Monty Hall, Monty Fall, Monty Crawl

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It took me a long time to figure it out (after being told I was wrong the first time) but I finally decided this was a good way to think about it:

The key is that the host CAN NOT open the same door you chose. His or her hand is forced if you chose wrong.

2/3 of the time, your initial choice will be wrong, so 2/3 of the time the host’s choice is forced.

In those cases, the host MUST open the other wrong door, leaving the third door as the right one. So, 2/3 of the time, the switch will be beneficial.

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Is there a subtle difference between the questions for A. The Unfinished Game (answer = 2/3) and B. Nebulous Neighbours (answer = 1/2)

A. “Let’s say, hypothetically speaking, you met someone who told you they had two children, and one of them is a girl. What are the odds that person has a boy and a girl?”

B. “Your new neighbours have two children of unknown gender.
From older to younger, they are equally likely to be girl-girl, girl-boy, boy-girl, or boy-
boy. One day you catch a glimpse of a child through their window, and you see that
it is a girl. What is the probability that their other child is also a girl?”

http://www.probability.ca/jeff/writing/montyfall.pdf

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Hypothetical Situation:

The Monty Maul problem. There are 1 million doors. You pick one, and the shows host goes on a bloodrage fueled binge of insane violence, knocking open doors at random with no knowledge of which door has the car. He knocks open 999,998 doors, leaving your door and one unopened door. None of the opened doors contains the car.

Are your odds of winning if you switch still 50/50, as outlined by the linked Rosenthal paper? It seems counter-intuitive even for people who’ve wrapped their head around the original problem.

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Come ON!

There are two doors closed. One contains a car.
The fact a third door is opened that contains a goat is irrelevant.

This is NOT about your odds at the beginning, when you had more doors (conditional probability). The question is posed RIGHT NOW, with two doors from which to choose. Asking you to switch doors is just posing the question in a different way. Which door?

At that moment, there is a 50/50 chance, a .5 probability. All events leading up to this moment are irrelevant to the odds that the car is behind ONE of TWO doors.

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You call it the “Monty Fall” problem.

ABC calls it “Deal or No Deal”.

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@Ian - the subtle difference is that in “Nebulous Neighbours” one assumed that the child you observe is taken randomly from the two children. In “The unfinished game”, the parent tells you “at least one of my children is a girl”, and you can don’t assume that that information is randomly sampled (e.g. if the distribution is BG or GB, you don’t assume there’s an equal chance the parent would have said "at least one of my children is a boy),

Or at least, that’s my understanding.

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@Charles - I really hope you’re trolling.

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Oh, wow, I just failed hardcore. I put my e-mail address in the “website” line. Jeff, if you feel like going ahead and fixing that (I’ve got it linked correctly here), I’d be grateful. And if you don’t, no big deal!

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@DavidW
Now flip ten million coins and randomly distribute them in sets of various sizes. Can you guarantee that all the sets of 2 results are evenly distributed among all possible combinations?

It’s like if you were playing Blackjack. You know the possibility of getting every hand of 2 cards and then base your playing style off of that. But you are neglecting to take in account that cards already played affect the odds of you getting any particular 2 card hand.

Now, if we don’t the ratio of girls to boys of non 2 child families, we are not equipped to make assumptions based on the ratio of 2 child families.

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What’s wrong with goats? I’d take a goat. They make good pets.

Seriously, one of the problems people (like @mgb above) have with this is over-reliance on knowledge of the real Monty Hall’s motivations on TV. It’s unfortunate that the naming of this has contributed in part to the difficulty. If you ignore the preconception that comes with it and just look at the conditions as stated, it will help some of those people come closer to understanding. “It’s better to switch than fight.” (How’s that for some old TV?)

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This seems like it would be 1 in 3. Is this not correct?

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The Monty Fall PDF has a “Nebulous Neighbors” example which appears to be saying that the answer to the boy-girl problem is actually 50% after all.

Discuss.

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it’s a 33/33/33 choice and you drop one – how does that not make it a 50/50 choice?

That’s the mistake. You’re not dropping one, you’re getting both doors that you didn’t pick. He’s just opening one of them early. I didn’t understand this until I programmed up a simulation for myself.

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One way I like to think about the Monty Hall problem is to take out the player’s second choice. Since the problem’s question is ‘Is it better to always switch or always stick?’ then what we need is the probability of winning when one always switches, and the probability of winning when one never switches.

Then there is only really one choice; what door to pick to begin with (a priori, of course, all have 1/3 chance of being the winning door). Clearly there are two outcomes from this choice; you pick the right door (probability 1/3) and that you pick the wrong door (probability 2.3).

If you picked the correct door to begin with and ALWAYS switch, you will ALWAYS lose, if you picked the wrong door and ALWAYS switch you will ALWAYS win.

