The Monty Hall Problem

Forum Games
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It depends on whether you know Monty is lazy or not.

In this case we’re still relying on the knowledge of somebody about where the car is.

It becomes different if Monty rolls dice to decide which door of the two remaining to open (say, if he opens a car, you’ve lost, if he opens a goat he’ll allow you to switch, which then doesn’t affect your chances). Or if Monty stumbles and accidentally opens a door (which coincidentally happens to be a goat). Or if you for some reason get to peek behind one of the two remaining doors of your choice, and see a goat.

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62

I see your argument here, and apparently this turns out to be true in the Monty Hall Problem. But is the argument valid? It has been pointed out that, if the door that is being opened is chosen randomly from the three doors (independently and uniformely distributed), i.e. without consideration of which door has been chosen or which door contains the car. And it turns out that a different door than chosen is opened and it contains a goat, then the odds of the remaining two doors are 50-50, i.e. the odds that the door first chosen contains the car changed from 1/3 to 1/2.

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I think that yebellz refers to the chance a priori. Once you’re in the situation that a door has opened and by chance reveals a goat, you have to compute the probability of getting the car assuming a door with a goat opened. If the door with the car had opened, you would instead have 0 chance of getting the car by staying.

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There are 2 events to be considered:

Event 1: you choose what goat the host discovers. (or, you find the car, at which point event 2 calculation is moot anyway). The claim that you’re picking the door with the car looks reasonable, but that’s not what you’re really doing! Ergo: you have 0% chance to win a car in this event, 33% of picking the door without a goat and 100% certainty of revealing one goat, (via the host) :slight_smile:

Event 2: you have a 50% chance of winning the car. This is the only real event.

Additional Claim: 50% chance is 50% and not 50% + (say) 10% Host behavior bonus = 60% changing, 40% staying. We can only compute that which is quantifiable, and we have already agreed that there is a 50% chance with either door to have the car behind it, in fact, the bogus sales point of the argument to chance choice is that we go from 33% to 50% chance by changing door^W(what we’re measuring…cough cough).

The problem is phrased in a way that makes it look like something other than it really is and bamboozling the solver into thinking that Event 1 somehow defines Event 2 and that solver has consequential agency in event 1 that yields advantageous knowledge in event 2.

Event 2 is completely independent of Event 1 — it makes no different which goat you cause the host to reveal, and it doesn’t not give you any information about what door has the car at all, the second event is strictly a coin toss and by definition, no single choice here is better than the other in this case: you still must choose between the surviving goat and the car, and either is 50% to be revealed.

So I don’t think that this is a valid mathematical problem, it’s just a trick question.

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65

I’m not sure I follow any of your argument, but I’m pretty sure it’s wrong.

It feels a lot like you’re making the mistake of thinking that whenever there are two options available, then the chance is 50% for each option.

Event 1: you throw a die (hidden under a cup). Another person peeks under the cup and tells you it’s either a 6, or you’ve thrown less than a 5 (i.e., the other person tells you you haven’t thrown a 5).

Event 2: you have 50% chance of having thrown a 6.

(note that the chance is really 20%, but that’s because the information that you haven’t thrown a 5 reveals to you that we’re computing the chance of throwing a 6 assuming we have not thrown a 5, and not just the chance of throwing a 6. This would be the Monty Hall problem where the doors are opened by a random accident, instead of by Monty deliberately choosing a door with a goat)

As for the Monty Hall problem, you have 1/3 chance of picking the door with the car on the first guess, and 2/3 chance to pick a door with a goat. In the first case, if you switch, you’ll find a goat. In the second case if you switch, you’ll find a car. So in 2/3 of the cases you’ll find the car after switching.

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66

The problem statement is the issue here. It asks one thing and then answers another.

Numbers on a die are distinct objects but our prizes here are of 2 classes: a set of two goats and a set of one car, so our Monty Hall die has 3 sides, goat, goat and car.

All you know is that Monty eliminates one goat in the first event, or, ends the game immediately by opening the door on the car.

Hence, whether he knows where the car is doesn’t matter to event 2. If you get to event 2 for any reason, there is always the coin toss at the end.

The time at which you ask the question is important too — in your example, initially you’re correct, there is a 1/6th chance of rolling a 6. But event 2 narrows the universe of discourse and so, an entirely new decision must be made. Your old opinion of reality has no influence at all on the current facts – event 1 changed the premise.

In the goat example, the Universe of Discourse is not narrowed if the car is not immediately found by Monty, unless you wanted to make goat curry and hoped to win the fatter goat if you miss out on the car. Whatever happens, iff you get to make the choice between car and goat, it is a coin toss between selection from 2 sets: goats and car.

So, you have a 1/3rd chance of divining the correct door in the show in event 1, and a 1/2 chance of winning the car in event 2.

The slight of hand occurs when the two questions are conflated and one thing is asked and another is answered — this is a good exercise for the Politics 101 course and, a good example of why being extra picky in framing formal problem statements is very important.

Consider this classic joke:

" Sheep in Scotland

A mathematician, a physicist, and an engineer are riding a train through Scotland.

