## Within the well-known Monty Corridor downside, how do the possibilities change if the host opens one of many two remaining doorways at random and it occurs to be empty?

As an alternative of the same old scenario of him figuring out which door has the automotive, and intentionally opening an empty (goat) door, think about he’s additionally clueless and simply opens one of many two remaining doorways at random and it occurs to be a goat.

Im fairly positive the scenario is now 50-50 so no profit in switching (versus 1/3 vs 2/3 in authentic downside), as a result of no ~~new~~ insider data is added however whats the proof?

For these unfamiliar: https://en.wikipedia.org/wiki/Monty_Hall_problem

Edit: to make clear on this hypothetical sport present the place the host can also be clueless, if he had opened the automotive door the sport would finish. Let’s not fear about that, simply deal with the scenario the place he opens a goat randomly (he did not realize it was going to be a goat both)

## Comments ( 6 )

Correct, the change in probability occurs *because* the host knows the location of the prize. If the host no longer knows the prize location, then the host opening (or not opening) one of the doors only effects your ability to open the same door. Your assessment is correct that this would lead directly from the percentages going from 1/3 to 1/2. Additionally, in your scenario, you could actually remove all the variables, including the host, and be left with basically the same experiment.

I think others have cleared it up, but it all comes down to the hosts knowledge. Another piece I see a lot in newer folks is that they think the probability is the same forever, for example Deal or No Deal they pick one case in say 25 and think they have a 1 in 25 chance of having it while there is a 24/25 chance they do not, once they open cases down to the end they think they have to trade because it is still 24/25 and not 50/50.

I don’t think it would change anything. Another way to think about the problem is to consider the option of picking ‘the other two doors’ where if the car is in either of those doors, you win. If you are the contestant and choose door number 1 for example and then are given the chance to trade your first choice with both doors 2 + 3, it’s pretty clear that you have double the chances of winning the car by switching to the two doors. You don’t need to know if Monty knew the goat was in the revealed door or not. In fact, they don’t explain this to the contestants in the game, he just always shows them the goat. For all we know, it was random the whole time and they just never aired the episodes where he randomly showed them the car! Like you said, let’s not worry about that.

It does change the probability, I think the intuitive way to understand it is if you draw out a tree of all the possible outcomes for the game, if Monty is opening the door at random then sometimes he opens the door with the car and the game ends, and that outcome needs to be included as a possibility for the game as well. In contrast, when Monty chooses the door to open then the probability of the game ending early is zero, and that changes the overall probabilities associated with all of the other outcomes for the game

You are right. Just list the cases with their probabilities.

1. You pick the car door, which means that the host will necessarily reveal a goat => 1/3

2. You pick a goat door and the host manages to reveal the other goat => 1/3

3. You pick a goat door and the host accidentally reveals the car => 1/3

If a goat happens to be revealed, we know you are not in case 3); only cases 1) and 2) remain as possibilities, and since they were equally likely, each must represent 1/2 of the new subset.

The other way you can corroborate this is writing a simulation.

I think that people who get this wrong is because they are carried away by the first impression and don’t take the time to analyze this thoroughly. Seriously, you can find easy analogies to see that the reasoning: “the host will not always reveal a goat, but as he did it this time, we must assign the same probabilities as he did it every game” cannot be correct.

For example, imagine that you are in front of three persons: Ben, Mark and John. You don’t know who is who, and moreover, you know that all are wearing blue jacket, so if one of them approaches you and you see that person wearing a blue jacket, that does not provide new information about who is that person. He would still be 1/3 likely to be any of the three men.

Now, suppose that instead of all wearing blue jacket, only Ben and Mark are wearing that color, while John is wearing a white jacket:

Ben -> blue

Mark -> blue

John -> white

If the person who approaches you uses a white one, you automatically know that he is John, but if who approaches you uses a blue jacket, would you say that it is as if all of them used a blue one and so you would still say that he is 1/3 likely to be, let’s say Ben?

You shouldn’t; seeing the blue jacket would restrict your possibilities to only Ben and Mark, so seeing that color would make it 1/2 likely that he is any of those two.

There is a difference between all wearing blue jacket, and only some of them wearing a blue one and who approached you just happened to be one from that sub-group, that are not all.

Suppose there are 3 doors and you pick one and the host randomly chooses one of the remaining two doors and it happens to goat. Let A be the event that you chose the door with the car and let B be the event that the host randomly chose a goat from the remaining two doors. Let X’ denote the compliment of a set X.

P(A|B) = P(B|A) * P(A) / P(B)

P(B|A) = 1

P(A) = 1 / 3

P(B) = P(B|A) * P(A) + P(B|A’) * P(A’) = 1 * (1 / 3) + (1 / 2) * (2 / 3) = 2 / 3

Thus,

P(A|B) = 1 * (1 / 3) / (2 / 3) = 1 / 2,

and also,

P(A’|B) = P(B|A’) * P(A’) / P(B) = (1 / 2) * (2 / 3) / (2 / 3) = 1 / 2.

So when the host randomly guesses and reveals a goat, then you actually have a 50/50 chance whether you stay or switch.

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This result really annoys me haha and at first I thought that it would still be 1 / 3 and 2 / 3 like in the original set up. But its fun to see using Bayes’ theorem that it is actually 1 / 2 and 1 / 2.