If you picked the correct door to begin with and NEVER switch, you will NEVER lose, if you picked the wrong door and NEVER switch you will ALWAYS win.

So when never switching, p(win in 2nd round) = p(win in 1st round) = 1/3. And when always switching, p(win in 2nd round) = p(lose in 1st round) = 2/3. QED

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This problem is neat. Yes you would want to switch. It’s a little more strait forward than the previous question, which had some ambiguity depending upon the semantics.

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I fail further. DoND is on NBC, not ABC.

Monty Crawl is interesting. Even more interestingly, even though it seems like the door you choose at the start should affect your odds, it doesn’t; I initially thought it did, until I realized it’s like the door I chose vanished and Monty is crawling between only two doors, and it really doesn’t matter which one is gone.

The neat part is that Monty Crawl, like the original Monty Hall problem, still gives you a 2 in 3 chance of winning. Even if Monty’s door-opening algorithm is completely deterministic, it gives you no bonus information over opening a non-goat door. Your answer- “always switch”- doesn’t even need to change either!

Let’s say you have pity on Monty, and take door 3. It really doesn’t matter for your odds. There are three possible universes for the problem, with equal probability:

1:3 - goat behind door 1 (losing) - universe 1
1:3 - goat behind door 2 (losing) - universe 2
1:3 - goat behind door 3 (winning) - universe 3

Any of these can happen with equal probability. So what does Monty do in each case?

universe 1- open door 2
universe 2- open door 1
universe 3- open door 1

Note that for universes 1 and 2, Monty has no choice; he’d go for the same doors even if he wasn’t exhausted. Only Universe 3 changes Monty’s behavior.

But you know more about Monty’s behavior! You know he won’t open door 2 in universe 3. So if he opens door 2, you’re in universe 1, and you must change doors to hit a goat. Universe 1 happens 1/3 of the time, so 1/3 of the time, you win the car because Monty’s behavior tells you 100% that the car is in door 1. Switch.

2/3 of the time, Monty will open door 1. Now you’re in universe 2 or 3. If he chose door 2, you would definitely have switched; now, the only information he offers is that it’s not behind door 1, because he essentially would have told you if it was. His answer, therefore, tells you about door 1 and provides no information as to whether the correct door is 2 or 3. Universes 2 and 3 are equally probable, and both have the same “symptom”; door 1 provides no evidence as to whether you should switch or not, as there’s a 50% chance on either door. If you switch, you get a 50% chance; if you don’t, you also get a 50% chance. “Always switch” still works, although it’s not necessary if door 1 opens.

Interestingly, this version of the problem is more obviously a 2/3, because it’s conceptually identical to getting two chances to choose a door. You make your first choice; then Monty tells you flat-out whether or not the car is behind the lowest-numbered unchosen door. If it is, you win. (odds 1/3.) If it isn’t, you get a second pick from two doors- which gives you 1/2 odds, but it only happens 2/3 of the time, so this contributes 1/3 to your chance as well.

I’m not completely sure I’m right on this. I haven’t read the paper for its answers yet.

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This is by no means the same as the boy-girl or flip-a-coin choice. BTW, it obeys the stats laws all the way. It simply says that you are more likely to pick the wrong door to the odds of 2 out of 3 times. Hence if you assume you picked the wrong door and the host shows you the other wrong door then then switching should give you 2 to 1 odds of being right. Buyer beware. What all the mathematicians are trying to say is if you have a ‘truly’ random chance of winning that no matter how many times you repeat a result over and over again, next result is the same chance as the first try. So if you flip a coin ten times and get heads as in HHHHHHHHHH the chance of getting another head is still 50%. That is non-intuitive not this half-baked example. Two seconds of “real” thought solves the door issue. I personally forgive the PhD’s as the contrived"ness" of the example was very poorly worded. Hence the confusion. For the record, if there is a million doors and you eliminate all but one and your door and there is only one winning door you would want to switch even more because there is only one in a million chance that the door you chose is the right door.

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Case 1: 33%
Goat - you open this one
Goat - host will open this one
Car

Case 2: 33%
Goat - host will open this one
Goat - you open this one
Car

Case 3: 33%
Goat - 50% host opens this one
Goat - 50% host opens this one
Car - you open this one

In the first 2 cases (66%), you get the car if you switch. In the last case (33%) you get the car if you don’t switch.

The host tells you something by his choice of which door to open.

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Ok, I see it like this. You pick a door at random, the host offers “Your door, or these other two doors?”. Do you switch? According to the game rules, at least one of the two doors contains a goat: you don’t have to be surprised by this.

Probabilities are tough. But then who seriously claims truth must be intuitive; does the sun actually rise?

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I always found the proof of exhaustion really solidified it for me. The wikipedia article has the diagram proving that changing your solution wins 2/3 of the time, but it’s trivial to do the same showing that not switching loses 2/3 of the time.