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Comment by antris | Hacker News Reader

The fact that the show host knows what's behind the doors and never chooses a door with the prize behind it is crucial. The host leaks information about what's behind, as he chooses a door.

I was told about this problem and the person telling it didn't say anything about that, just that you pick a door and the host opens another one which doesn't have a prize behind it. So I thought the probabilities don't change at all because the host just picked a random door. Then everyone who heard of the problem before piled against me like I'm a dumbass, but never really explaining why I was wrong.

Yes, I'm still salty about it.

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I don't think the host's knowledge is important. Consider the alternative where:

1. You pick a door.

2. The host reveals a different door at random. The host has no knowledge.

3. The door the host revealed is a goat.

Should you switch or stay?

The puzzle doesn't work without the host always revealing a booby prize.

In the example you gave it's just random chance. In the Monty Hall puzzle it works out that statistically you should swap.

You should also swap in the example I gave.

Nope. That version is listed here: https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_...

"Monty Fall" or "Ignorant Monty": The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car. Switching wins the car half of the time.

Swapping doesn't affect the outcome, it's a 50/50 whether you swap or not.

This is what I thought at first but simulating it out, it’s not true. The strategy of switching doors works 2/3rds of the time but only if you allow picking the opened door (when it has the prize). That’s easy to see since it works whenever the prize isn’t behind your first choice. Conditioning on the opened door revealing a goat gives that strategy only 50% chance. After all, the opposite situation happens 1/3rd of the time and has a 100% success rate (since you can see the car), so this other case (opened door has a goat) has to have a 50% chance to add up to the unconditioned success rate. It’s still unintuitive to me though.

"Knowledge" is some physical state of the brain. I don't know exactly, connections in the brain, activation of neurons, different chemicals or electrical signals - it's something, some physical actual state of the brain. How could it be possible for Monty's brain-state to have an effect on the problem?

The obvious answer is the right one - it's not possible and Monty's knowledge has no effect on the game or the strategy the player should use. Monty knowing where the prize is and opening a bad door is the same as Monty not knowing where the prize is and having randomly opened a bad door. In either case, you should switch, Monty's knowledge doesn't change things.

Another way of thinking about it - how would the participant know if Monty opened a door at random or knowingly opened a goat door? What if they thought Monty knew, but Monty had actually forgotten and just coincidentally opened a goat door? Does any of this matter? No, because Monty's knowledge doesn't effect the game or strategy.

I also wrote a quick simulation in my Javascript console which confirms what I'm claiming here.

https://pastebin.com/y5G4PE75

The "pick and stay" strategy wins about a third of the time while "pick and switch" wins about two thirds - same as the original problem. Writing the code emphasizes that it is basically the same thing - Monty coincidentally reveals a goat versus knowingly reveals.

Strictly speaking it's not the hosts knowledge that changes the problem, but the fact that as a contestant you know how the host operates. This makes you able to extract information about the choice the host makes.

Different ways of host operation changing the outcomes are listed here: https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_...

>"Monty Fall" or "Ignorant Monty": The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car. Switching wins the car half of the time.

You stimulated out the situation where Monty Hall always picks the door without the prize. That’s exactly the standard Monty Hall problem. Change your code to instead allow him to choose the prize door (but never the player’s door). Then condition on him picking the goat door by dropping all the cases where he picked the prize door.

You’ll see that that changes it by discarding scenarios where switching is good (the prize is shown to you) but not ones where staying is bad.

I always believed that saying that Monty knowledge affects the odds is just more people friendly way of saying that those are not independet events and simple probabilistic model does not work when Monty is not allowed to select prize when revealing gate - You need something fancy (like a bit of Bayes conditional probability)

The choice happens after the door is opened. If in this variant you are allowed switch to the revealed prize door, you win 100% of the time the prize turns out to be behind the door they open. If you aren't allowed, you lose 100% of the time that happens. So that new "branch" doesn't affect any decision you could make.

But given they do open a door with a goat, the probability is the same as the regular Monty Hall problem. So the parent's point stands: it's not the host's knowledge that makes the game "work". He's leaking information by not opening your door, thus "concentrating" the 2/3 chance it wasn't behind your door into the one remaining door.

