Hacker News Reader

I think this is the original "Ask Marilyn":

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 #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat. He says to you, "Do you want to pick door #2?" Is it to your advantage to switch your choice of doors? Craig F. Whitaker Columbia, Maryland

Yes; you should switch. The first door has a 1/3 chance of winning, but the second door has a 2/3 chance. Here’s a good way to visualize what happened. Suppose there are a million doors, and you pick door #1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?

-----

Yes, the question as posed by a reader was ambiguous but Marilyn made a very reasonable interpretation based on what someone at that time would probably tend to know about game shows. And she even gave an explanation (using more doors) that I find tends to work with folks combined with the explicit assumption that the host will always pick goats.

Yes, I'm literally talking about this sentence: "host, who knows what’s behind the doors, opens another door, say #3, which has a goat."

I parsed that as "50% of the time, monty opens another door and it has a car and you win immediately, and 50% of the time, monty opens another door and it has a goat". In retrospect I think my brain just sort of pictured that and proceeded to assume there was no reason to switch, and it wasn't until I read some answers (not her explanation) that I understood I was wrong (and I went through a lot of anger, like many of the commenters, and thought she was very wrong before that).

If there had been exactly one more sentence saying "monty always opens a door showing a goat", I'm pretty sure I would have recognized that. BTW I was familiar with the game (I liked Jeopardy better, as Let's Make a Deal was a fairly dull show) and I don't think they had a game that was identical to the problem as described.

> 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...

What astonishes me is not that people interpret the scenario differently, but that they are so quick to conclude that others are doing the math wrong and feel superior or even angry, rather than thinking about whether you might be talking at cross purposes about different scenarios.

So I see the second order problem a bit differently: it's not "was the statement unambiguous", it's about what you do with the (possibility of) ambiguity: do you assume you can't be wrong and therefore others must be, or do you try to figure out whether you were solving different problems?

Humans don't backtrack well :)

It's kind of like the xkcd airplane on a treadmill blogpost:

https://blog.xkcd.com/2008/09/09/the-goddamn-airplane-on-the...

There appears to be problems that make it easy for people to assume the worst of others.

And on a side note, I discovered about 6 months ago that mythbusters actually did this experiment. At the end of the day, few things work better than seeing something happen in real time.

Vos Savant mentions this, and she's kept track of who did and didn't understand the conditions:

> And a very small percentage of readers feel convinced that the furor is resulting from people not realizing that the host is opening a losing door on purpose. (But they haven’t read my mail! The great majority of people understand the conditions perfectly.)

https://web.archive.org/web/20181118225305/http://marilynvos...

Nice find! I have wondered this. It's easy to say after you have come around that you had first misunderstood the scenario and done the math right in that context, even if in truth you (implicitly) understood the scenario but just neglected to account for the constraints.

As someone who misunderstood the problem as you describe, I'm much more invested in this version of the problem :)

The problem statement is just a mess:

> You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which has a goat.

even tiny edits like:

> You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door which has a goat, say #3.

or even

> You pick a door, say #1, and the host, who knows what’s behind the doors, opens another door, say #3, which he knows has a goat.

make it clear that the last phrase is describing an invariant of the problem, not an artifact of the illustrative hypothetical.

That's much better, but still not enough. We also need to add that the host will open a door regardless of which door I initially choose.

Ah yeah, good point:

> You pick a door and the host, who knows what’s behind the doors, opens exactly one of the two remaining doors according to the following algorithm: if both doors contain goats, he picks a door at random, else he picks the door he knows to contain the goat.

That interpretation requires believing that the host, who knows what’s behind the doors, would sometimes open the door revealing the automobile and ask the contestant if they want to switch to that door. It doesn't really make sense, so that should tip off readers that they've misunderstood the show format.

Marilyn even made this explicit in her answer, writing that the host "knows what's behind the doors and will always avoid the one with the prize" [0]. So readers were arguing her answer was wrong even given this understanding of how the show works.

[0] https://www.newspapers.com/article/the-missoulian-the-monty-...

> That interpretation requires believing that the host, who knows what’s behind the doors

The host has to always open the door, too. Else-- if your strategy is always to switch when shown a goat, the host can choose to only show you a goat if you have already picked the car (which would cause you to lose 100%). Or various mixed strategies.

This is one of those problems that is hard to grasp at baseline, but a little harder to grasp because the situation isn't fully specified. As a result it tends to create a lot of controversy, like the one about an airplane on a treadmill.

I assume that at the time of writing, most readers would have been familiar with the actual game show and the fact that the host always opens a door with a goat?

The show uses a different game.

I had never heard of the show when I was introduced to the problem so all my knowledge comes from the problem description.

As Monty Hall pointed out in interviews, he was not obligated to offer the opportunity to switch and sometimes did not. So in this case knowledge of the actual show may have contributed to confusion.

https://en.wikipedia.org/wiki/Monty_Hall#Monty_Hall_Problem

(But vos Savant stated in later columns that most of the critical responses she received assumed that the host was obligated to offer the switch, so they were genuinely confused by the paradox.)

Your re-framing is actually not quite right either (or at least, not complete). Key point: if the host picks the car, the game essentially re-sets.

Imagine the extreme scenario with 999 goats and 1 car. You select the first door, the host opens 998 doors at random, leaving your selected door and one other. Two scenarios are now possible:

1. There was a car in the 998 doors that got opened. Tough luck, you lose the game automatically now (game resets).

2. There was not a car in the 998 doors. In this scenario, you are still advised to switch.

Edit - I guess I misremembered a simulation I once did about it.

Also you're picking between Reward and Bigger Reward. Positive EV no matter your choice so eh don't worry about it too much :)

A nice goat fetches up to $1000 to according to a quick google.

A nice goat, you say? How much for a mean goat, jaded from years of being the spoiler prize that nobody wants?

Whenever someone talks about mean goats I think of only one thing: Buttermilk the baby goat, being a jerk.

https://www.youtube.com/watch?v=AWvefaN8USk

Enjoy.

I am suddenly reminded of the clever, morally educational joke about the dead horse raffle.

Nope. Write the simulation for 2.

Or look at this case analysis for the three element case. https://news.ycombinator.com/item?id=24714612

You can get intuition by understanding that the cases where the game is ruined are only the cases where you didn't choose the car.

As I explained in my reply to your comment on mine, this analysis doesn't work out, even in this very-lopsided case. If the game resets when a car is revealed, then games where you first choice happened to be a car will always complete, and games where your first choice was a goat will almost always reset. On average, of 1000 games, your first pick will be a goat 999 times, and of those, 998 will be reset because MH opens the car and 1 will complete and give you the car if you switch, whereas in 1 in 1000 you will pick the car first, in which case the game is certain to complete but you will lose the car if you switch. Therefore, most games will reset, but of the ones that complete, there is no advantage to switching.

> the problem doesn't actually allow you to interpret otherwise

How so? The problem is typically presented as a single trial, with nothing that clarifies the intent of the host.

No, there's not.

There's the Monty Hall problem and then there's the "can I get away with falling back on claiming the question was somehow wrong after embarrassing myself by getting the Monty Hall problem incorrect" problem. No one who claimed it was unclear or ambiguous did so before getting it wrong.

It's really tiresome to have to always pretend to assume good faith when it's really clear what's going on.