If you are not already familiar with the Monty Hall Problem, here it is:
It appears like this is a fraud. I find it difficult to believe how it became so popular.
The confusion in the argument is that it doesn't consider the dynamic of time.
The moment something changes effecting the probability of the outcome, the entire decision should be reconsidered.
While it is true that at the time when the initial choice was presented and the first door was chosen, the odds of that door belonging to the car is 1/3. However, the moment the host opened a door to reveal a goat, the original decision must be reconsidered as the odds of the door chosen hiding the car are now 1/2. Therefore, it should make absolutely no difference if you remain with your original decision (which now has a 1/2 chance of belonging to the car) or switch (which also has a 1/2 chance of hiding the car).
Why do so many people fall for it? Can someone please explain?
As per Eli's suggestion (in the comments) I experimented with this here: http://www.nytimes.com/2008/04/08/science/08monty.html
here are my results:
but I can't see the programming, so it's not really a perfect experiment.
Here's the explanation provided there, for the first time it's beginning to make some sense:
"If your strategy is to always switch doors, you will lose only if your initial choice is the door with the car, which is a 33.3 percent chance. In the other two cases (66.7 percent of the time) you will switch to the car and walk away a winner."
In other words, sequence makes all the difference. After you have made a 1/3 chance choice, the game show host points you closer toward the more likely option by showing you which of the remaining two it is not.
Why do the odds of "not switching" not change now that one of the doors hiding the goats has been revealed? Why is it that if someone were to come and make a fresh choice with one door open revealing a goat and two doors closed one hiding a car and one hiding a goat, the options would be equal?
Consider it from this perspective: The game show host had to open a door revealing a goat. There is a 2/3 chance that you limited his choice to one door (if you chose a door with either of the two goats you would be forcing him to reveal the only other door with a goat), so there is a 2/3 chance that he revealed the other door with a goat, making it a 1/3 chance that you win if you stay with your choice but a 2/3 chance to win the car if you switch.
If you are presented with a first choice with one door open already revealing a goat, then your chances are 50/50, since there is no 2/3 chance that you forced the host to open the door with that goat and he could have just as well opened the door with the other goat.
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ReplyDeleteEli, I'd love to be wrong, and I'm sure I'll be surprised if I tried the experiment (I've always had bad luck).
ReplyDeleteThe Problem is the explanation, I don't get it.
I agree that it's difficult to comprehend and that it's "counterintuitive", but calling it a "fake" and a "fraud" is not gonna change the fact.
ReplyDeleteI'll try a slightly different angle:
In your first choice - you're choosing "against" 66%, so the chance that it's behind one of the "other 2" doors is 66%. So all he did by opening one of the other ones - was moving over the 33% of that door to the third one. It doesn't move over to the first one because the first one was taken out of the "race" already with a 1/3 chance, and 2/3 against it. He's just giving you a "tip" regarding the other 2/3 but there are still another two thirds. By him opening the 2nd door he's not telling you "anything" in regard to your first choice. He's only telling you in regards to the "66%" that stands against you.
In other words, after opening the second door, you could then "confidently" tell your self, "well, if the odds are against me (which they are because of the 66%) then it's not in door 2 so it must be in three".
I can try to go on and on, it's very difficult to understand, let alone explain. What's more "shocking" is that an outsider who wasn't given the first choice - indeed has a 50/50 chance.
I'm not sure what would happen if they would do it together.
Last hack at it:
Suppose you took the same scenario only this time you were told to try and hit a "goat".
You choose door 1, and then he tells you "well, the goat you "didn't" choose is behind door 2 and now that's out of the equation, wanna switch to 3 or stay with 1?"
I'd say stay with one. Similar reasoning. Your first choice had a 66% chance. why change it?(I'm not sure on this one though).
Good luck and correct your other post as well (check my comment there).
Thank you for citing the classic explanation. What I don't get is why you say:
ReplyDelete"the first one was taken out of the "race" already with a 1/3 chance"
The moment the goat was revealed in one of the doors, the probability of the first option change from 1/3 to 1/2.
Thank you for citing the classic question.
ReplyDeleteI don't claim to "comprehend" this problem very well.
But since the facts prove that it does "remain" a 1/3 and not 1/2 chance, I'm compelled to "admit" that it's correct rather than call it a fake and fraud.
One cannot argue with facts. It may "sound" even outrageous, but it doesn't change the fact.
Sorry I can't do better explaining. I don't know if there is a better explanation to be given.
What are you going to tell yourself when you actually "try" it and see you're wrong?
My guess is, you'll try to tell yourself all of my above arguments. :)
By the way, did you ever study "Probability"?
ReplyDeleteIt may help you understand.
You can ask on the very existence of "Probability" as a math subject. Isn't "Math" by definition a precise science? What has probability to do with it?
Good question - but I can't say before I study probability. Neither can you. (I'm guessing that if you had, it would be more easy for you to grasp this).
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ReplyDeleteWhy do the odds of "not switching" not change now that one of the doors hiding the goats has been revealed? Why is it that if someone were to come and make a fresh choice with one door open revealing a goat and two doors closed one hiding a car and one hiding a goat, the options would be equal?
ReplyDeleteConsider it from this perspective: The game show host had to open a door revealing a goat. There is a 2/3 chance that you limited his choice to one door (if you chose a door with either of the two goats you would be forcing him to reveal the only other door with a goat), so there is a 2/3 chance that he revealed the other door with a goat, making it a 1/3 chance that you win if you stay with your choice but a 2/3 chance to win the car if you switch.