It’s not the same. Try stimulating it out if my argument above didn’t convince you. WLOG if you pick door 0 out of 0,1,2 and host picks door 1 always, then the the relevant value is in numpy:

correct = numpy.random.choice(3, 10000)

(correct[correct != 1] == 0).mean()

This gives 0.49 or similar when I run it.

Hmm, you are right. If the host chooses randomly, we get this tree:

- 1/3 it is behind your door

  - 1 host picks a goat

    - switching always loses -> 1/3 chance of losing

- 2/3 it is behind one of the other two doors

  - 1/2 host picks the goat

    - switching wins always -> 2/3 * 1/2 = 1/3 chance of winning

  - 1/2 host picks the prize

    - result undefined -> 1/3 chance of ending up here

So you get 1/3 wins, 1/3 losses, 1/3 undefined. So given the goat is revealed, you have a 1/2 chance to win by switching.

On reflection, this makes sense. The host is indeed leaking information by not be able to pick the around the prize. If the prize is not behind your door, his choice is constrained. So his revealing it supplies information about that constraint. If he is not constrained, then you are both picking doors at random. I'm sold.

That person, and, sadly, now you too, are completely wrong. What the host knows or does not know is completely irrelevant. Imagine:

You are playing the classic Monty Hall problem. You pick door 1. Monty reveals a goat behind door 2. Monty asks if you want to switch. You know the correct strategy is to switch, so you are about to say "Yes - switch" when suddenly, the lights dim and announcer's voice booms over the speakers "You thought Monty was knowingly revealing something to you, but actually Monty just revealed a door at random. He had no foreknowledge." Are you now ambivalent about switching or staying?

If you still want to switch, and you should, that's because obviously it does not matter what Monty knew going into the problem. If you don't want to switch, please explain how the contents of Monty's brain affect the probability of which door conceals a car and which a goat.

You make the decision based on your priors: the conditional probability of the prize being behind a certain door is updated by the new information. The information content of that update can certainly be affected by your knowledge of his constraints. In the original Monty Hall problem, that knowledge you have is “he can’t reveal the prize”. There is nothing magic about “the rules were x, and given that the rules were x, my update of the probabilities is y”. It has nothing per se to do with his mental state; it has to do with the rules he had to follow, and what you can infer from them.

Let’s turn this around: explain why I should switch doors, but starting from scratch with the new problem, instead of by reference to the original. I think you’ll end up thinking the original solution is wrong, based on your no-telepathy rule, or you’ll see how they are different.

Or try this out. Let’s have another variant: before you pick the door, the host picks one that turns out to be a goat. Now you pick a door, and _then_ you have the option to switch. Do you still switch? Does that make sense? The situation is exactly the same (a goat behind one door and two closed doors). If both actions so far are random, it doesn’t matter what order they go in.

Then you aren't simulating the problem statement.

This reminds me of the two envelopes problem

https://en.wikipedia.org/wiki/Two_envelopes_problem

Here is a simulation in Clojure showing that the host's knowledge is important.

    (def doors {0 :goat 1 :goat 2 :prize})
    (def choose (rand-nth 3))

    (defn game []
      (let [choice (choose)
            remaining (dissoc doors choice)
            monty (rand-nth (vec (keys remaining)))]
        {:choice (when (= (get doors monty) :goat) (get doors choice))
         :monty (get doors monty)
         :switch (when (= (get doors monty) :goat) (first (vals (dissoc remaining monty))))}))

So Monty is choosing a random door that you have not picked. When he picks the prize, the game is over and you lose.

    (def ten-thousand-games (repeatedly 10000 game))
    (frequencies (map :choice ten-thousand-games))
    ;; {:goat 3309, :prize 3307, nil 3384}
    (frequencies (map :switch ten-thousand-games))
    ;; {:goat 3307, :prize 3309, nil 3384}

Can you elaborate on why you feel this proves that the host's knowledge is important in deciding on whether to switch after a goat is revealed from one of the two doors you haven't chosen? As far as I'm aware (but it's been a while), the host's knowledge is only important up to the point where the first door to be revealed contains a goat. All scenarios where the host reveals a prize on the first door to be opened are discarded for the sake of the problem statement (when dealing with the host being ignorant).

Sure. This simulation shows the scenario OP described in his misunderstanding of the MHP, which is that he didn't realize the host was picking with foreknowledge of which doors were which. So if the host doesn't know that and is himself simply picking a random door (and 1/3 of the time, revealing the prize), then switching does not improve your odds.

