t gain any knowledge from his reveal.
The typical proper Monty hall formula states this assumption clearly that he will always open a goat door. The original one does not state this.