If you are presented with a first choice with one door open already revealing a goat, then your chances are 50/50, since there is no 2/3 chance that you forced the host to open the door with that goat and he could have just as well opened the door with the other goat.
The question is what would happen if you had the two combined.
ReplyDeleteOne guy was guessing in the mony-hall case, whilst another guy would pop into the room every time "after" a door was opened.
Suppose guy 1 consistently chooses door 1 and doesn't switch. He'll hit the car 1/3 of the time. So if it's done 100 times, the car will be behind door 1 only 33% of them.
Guy 2 watching, consistently chooses door 1 as well. If for him it's 50/50, why won't he get his full 50% on his consistent guessing of door 1?
Regarding your NYT test - that's why I say try it out with a friend. No "behind the scenes" there. Have him write it down if you'd like.
If guy #2 a) knows that the host must reveal a goat and b) he knows which door guy #1 picked first, his guessing is identical to guy #1 and the probability is the same (1/3 car if not switching, 2/3 car if switching).
ReplyDeleteIf guy #2 isn't aware of either detail, he has 50/50 chance, because the door with the open goat is meaningless to his chances of guessing the door with the car since it doesn't reveal any clue as to the 2/3 chance of the whereabouts of the first goat as in the case of guy #1 and the hosts revelation. What is important to remember when thinking about this, is that the host opening a door that must be a goat is a clue that says there is a 2/3 chance that the first door chosen was a goat and you should switch to up your chances of getting the car. Without such a clue, chances remain 50/50.
I'm talking about a scenario where he's unaware.
ReplyDeleteI spelled it out in my question. Again - they are both betting on the same car. The game show host will ask guy 1, and then guy 2. Say they're both consistent about choosing door 1.
Guy 1 will be correct 1/3 and guy 2 1/2?
That's impossible. We're talking about the same door/car/goat. Is it or is it not behind door 1 the remaining 1/6 of the time?
To answer a question with a questions. What are the chances of them consistently choosing door 1 without sharing information.
ReplyDeleteIf guy #1 consistently chooses door 1 without switching, his chances of winning are 1/3 of the time, since he chooses between three options.
In this case, guy 2 will also have a 1/3 chance of winning, because his choices are based upon guy 1's 3-option choice, since the door the host opened was dependent on the choice of guy 1 (leaving door 1 consistently open).
You seem to be implying that there is a difference what guy 2 chooses.
ReplyDelete50/50 would mean the 50% of the time the car is behind door 1 and 50% of the time it's behind door 2 (which is door 2 or 3 for guy 1), regardless of what guy 2 actually chooses.
What I'm not understanding is how come in reality, whether or not he chooses door 1 - it will be behind door 1 only 1/3 of the time.
I'm again talking in a case when he has "no clue" of guy 1's choices. He comes into a room, sees 2 doors, #3 has been opened and now taken away - so that it's totally not in the equation for guy 2, so practically speaking he only has 2 choices. But instead of it being behind door 1 50% of the time and behind door 2 50% of the time, it'll be 33/66 (somehow, guy 1 still had an effect on guy 2's outcome, even though guy 2 has "no clue whatsoever" of guy 1's case).
If there were 100 doors, and 98 were then opened, guy 1 would essentially find the car behind door 2 99% of the time.
ReplyDeleteThe same would be for guy 2. So for guy 2 it's nothing 'near' 50/50.
(Yeah - mathematically this post doesn't add anything :), but just to perhaps better express the issue at hand).
Are the two guys in the same game show? If so, guy 1's choice of door 1 forced the game show host to open a door with a goat making the switch door 2/3 chances of it being a car. Yes, guy 1 influenced which door would be open when guy 2 walks in, so if guy 2 choses door 1, he will have 1/3 chance of being correct (since door 2 now holds the 2/3 chance thanks to the open door 3 that was influenced buy guy 1's choice of door 1).
ReplyDeleteFrom guy 2's perspective, he sees two doors and might not know that it would make any difference which door he chooses (if he didn't know about guy 1's choice of door one and that the host's decision about which door to open was based on guy 1's choice) but the reality is that now there are odds are 1/3 car for door 1 and 2/3 car for door 2. And if guy 2 keeps on playing he'll notice that the car is behind door 2 2/3 of the time (since guy 1 keeps on choosing door 1).
Indeed...
ReplyDeletemr zalman, i too fried my brains trying to prove to others and myself that this problem was flawed. trying to do that, i convinced myself otherwise... that the answer is correct. i tried to use a lot of math and stuff to understand it, and didnt knew why it didnt made sense to me.
ReplyDeleteat the end, the most simple facts helpep me understanding it: everytime you choose a goat door and switch, you win. everytime you choose the car door and switch, you loose (these are absolute truths). if the chances of choosing a goat when the 3 doors are closed is 66%, if your strategy is to switch all the time, you will win 66% of the time.
the fact that the host knows where the goats are, and always has to open one of them, changes all. amazing and very odd to our eyes.
the thing is, it will keep sounding like fake to you untill you can visualize it. the way the monty hall is proposed is very dificult for most of us to do that, bus this video has an explanation at 2:20 that makes it very easy, doing it by exageration.
ReplyDeletehttp://www.youtube.com/watch?v=koPBkK_Ra-k
Guys I’ve tested this 1000 times using flash cards and I come up with about 50/50. Monty hall is wrong! The doctor who came up with this is gay and doesn’t understand how probality works.
ReplyDelete