Foreknowledge has absolutely no effect. Again, go back to the problem I posed where Monty reveals a random door and has revealed a goat. Should you switch or not? The correct answer is that you should.

Monty's knowledge has no effect on the problem or the player's choices. What has an effect is the information the player learns from seeing the revealed door. If the player sees a goat behind the revealed door - that is the relevant information. Whether Monty was surprised or not by the revealed goat is irrelevant.

This is just asserting the same thing over and over again. You can go upthread to the simulation to see that when you are in the "Monty showed you a goat" scenario, the results for staying and switching are equal. If you rewrite the simulation such that Monty always shows you a goat and look at the scenarios where "Monty showed you a goat" (i.e., all of them), you will see that switching wins.

I'm not familiar with Clojure, so I can't point out the error in your simulation, but there undoubtedly is one. Here is a simulation in javascript that supports my point - https://pastebin.com/y5G4PE75

And yes, your simulation should include that Monty always shows a goat. That is given in the problem statement.

> And yes, your simulation should include that Monty always shows a goat.

I feel like I've said this thirty times now, I'm specifically discussing a misinterpretation of the problem where Monty does not know what he's opening, so he sometimes opens a prize door. A game where Monty always opens a goat door is the original, correct Monty Hall Problem, and switching is the correct strategy.

The parent comment that starts this thread[1] is antris writing "The fact that the show host knows what's behind the doors and never chooses a door with the prize behind it is crucial." (Emphasis added). Then, I reply[2], saying "I don't think the host's knowledge is important." I include an example where the host has no knowledge, but has coincidentally revealed a goat, and the best strategy remains switching. In other words, antris says the host's knowledge is crucial and I'm saying the host's knowledge is irrelevant, what matters is what was revealed.

If you are talking about some other scenario - then why? I've just explained what this thread is about. Why would you invent an alternate scenario, not tell anyone about your alternate scenario, and argue that correct interpretations of the scenario I laid out in [2] are wrong because they don't fit with your interpretation of your private variation?

1 - https://news.ycombinator.com/item?id=31684653

2 - https://news.ycombinator.com/item?id=31683850#31684801

You are misunderstanding conditional probability. You keep doubling down on the idea that the alternative scenarios don’t matter, but this is mistaken. They very much matter. They are the crux of how Monty Hall “works”.

Bayes’ theorem is the relevant piece of information here. It tells us the relationship between a conditional probability and the information you have. Specifically, the probability of P(door 3 has the prize | host opened door 2 and found a goat) depends in a very important way on P(host opened door 2 and found a goat). You are acting as though it does not, that only the specific current state matters. This is the cause of your confusion: everyone else is working out the conditional probabilities and you are declaring them irrelevant.

The key is that if the host doesn’t know where the prize is, he can’t avoid picking the door its behind. This changes the probability that he would open that door and reveal a goat in the first place, and how that relates to the various places the prize could be. It would be really strange if the host’s behavior were in some way controlled by the location of the prize and this didn’t give you additional information on that location.

To spell it out:

You choose door 1, the prize is behind door 3, host opens door 2

Prop A: the prize is behind door 3

Prop B: the host opens door 2 and found a goat

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

P(A) = 1/3 in either scenario

In regular Monty Hall, the host will open door 2 always if the prize is behind 3, and half the time it is behind 1. So:

P(B) = (1/2)(1/3) + 1/3 = 1/2

P(B|A) = 1

Thus P(A|B) = (1/3)/(1/2) = 2/3. You’ll recognize this as the answer to Monty Hall. Note that conditional probability is foundational to how we got that answer.

Let’s try with the new rules. The host can pick either door, whether it has a goat or not. He can do this because he doesn’t even know which door is which. The host opens 2 half the time. 2/3 of the time he picks that, the prize is not behind it, so he finds a goat.

P(B) = (1/2)(2/3) = 1/3

P(B|A) = 1/2

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

You should update the Wikipedia article that is linked as the OP, since you disagree with its conclusions: https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_...

If you scroll down to "Monty Fall," the description of the problem is "The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car" and the solution is "Switching wins the car half of the time." (This is the result demonstrated in my simulation)

So if the host can reveal a prized door your odds are 1/3 prize, 1/3 goat, 1/3 redo. So each door has a 50% chance of having the prize after Monty reveals a goat door. Basically the game is just reduced to having 2 doors instead of your choice vs. all other doors.

