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> 50% of the time, monty opens another door and it has a car and you win immediately

Why would you think you won? You didn't choose that one with the car and you aren't offered the opportunity to switch to the open one, only the closed one. The only interpretation which makes sense is that he always shows you a goat.

I think they meant you lose immediately.

Not to mention Monty always showing a goat is what adds the tension and interest a game show needs.

I disagree. If Monty always reveals a goat, there is no tension. You know exactly what is going to happen. Maybe it could be used a way to pad the running time of the show, but it would not add tension. "Coming up next: Monty reveals what's behind one of the other doors!" seems like something a game show would do. Whereas "Coming up next: Monty reveals a goat!" does not.

Yeah, you make a good point. I can sort of see it both ways, I think it would depend on how the show handled it. Leaving it across a commercial break then yeah it probably does make more sense the door is random.

counterintuitively, it doesn't actually matter whether monty knew there was a goat before he opened the door; what matters is that you observed the goat, so you know you're not in one of the possible worlds where he opened the car door

edit: this is wrong, dekhn was right, see below

If you want the result where there is a lower conditional probability of your originally selected door having the prize versus the remaining unopened door, then it completely does matter to have the setup be "Monty always opens a door with a goat" instead of "Monty opens a door at random, and in this particular case, it happened to have a goat behind it".

nope, do the math

edit: i ran a monty carlo simulation¹ and i was doing the math wrong. it really does matter if monty knows or not. here's the simulation where he knows:

    In [15]: non_censored_trials = got_car_trials = 0

    In [16]: for trial in range(100_000):
        ...:     car_door = random.randrange(3)  # the other two doors have goats
        ...:     your_door = random.randrange(3)
        ...:     monty_door = random.choice(list({0, 1, 2} - {your_door, car_door}))
        ...:     if monty_door == car_door:
        ...:         print("Monty showed the car, never mind")
        ...:         continue
        ...:     non_censored_trials += 1
        ...:     your_new_choice = next(iter({0, 1, 2} - {your_door, monty_door}))  # you change your choice
        ...:     if your_new_choice == car_door:
        ...:         print(f"You got the car because you changed from {your_door} to {your_new_choice}")
        ...:         got_car_trials += 1
        ...:     else:
        ...:         print(f"Too bad you changed; you should have stuck with {your_door}")

    (...output omitted...)

    In [17]: got_car_trials / non_censored_trials
    Out[17]: 0.66921

so in ⅔ of the cases, switching doors gets you the car. by contrast, if monty didn't know which door would reveal a car, it's only ½ of the cases:

    In [21]: non_censored_trials = got_car_trials = 0

    In [22]: for trial in range(100_000):
        ...:     car_door = random.randrange(3)  # the other two doors have goats
        ...:     your_door = random.randrange(3)
        ...:     monty_door = random.choice(list({0, 1, 2} - {your_door}))
        ...:     if monty_door == car_door:
        ...:         print("Monty showed the car, never mind")
        ...:         continue
        ...:     non_censored_trials += 1
        ...:     your_new_choice = next(iter({0, 1, 2} - {your_door, monty_door}))  # you change your choice
        ...:     if your_new_choice == car_door:
        ...:         print(f"You got the car because you changed from {your_door} to {your_new_choice}")
        ...:         got_car_trials += 1
        ...:     else:
        ...:         print(f"Too bad you changed; you should have stuck with {your_door}")

    (...output omitted...)

    In [23]: got_car_trials / non_censored_trials
    Out[23]: 0.49987257709086

so if monty picked the goat door on purpose, you do gain by switching. but if he just got lucky, you don't

______

¹ thank you, manoj

I dunno about anybody else, but to me teaching people how to convert word problems into code/simulations and how to interpret the results is one of the most important things a country that wants to be wealthy and powerful in the future should be doing.

I've always admired folks who can do these sorts of things in their head, while it takes me a bunch of time to inspect the code and convince myself it's an accurate representation of the word problem.

in this case I couldn't verify it against an actual experiment, which i think is good practice for newly programmed simulations, but when I saw that removing car_door from monty's choices made the probability go from 50% to 67%, i was reasonably sure that i hadn't fucked up the code, just the stuff i did in my head

One way to think about is, suppose you switch -- why not switch back?

In the random-open case, you really know nothing new about either of the closed doors. If you can talk yourself into switching, you could make an equally good argument for switching back.

In the Monty-knows-and-always-shows-goat case, you have gained information about one of the closed doors. You haven't gained any information about your initial pick door. But the other remaining door, you know there's a 2/3rds chance that Monty was forced to avoid it so as not to reveal the car. Only in the 1/3rd case where you were already on the car does Monty have freedom to open either door willy nilly.

that's all true, but i had concocted explanations that sounded equally convincing to me of why it was wrong

You can tell because only trials where you chose a goat in the first round can be "censored". Your first pick is still twice as likely to be a goat as a car, but half of the times you do choose a goat on round 1 you won't get a chance to switch. Whereas if your first pick was a car, the game is guaranteed to complete (and switching is guaranteed to lose).

this is an excellent insight; thank you

But if the car door is picked, there's no further game to play. Surely the only interesting thing to ask is, conditioned on seeing a goat, what's the probability the third door contains a car. The cases of observing a car are irrelevant since that's not the scenario. I'm still not convinced it's any different whether Monty Hall knows or not so long as the goat door is opened.