The engineer looks out the window, sees a black sheep, and exclaims, “Hey! They’ve got black sheep in Scotland!”

The physicist looks out the window and corrects the engineer, “Strictly speaking, all we know is that there’s at least one black sheep in Scotland.”

The mathematician looks out the window and corrects the physicist, " Strictly speaking, all we know is that is that at least one side of one sheep is black in Scotland.""

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Your old opinion of reality has no influence at all on the current facts – event 1 changed the premise.

Actually, the funny thing about games of imperfect information or partial-observability (of which Monty Hall is a very simple one) is that your old opinions and the manner in which you arrive at a later situation DO affect the probabilities of events. This complexity underlies a lot of the theory for the algorithms that are used nowadays for approximately solving games like Poker - and is why you have to apply such algorithms as counterfactual regret minimization that carefully manage the relative probabilities of events rather than simply MCTS or other tree-search that works Chess or Go or even games like Backgammon, which have randomness but aren’t imperfect-information.

It should be obvious if you think about it, that you can have two observationally-identical situations in poker - e.g. you’re facing the same decision as to call/bet/fold, the pot is the same size, the cards on the flop are the same - and yet, depending on your beliefs about your opponent’s behavior and how aggressive they are in general, and the past events that led to that point, you could legitimately have very very different beliefs about the opponent’s likely hole cards.

The same is true in Monty Hall, for different variations of how Monty behaves. For example, the following are true, and you might enjoy working out for yourself why, if you haven’t spoiled yourself in the thread above:

If Monty behaves in certain alternative ways, it’s possible that conditional on reaching event 2, switching causes you to always lose and staying cases you to always win. How?

Suppose Monty only ever offers you the option to switch in the first place when you initially pick the car. If you pick a goat, he never offers you a choice to switch. Then conditional on finding yourself being offered a switch, switching will definitely cause you to lose.

If Monty behaves in certain alternative ways, it’s possible that conditional on reaching event 2, switching causes you to always win and staying cases you to always lose. How?

Suppose Monty only ever offers you the option to switch in the first place when you initially pick a goat. Then conditional on finding yourself being offered a switch (and with the other goat revealed), switching will definitely cause you to win.

Okay, those are pretty extreme differences in Monty’s behavior. But they established the general principle, that “old opinions” do matter. In all cases you picked a door, then a goat door was opened, then you got a choice, but even though every observation you saw was identical, just like in poker the probability still varies even after that depending on the other person’s behavior in ways you don’t directly see.

More subtly, these two behaviors are also different:

  • Monty opens a door you didn’t pick uniformly randomly and only offers you the choice to switch if he happens to open a goat door. The game ends if this random choice makes him “accidentally” reveal the car.
  • Monty opens a door you didn’t pick uniformly randomly except he uses his knowledge of where the car is to always avoid the car door. (So if there’s two goats, he picks randomly, but if it’s a goat and car, he avoids “accidentally” revealing the car).

And just like the extreme differences in Monty’s behavior led to extreme differences in the probability, here this subtle behavior difference leads to a subtle difference in the probability. For one of them, conditional on reaching “event 2” - that is, finding yourself in a situation where you do see a goat revealed, and where you do then get a choice - the probability of winning when switching is 1/2, and for the other one, the probability is 2/3. Can you math out which is which and why?

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68

It certainly matters: if he knows where the car is, he will never open the door with the car. If he doesn’t know where the car is, he could open a door with the car by chance (in 1/3 of the cases).

It does indeed matter how the problem is stated. If it’s stated that Monty opens the door and offers you a switch, there is the ambiguity that we don’t know whether Monty opened the door with the goat knowing it was a goat. It is still quite “likely”, from our general understanding of the world, that the quizmaster knows where the car is; he’s the quizmaster after all, quizmasters tend to know where cars are hidden.

However, if the problem states that Monty declares that he will open a door with a goat and offer you the choice and then open the door, then we can be sure that Monty knows where the car is (assuming he’s not lying).

But it does! That’s the whole point of computing probability under an assumption. Could you show me how it doesn’t matter?

For example, whether it rains outside or whether it doesn’t rain outside does not matter, since the chance that Monty opens a door with a goat is not changed at any point in time by whether it’s raining (presumably).

But, the door that you choose as a first option does matter to whether Monty can open a door with a goat:
Let’s say Monty doesn’t know where the car is:

  • if you choose the car (unknowingly), then Monty cannot open a door with a car.
  • If you choose a goat (unknowingly), then Monty can open a door with a car (it just turns out he didn’t).

Or let’s say Monty does know where the car is:

  • if you choose the car (unknowingly), then Monty has a choice between two doors.
  • if you choose a goat (unknowingly), then Monty has no choice but open the remaining door with a goat.

Hence your old opinion of reality (the chance of picking the car initially & on Monty’s knowledge) has influence over the likelihood of the current event (which door Monty opens).

It’s a good joke, though. :slight_smile:

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I think you’ve misunderstood the problem setup. That’s my fault for not being clear in my presentation. In my original post, I mean to convey the typical version of this problem where Monty always opens a door to reveal a goat.