The problem statement doesn't permit him to choose the prize. He is explicitly stated as having revealed the goat. Given that, do you switch? Even if he could have revealed the prize (due to lack of knowledge or care), he does not reveal the prize so why are we considering that circumstance?

>The problem statement doesn't permit him to choose the prize

Doesn't disallow it either. That's the problem: if you describe the problem without all necessary details, it doesn't work like the Monty Hall Problem is supposed to. If you fill in the details yourself (host never showing the prize in this case) then of course the problem becomes complete and then works like Monty Hall Problem again.

The statement is that he revealed the goat, in what world could that be interpreted as him revealing the prize?

"But he could have" but he didn't.

"But does he always" but he did.

What he always does or could do in an alternative scenario are irrelevant, you are given one scenario: He reveals the goat. Don't overthink it, it's not hard, switch and double your odds or stick with it and keep the 1/3rd chance.

This has already been explained elsewhere in this thread, but here it goes. The original problem formulation in Parade magazine is as follows:

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Assumptions about the strategy or decision rule of the host are not irrelevant, but crucial. Suppose the hosts' decision rule (unknown to the candidate) is

IF (candidate's initial choice is correct): open one of the remaining goat doors ELSE: don't open anything

then it's of course not advantagous to switch. The problem with the original Parade formulation is that it only describes what the candidate sees and what the host actually did in one particular scenario, not what rules the hosts must obay. So you don't know if the hosts is playing adversarial games, but you can't rule it out either.

That's why (as per linked article) in later formulations the "standard assumptions" (that the hosts will always open a door with a goat) are usually given explicitly (in which case the solution is also a lot more obvious).

It's entirely possible the host is a cheat and the prize is always moved to whichever door you didn't select so you only can ever get a goat. So don't play the game.

We can come up with a million what-ifs, but none have a basis in the problem statement so there's no point in them.

The original Parade problem is fairly ambiguous and incomplete. Therefor there is no "correct solution". Other formulations typically include the conditions that the host will always open a door, and that door will always contain a goat as explicit assumptions. In that case the math and logic are pretty straightforward.

But in the setting of an actual repeated game show, I think the more realistic case to keep the show interesting is that the host will mix up his strategy, i.e. sometimes he opens another door, sometimes he doesn't. If it's a long running daily or weekly show, if he always opens a door with a goat, the public will work out pretty quickly that it's advantageous to always switch.

The original formulation was probably written under the assumption that the readers were familiar with the game show it was drawn from.

As TFA notes the Parade magazine formulation is only "loosely based" on the actual show Lets Make a Deal" hosted by Monty Hall. The real Lets Make a Deal* is a bit more complicated[1], and AFAICT there is no easy analysis or solution for an optimal strategy.

The game described in Parade magazine is different from the TV show game, and the assumptions (about the host's behavior) are incomplete/ambiguous enough that switching is not necesarrily the optimal strategy. This, like almost everything else you can think about the Monty Hall Problem, is also described in TFA[2].

[1] https://en.wikipedia.org/wiki/Let%27s_Make_a_Deal#Format [2] https://en.wikipedia.org/wiki/Monty_Hall_problem#Other_host_...

If the door the host chooses is random, and the door you choose first is random, then it's 50/50, because there's only two doors left and neither door choice was made with any information of what's behind the door.

Your chances improve from 1/3 to 1/2 when the host opens the empty door, but whether you stay on the first door or not doesn't matter.

If the host randomly selects the goat or not doesn't change the odds. You should switch. The statement is that he does reveal the goat. Why or how is irrelevant.

One of three doors is selected.

One of the other two is opened, revealing a goat.

Should you switch to the unopened door? Yes, it doesn't matter how that goat was revealed, it could've kicked the door open itself. The odds are better to switch.

You're the one who's saying the odds change when the door is opened. I'm the one saying that the odds remain the same (or rather, increase to 50/50) if the doors picked are random.

If the host knows where the prize is, the odds change when the host picks a door, because we know the hosts strategy. If it's just random, then the odds don't change and changing the door doesn't matter, because each unopened door has the same chance of revealing the prize (1/3 before a door is revealed, 1/2 after).

