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Marilyn vos Savant and the Monty Hall Problem (2015)

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Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#161
Wow, I knew about the problem and was vaguely aware that it had generated some controversy amongst statisticians but I had no idea about the insanely arrogant and obnoxious (not to mention wrong) abuse Marilyn had received from the "intellectual elite".

I have little sympathy for the "ambiguous question" defence. Not only is Marilyn's interpretation grammatically valid, it just wouldn't make sense in the context of the game show for the presenter (who knows where the goats are) to ever open a door to reveal the car and give you the option to switch. The game would be reduced to "does Monty feel like gifting you a car or a goat today".

Furthermore, Marilyn's own explanation of her answer makes it very clear that she interpreted the question as meaning that the presenter will only ever open goat doors. And all the attacks seem to focus on her grasp of the math rather than her interpretation of the problem question.

To be clear, I think it's perfectly reasonable for a person to interpret the problem the other way without thinking too much about it, and casually throw out an incorrect answer. But I would expect an academic to devote at least another 30 seconds of thought to the problem before publicly excoriating a woman known for her high intelligence.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#162
post #93

There's the Monty Hall Problem and then there's the Second Order Monty Hall Problem Problem. That is, did the wording as it was originally printed unambiguously define the problem to be solved, or was it ambiguous enough that some people who got it wrong got it wrong for the right reason? Like many others, I completely missed that when Monty opens a door to show you a goat, he always shows you a goat. I think under c…

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 ga…

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.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#163

Earlier quoted context omitted.

> And this is essentially what Monty is doing. He's retiring some doors and changing the odds. That still leads to the wrong answer, because he is retiring one of the losers, leaving a 50:50 chance of getting a prize.

You mean 1/3:2/3, right?

That's the correct answer. The wrong answer is 50:50.

The act of retiring leads to the wrong answer, so it does not aid in understanding.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#164
post #73

Has anyone else spent hours on understanding this and just accepted they’ll never accept it? I completely get all the explanations but it just feels too weird.

Based on other comments, it seems the people that refuse to accept the answer that it's better to switch and not 50/50 are nitpicking how the game is described.

The idea is supposed to be that Monty always reveals a goat. He does not pick a door at random to reveal before offering the choice to switch. If you pick door A, and the car is in door B, he will reveal door C and ask if you want to switch. If you pick A, the car is in C, then he will open door B and ask if you want to switch.

Expanding it to 100 doors with Monty opening 98 of them and then offering you the opportunity to switch should make it very intuitive. Let's say you pick door 7, and Monty then opens every door except your door and say, door 48, all revealing goats, then it should make you think "Wow...very odd that he didn't reveal door 48...there must be a reason he skipped that specific door".

Make it 1,000,000 doors. You choose door 1. Monty opens every door except number 423,901, and all the doors he opened had goats. You're tired and hungry because all you've done for 4 days is watch Monty open doors to goats, but you should certainly be thinking it's odd that he skipped that one specific door. You should probably switch, though maybe take a nap before claiming your car.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#165
post #160
post #112

Earlier quoted context omitted.

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, an…

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".

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#166

This is something I remember from the time. I could not grasp the probability maths (still shit at it, sorry Ms Von Savant) but I could write a program to experimentally demonstrate it, so I did. In QBasic. Because that's what we had and we were happy.

Ha, I did the same thing after first reading about the problem. I understood the probabilities from her explanation, but it felt more indisputable after seeing the results play out after a thousand tries. Guess you could call it a "Monty" Carlo simulation.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#167
The asymmetry is that if you pick the one with the car first (only 1/3rd chance) then Monty Hall can show you either of the other doors. If you pick one without a car first (2/3rd chance) then Monty Hall must show you the door that doesn't have a car and the remaining door always has the car.

So 2/3rd of the time you have a 100% chance of a car if you switch, 1/3rd of the time you have 0% chance of a car if you switch. So switching is a 66.7% probability of a win and is the 'inverse' of if you picked the car right on the first guess or not.

The choice of which door Monty Hall shows you is constrained and isn't an independently random event since he never shows you the car when he opens a door.

You can also laboriously go through every event and "bifurcate the world" and then add up the probabilities, but you need to be careful about how you do it:

    1.  assign the car to a door (there are three possibilities)
    2.  have the contestant pick a door (there are now nine possibilities)
    3.  handle which door Monty Hall shows (this is tricky because for the three permutations in #2 where the contestant picked the car, then the door Monty Hall picks is unconstrained and so the world splits, but you are slicing a 11.1% chance in half for ~5.6%, while for the six permutations where the contestant didn't pick the car, Monty Hall is constrained so those remain 11.1% chances (at every level the probabilities have to add up to the probabilities before them and you can't violate 100% when you sum across all of them).
    4.  then have the contestant either switch or not switch and cut all your probabilities in half again.
    5.  now color everything according to winning or not and sum it all up.
I call that the "Quantum Many Worlds Interpretation Solution to the Monty Hall Problem" and if you do step #3 correctly you should get vos Savant's answer.

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#168
post #93

There's the Monty Hall Problem and then there's the Second Order Monty Hall Problem Problem. That is, did the wording as it was originally printed unambiguously define the problem to be solved, or was it ambiguous enough that some people who got it wrong got it wrong for the right reason? Like many others, I completely missed that when Monty opens a door to show you a goat, he always shows you a goat. I think under c…

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?

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#169
post #56
post #39

The intuitive way for me to understand the Monte Hall problem is to pretended there are 1000 doors. You pick 1. There’s a 1 in 1000 chance you get it right. The host then opens 998 doors that don’t have the prize. Do you keep your original or do you switch? Are the odds 50:50?

I think the reason it's so hard to understand is that we try to break the problem down into more digestible problems, and in this case, if you only consider the last stage of the game, it certainly seems like it should be 50:50 odds. In other words, at the final round, you are presented with two doors. Ignore the fact that you already chose one, and assume you were starting at this point: you enter the game, see two…

That's especially tempting since we're used to probability problems often being set up where considering previous state is the sucker's answer. E.g. a fair coin has landed on heads three times in a row, what are the odds of it being tails on the next throw?

Re: Marilyn vos Savant and the Monty Hall Problem (2015)

#170
post #160

Earlier quoted context omitted.

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

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