Edit: having written this, thinking about the 100 door case. If 98 random doors open (that aren't the one I picked) then the fact they all contain goats is pretty suggestive that I have the car, or at least 50/50. I'm not convinced by the code example though.

your first paragraph is what i thought before running the simulation, but it turned out to be wrong. if my simulation isn't convincing to you, try writing one that is

I'm convinced i think, just not initially by the simulator! I needed to reason my way separately but I think I'm happy with the sim now.

Yes, it's exactly 50/50. It's not hard to work out the details, as I did in another thread elsewhere on this article. Here goes:

-There's a 1/100 chance my door is a car, in which case it doesn't matter which of the others stays closed, it will always be a goat. The game will proceed, and switching will lose.

-There's a 99/100 chance my door is a goat, in which case the car is behind some other door. Choosing 98 out of 99 doors to open at random is the same as choosing 1 out of 99 doors to leave closed at random. So the chance that the car stays hidden in this case (so that switching will win) is 1/99, and 98/99 that the game ends early because the car is revealed.

-Adding it up, the game ends without the chance to make a choice 99/100 * 98/99 = 98/100 of the time. Of the remaining 2%, 1/100 comes from the first case (switching loses) and 99/100 * 1/99 = 1/100 comes from the second case (switching wins). The strategies are equally effective.

> the fact they all contain goats is pretty suggestive that I have the car, or at least 50/50

That's exactly the point. If he doesn't know, then it's exactly 50/50 and there is no reason to switch. If he does know, then it's 1/NUM_DOORS versus NUM_DOORS-1/NUM_DOORS, so you'd be crazy not to switch.

The point is, if he picks at random, in the 100 door case, the vast majority of the time he will open the car door while opening those 98 doors. The case where you get to pick again would be exceedingly rare. Conversely, if he only opens goat doors, you will get your second pick 100% of the time.

This is why one can't ignore the car picked cases, so the simulator is correct.

I believe the confusion with "ignoring the car picked cases" comes due to thinking that they are additional games to the original ones, instead of being part of those that constitute the 2/3 in which the player starts picking wrong.

So you thought that eliminating them you were left with the original 1/3 vs 2/3, when you actually removed half of the 2/3.

I don't think your interpretation of the sentence is sensible. The sentence mentions that the host knows what's behind the doors. So, if he is allowed to open the door with the car, the problem would become insoluble and would just be about speculating on the host's personality. And it definitely doesn't support the conclusion that the probabilities become 50/50.

it's pretty counterintuitive that his personality enters into it, isn't it?

If the format of the game allows him to show the car to the player, how could his personality not enter into it? In every case where the player picks a goat-door, the host will be presented the option to either reveal the car or the goat. I mean, one can imagine various complicated scenarios in which the host might reveal the car exactly 50% of the time in such cases, but none seem like they can be reasonably arrived at.

it turns out that there is in fact no way that his personality could not enter into it, but that was not obvious to me until i did the simulation. even if he chose to reveal the car exactly 50% of the time in such cases, that would be a result of his personality, wouldn't it?

You can imagine factors other than his personality, but they're all equally as speculative. Additional rules to the game, for example.

> opens another door, say #3, which has a goat

There is no randomness in this sentence. The door has a goat, not a car.

Then again, you might read it as an example outcome. But even if the door was chosen randomly, it would not justify the 50%/50% answer.

Independently of that, assuming that the host chooses the door with the car and the goat with 50% probability each is the same mistake that confused so many, you cannot always assume uniform probability only because there are 2 options.

(I wanted to point out the flaws in your thinking because this has also been a confusing problem for me)

I think you are correct about my thinking: at the time, I assumed it was an example outcome and my brain turned off at that point and simply assumed the probs were still 50-50. In some sense I was both intellectually wrong (in my assumption about it being an example) and intellectually lazy (in my failure to work through the implications of my assumption).

Personally the whole thing taught me is one true hallmark of intelligence is the ability to eliminate unnecessary ambiguity and find the "right" answer.

This was an interesting discussion when I read about it long ago.

I don't think it's obvious that Marilyn's interpretation is the correct one. Two possible interpretations could be equally valid.

In law we have the "rule of the last antecedent" that says descriptive clauses modify the nearest antecedent noun.

Under this interpretation "which has a goat" modifies the exemplary door "say #3". Same as saying host opens a door, for example door #3 containing a goat. It could have been say door #2 containing a car.

Marilyn's interpretation is that "which has a goat" modifies "host opens another door." Same as saying the door opened by the host always has a goat.

Language is inherently ambiguous. I don't think either interpretation is unreasonable...