Look up "other host behaviors" on the Wikipedia article, especially the table with "Ignorant Monty", if you don't believe me.

> You're the one who's saying the odds change when the door is opened.

No, I said the player's odds of winning improve if they switch (to the unopened door).

If the host randomly revealed a goat or deliberately revealed the goat is immaterial. A goat was revealed, that's all that matters.

The player has 3 doors to choose from. 1 is selected, 2 are not. That means the other two have a combined 2/3rds chance of holding the prize. The selected door has a 1/3rd chance of holding the prize. A goat is revealed behind one of the two unselected doors. The remaining closed, but unselected, door has a 2/3rds chance of being the one with the prize.

The selected door's odds of being the winning door remains 1/3rd after the reveal. The player's odds of winning improve if they switch.

I feel like this is just going around in circles, but let me take a stab at explaining the flaw in your reasoning.

Let’s try to simplify the problem. Let’s assume there are three numbered doors: 1, 2, and 3. A car is randomly placed behind one door, and goats behind the other two.

To make it easy, I will always guess door 1. In the ‘ignorant host’ scenario, we can just have the host always reveal door 2 (the math doesn’t change if he randomly selects which door to open or always opens the same door, so keeping it always door 2 makes it easier to understand).

Ok, so we guess door 1, and the host opens door 2.

1/3 of the time, the host reveals a car. Obviously we would switch in this case, since we know exactly where the car is.

2/3 of the time, the host will reveal a goat. When this happens, 1/2 the time the car will be in door #1 (our door) and 1/2 the time the car will be behind door #3. If the car is not revealed behind door #2, we have a 50% chance whether we switch or not.

With this random reveal style, we end up winning 1/3 (the chance of revealing the car behind #2) + (2/3 * 1/2) = 2/3rds chance of winning… which makes sense, because we get our original 1/3 chance plus the free 1/3 chance it is behind the revealed door. Crucially, though, it doesn’t matter if we switch or not. You always win if it is behind door #2 (because you are shown it) and win half the time if it is behind #1 or #3.

Contrast this with the case of the host always revealing a goat. In that case, you again have a 1/3 chance of it being behind your door #1 choice. However, in this case, you can essentially guess that your choice was wrong (2/3rds chance) by switching your guess, which will make your guess correct if it is either in door #2 or #3 (since the host will reveal the one which it isn’t behind if it is one of 2 or 3.)

In both the ignorant host and non-ignorant host situations, you end up with a 2/3rds chance of getting the car. However, in the ignorant host situation, you don’t need to switch unless the host shows the car in his reveal to get the 2/3rds odds.

Of course, it never hurts to switch in either case, so always switching is not a bad strategy, just not necessary if the host knows nothing.

From the article:

The original problem:

>The host acts as noted in the specific version of the problem.

>Switching wins the car two-thirds of the time.

The random door pick:

>"Monty Fall" or "Ignorant Monty": The host does not know what lies behind the doors, and opens one at random that happens not to reveal the car.

>Switching wins the car half of the time.

> The problem statement doesn't permit him to choose the prize.

I agree, but this subthread is discussing a misinterpretation of the problem statement, and asking the question about whether switching still matters if the host is also picking randomly. This is showing that it does in fact matter whether the host is picking deliberately or randomly.

OP for the thread:

> I was told about this problem and the person telling it didn't say anything about that, just that you pick a door and the host opens another one which doesn't have a prize behind it. So I thought the probabilities don't change at all because the host just picked a random door. [emphasis mine]

It doesn't matter if it was at random or deliberate. It matters that a non-prize (goat in most versions) is revealed. The question is about one very specific scenario, even in what antris was originally given: The goat is revealed. In that scenario, you switch.

The OP misunderstood the problem because they forgot (as many others have in this discussion) that it's about one scenario, not about all times. If Monty selects a door at random and reveals the prize, then it's not this same scenario anymore so the odds of that situation are necessarily different over all scenarios. And if we include all possible scenarios (really, two) then the odds of winning drop, but only because you have a 100% chance of losing when he reveals the prize (or 100% chance of winning and the odds increase).

This is the fundamental misunderstanding in many of the comments. If I tell you, "We select, each day, either a prize or a tiger to put behind this door at random. Today it's the tiger. Do you open it?" You'd be a fool to open it. And over all scenarios you'd have a (if chosen uniformly) 50/50 chance of getting the prize or the tiger. But if today I tell you, "It's the prize" should you open it? There's a 100% chance it's the prize, I've told you it's the prize.

Now we can get into adversarial situations, whether I'm honest or not. But that's diving into something entirely different and makes it impossible to consider the problem statement because we'll be chasing nonsense after a few minutes. I've just told you, 100% chance it's the prize today. You should open it.

The Monty Hall problem is the same. It is one scenario, a goat is revealed. The reason or mechanism of that reveal is irrelevant. He could know it, he could not know it. He could have revealed the goat by dumb luck. Or maybe because it started kicking at the door and was too obvious not to reveal. But today, for this game show he revealed a goat. So your odds are better if you switch. And even antris' original statement presented that scenario, they just answered a different question.

> It doesn't matter if it was at random or deliberate. It matters that a non-prize (goat in most versions) is revealed. The question is about one very specific scenario, even in what antris was originally given: The goat is revealed. In that scenario, you switch.

You can just go back to the simulation numbers and check. If you eliminate the games where Monty "wins" (i.e., reveals the prize), you are in the specific scenario you described. You have chosen a door and Monty has revealed a goat, and you win 3309 times by staying and 3307 times by switching.

If you rewrite the simulation to follow the correct problem statement (that Monty knows and always picks a goat), you will see that switching is in fact the correct strategy.

If the host has knowledge what matter is whether the host has the freedom to not open a second door (which they often have in TV shows where time is of the essence), and deprives you of a second round. If the host has this freedom, and he is adversarial and assumes you know the monty-hall paradox, then the only time where he will offer you to show a second door is if you have guessed right to make you change.

In this version it doesn't matter. #3 eliminates the 1/3 chance that the host reveals the prize by accident, so you're left with the 1/3 chance that you chose the prize, or the 1/3 chance that you didn't choose the prize and the host also didn't choose the prize.

if it's opened randomly then it doesn't matter, it is 50-50. You have to account for the situation where the host might reveal the prize in his random pick, and by not doing so you learn that your choice is more likely to be the prize than initially thought.

If you picked the prize P(A = 1/3), then the random open is going to be Goat P(1). if you didnt pick the prize P(B = 2/3) then the random open is going to be Goat P(1/2) or Prize P(1/2)

P(Goat reveal) happens (1/3)(1) + (2/3)(1/2) = 2/3 times

P(Prize reveal) happens (1/3)(0) + (2/3)(1/2) = 1/3 times

P(you picked prize, given goat reveal) = (1/3)(1) / (2/3) = (1/2)

You don't have to account for the chance of him revealing the prize because he does not reveal the prize, he reveals the goat. So switch because it doubles your odds. Or don't, you're a free person. But it would be foolish.

If he's guaranteed to reveal the goat then he knows which door has the prize. This situation he isn't guaranteed to reveal a goat, he just did it this time by chance.

It doesn't matter if he's guaranteed to reveal the goat or not. He did reveal the goat, that's the scenario that's been described. So given that he did reveal the goat, should you switch? The answer is yes, from the information given, you should switch.

Everything else is overthinking.

>It doesn't matter if he's guaranteed to reveal the goat or not

Yes, it does. this is the entire point behind the popularity of the Monty Hall problem. It is counter-intuitive and feels like overthinking, but it is truth.

I described the situation pretty clearly (imo). I did not need to think about it much at all, though it did take some thought to describe it clearly.

I can tell you, without much thought, having been trained thoroughly in probability and statistics, it is 50-50 if a door is opened randomly to reveal a Goat. You can switch if it makes you feel better - because it is 50-50 so you wont be hurting yourself.

https://en.wikipedia.org/wiki/Conditional_probability

there are other examples that make it more clear why condition matters.

Whether he reveals the goat by luck or because he knew which door had the prize is the entire point of the problem.

No, it's not. The point of the problem is that it's better to switch than not given the scenario presented. Whether he knows or not doesn't actually change anything, he does reveal a goat. Whether it was luck or deliberate makes no difference.

Switching doesn't change where the prize is more likely to be. That's not how the Monty Hall Problem works. It's the host giving you information about what's happening behind the scenes, by choosing a door to open and you knowing how the host operates.

Whether you end up switching or not, the other opened door is 2/3 likely to contain the prize and the door you picked originally has a 1/3. Whether you switch or not, doesn't change the odds.

This is indistinguishable from the host having secret knowledge; without knowledge the host would open the door with the prize a third of the times.

Yes, that's the point, that it is indistinguishable. The host's knowledge doesn't matter. The host might know or not know. What matters is that the host has revealed a goat.

But the only way to always reveal a goat is to have knowledge of the doors.

I would switch. The question is what happens if the door is the prize?

I agree. I think the way the problem is often described and explained is terrible.

Why would the game show host want me to win? or want me to lose? What's his motivation behind giving me the option to switch? What knowledge does he have?

Well if the host opens a door and it never has a prize behind it they have to know which door has the prize right?

It depends on if that's the scenario explained, or you have just one example (not knowing if the host will always show a goat, or if the host will sometimes show you a car either intentionally or by accident).

The part about never having a prize behind it was not explained to me.

It was just "you pick a door, host opens a door without prize". That doesn't imply that there's never a prize behind the door that the host opens.

> That doesn't imply that there's never a prize behind the door that the host opens.

The entire activity is meaningless if the host opens the door with the prize—any questions about switching or not are pointless, since obviously both closed doors are worthless. So that can’t be the interpretation of the puzzle.

What else does "host opens a door without prize" mean?

The difference is "host opens a goat door because he knows which one has the car and chooses to not open it" versus "host opens a goat door by random chance".

For the purpose of the problem there's no difference between the two cases. The only thing that matters is that the host (whether by knowing or by chance) opens a door with a goat behind it.

There is a big difference, "Ignorant Monty"/"Monty Fall" makes switching 50/50 instead of 67/33. The main change is that the door could have had the car, but didn't by chance, and we get to ignore the (one third of the) times he would have revealed a car by mistake. The original Monty never shows you the car because he always picks right, so those possibilities could never happen.

That the host opens a door without prize.

I think you're right.

When you hear about the problem the first time, they just tell you "and then the moderator opens a door with a goat behind it". So you assume that the moderator usually opens a door - where in your case just happens to be a goat behind. You don't know about the show, so you assume it could've been the car, but just wasn't. So, with this assumption, it doesn't matter whether you switch. And then they tell you that the moderator always shows you a goat, but the damage is already done, your mind has made the decision that it doesn't matter.

You start with a 66% chance of choosing a goat. The host opens a door and shows a goat. If you switch, you turn that 66% chance of choosing a goat into a 66% chance of choosing the prize. The fact that the host knew or didn't know what was behind the door is irrelevant.

Also, if the host opened the door with the prize, what would happen then?

In your specific scenario it doesn't (I think!). However, if the host doesn't know what door the prize is behind then there is a 1/3 chance of picking the prize and you instantly loosing because both other doors are goats and you must choose one.

I don't think he leaks information. He invites you to switch from a game with a 1-in-3 chance to a game with a 2-in-3 chance. If you stick, you are staying in the 1-in-3 game, but if you switch then you are changing to a 2-in-3 game. That's all there is to it.

Edit: I should have checked before I posted, obviously it's 2-in-3 when you switch. But it's same idea, you have better odds when you switch because then it's like you are entering a new game with only 2 doors.

Still hard to believe that Erdos wouldn't believe this. He spent his life couch surfing at other mathematicians but not could convince him until he saw a monte-carlo simulation, apparently.

A common way to explain this is to increase the number of doors.

You have 100 doors, select one blindly, and have a 1% chance of being correct.

The host removes all but one other door. Do you stay with the door you chose, a random guess of 100, or the one remaining after the host has removed all others? A door that is now statistically much more likely to be the correct choice?

Good idea. That makes it extremely obvious. Hard to believe that Erdos didn't believe this until he did monte carlo simulations.

No, that's not right.

You need to read up on it more as it's clear you haven't truly understood the puzzle.

The host knows what's behind the doors so he'll always eliminate a wrong answer.

In the opening, you pick one door. That's a 1 in 3 chance of being right.

So there's a 2 in 3 chance it's in either of the two remaining doors. He simply eliminates a wrong answer.

Thus, 2 out of 3 times you win by switching. You win 1 in 3 times by staying with your original choice.

If you switch, your odds of winning are 2/3.

Oh yes, thanks.

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