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Thread: The Monty Hall Problem

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    Default The Monty Hall Problem

    The Monty Hall problem, or the Monty Hall paradox is a probability puzzle which originated from a letter to the American Statistician magazine regarding one of the tasks in the American game show 'Let's make a deal' which is still running

    Quote Originally Posted by Steve Selvin
    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 No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

    Vos Savant's response was that the contestant should switch. If the car is initially equally likely to be behind each door, a player who picks door 1 and does not switch has a 1 in 3 chance of winning the car while a player who picks door 1 and does switch has a 2 in 3 chance, because the host has removed an incorrect option from the unchosen doors, so contestants who switch double their chances of winning the car.

    Many readers refused to believe that switching is beneficial. After the Monty Hall problem appeared in Parade Magazine, thousands of readers including 1,000 PhDs, wrote to the magazine claiming that Vos Savant was wrong.

    Decision scientist Andrew Vazsonyi described how Paul Erdős, one of the most prolific mathematicians in history, remained unconvinced until Vazsonyi showed him a computer simulation confirming the predicted result.

    Opinions on this paradox? What does it tell you about the human psyche? Have you heard of this before?

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    its because of the conditional probability.

    You are making two decisions back to back instead of a single decision so the consequences are different.

    Had to solve this problem in probability, wasn't a fan of it but it gets a lot of recognition as a popular problem.

    Recently I've been investigating predator and prey relationships.

    The take away I guess is that its better to modify one's guess after more information comes in rather than staying the same. In reality iterative systems of guessing could be considered, in which over time options are eliminated. In such a case its advantageous to change one's guess as options are eliminated.
    Last edited by male; 02-05-2013 at 05:19 AM.

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    this is interesting. i keep re-reading trying to figure out what i'm missing because it doesn't make any sense.

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    It's never made sense to me, it feels like when people play around w statistics, twisting things and playing w words...in this case numbers and operations.

    It's 2 choices, not 1.
    First choice has 1 in 3 chance to be correct.
    Second choice has 1 in 2 chance of being correct, whether they switch or not, the choice is still between 2 options, not 3. Choice number 1 has nothing to do with the second choice.

    If ya want to insist on keeping 2 out of 3 chances, then both doors left over have the same 2 out of 3 chances.

    Showing it on a computer simulation says little as the programmer was able to play with numbers and calculations to force his point. Let's see the experiment occur irl, w/o a computer. THEN show those results.
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    Its legitimate, just take time to think about it lol. If you prefer a world in which things don't depend on history then you could also study Markov Chains http://en.wikipedia.org/wiki/Markov_chain#Physics

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    Consider the hypothesis of increasing the number of doors.

    Instead of 3 doors, consider 10,000 doors.

    You open 1 door, the host opens 9,998 doors and shows goats.

    Would you switch and why?

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    Quote Originally Posted by Shang Tsung View Post
    you are comparing make a single decision (1/3 chance of success) against making two decisions given the constraint that the first decision is a failure (1/3 chance of immediate success + 2/3*1/2 chance of success on the second decision = 2/3).

    2/3 > 1/3 so its better to modify as a result of failure than not to
    doesn't this assume that the first decision was a failure?

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    Like i said...show me the experiment that didn't use a computer simulation where simple playing with programming would lead to the desired output.

    Otherwise, it's little more than another Ti game, using math.
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    Quote Originally Posted by anndelise View Post
    Like i said...show me the experiment that didn't use a computer simulation where simple playing with programming would lead to the desired output.

    Otherwise, it's little more than another Ti game, using math.
    Here is why switching wins, with images et al.

    There are effectively 6 possible scenarios/decisions for the 'entire game':



    In the opposite case,

    Sticking with the door wins
    Sticking with the door loses,
    Sticking with the door loses

    So, by sticking with the door there is a 2/3 chance of losing, whereas when switching door there is a 2/3 chance of winning.
    Last edited by InvisibleJim; 02-05-2013 at 03:32 AM. Reason: correction

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    Quote Originally Posted by anndelise View Post
    Like i said...show me the experiment that didn't use a computer simulation where simple playing with programming would lead to the desired output.

    Otherwise, it's little more than another Ti game, using math.
    A simulation can be made quite easily that is objective... just have a random number generator sample things thousands of times and compile the results from the two scenarios, as the number of samples approaches infinity then the results should approach the predicted mean as per the http://en.wikipedia.org/wiki/Central_limit_theorem, when their are less samples there will be a certain oscillations about the mean value, the more samples, the less oscillations. Hundreds of simulation runs for a given number of samples can even be averaged. Although in real life you won't have the resources to do an infinite number so its a matter of seeing this trend intuitively and only running the simulation enough for the desired accuracy.

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    Quote Originally Posted by InvisibleJim View Post
    Here is why switching wins, with images et al.

    There are effectively 6 possible scenarios/decisions for the 'entire game':
    [deleted img to save space]

    In the opposite case,

    Sticking with the door wins
    Sticking with the door loses,
    Sticking with the door loses

    So, by sticking with the door there is a 2/3 chance of losing, whereas when switching door there is a 2/3 chance of winning.
    I am fully capable of reading wikipedia, ty.

    1. What is the probability of winning the car by always switching? Answer: 2/3 (note: I would argue that it is also 2/3 chance of winning if always staying with the first choice..as that is still making a choice at that moment.)
    2. What is the probability of winning the car given the player has picked door 1 and the host has opened door 3? Answer: 1/2
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    Quote Originally Posted by anndelise View Post
    I am fully capable of reading wikipedia, ty

    (note: I would argue that it is also 2/3 chance of winning if always staying with the first choice..as that is still making a choice at that moment.)
    Try drawing out the full game, step by step.

     
    Last edited by InvisibleJim; 02-05-2013 at 04:06 AM.

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    Quote Originally Posted by InvisibleJim View Post
    Try drawing out the full game, step by step.
    Wow, the purpose of you editing my post to quote me as having made 4 angry emoticons was for?? Do you really think I'm emotional over something like this? Lol.

    I stated at the beginning that the puzzle had never made sense to me, and at least twice I stated that it feels like people manipulating words and numbers. I am obviously not a mathematician, nor claim to be one. I'm obviously not as invested in this puzzle as you two are. So you thinking I am being emotional about this is comedic. Ty, i needed the chuckle.
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    Quote Originally Posted by anndelise View Post
    Wow, the purpose of you editing my post to quote me as having made 4 angry emoticons was for?? Do you really think I'm emotional over something like this? Lol.

    I stated at the beginning that the puzzle had never made sense to me, and at least twice I stated that it feels like people manipulating words and numbers. I am obviously not a mathematician, nor claim to be one. I'm obviously not as invested in this puzzle as you two are. So you thinking I am being emotional about this is comedic. Ty, i needed the chuckle.
    Okay, come back and think about the problem again when you've calmed down.

    Otherwise, act snidey and people will point it out.

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    Quote Originally Posted by InvisibleJim View Post
    Okay, come back and think about the problem again when you've calmed down.

    Otherwise, act snidey and people will point it out.
    What you believe, and what's real, aren't necessarily the same thing.
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    Quote Originally Posted by InvisibleJim View Post
    Consider the hypothesis of increasing the number of doors.

    Instead of 3 doors, consider 10,000 doors.

    You open 1 door, the host opens 9,998 doors and shows goats.

    Would you switch and why?
    That's a great idea.

    Imagine there was a frequency to which these doors were revealed and a frequency to which one could change there guess and they may not ressonate. Modifying too much may disadvantage you, modifying at the right frequency would benefit you the most, and modifying at a slower frequency would benefit but diminish the further from ressonance. Thus one would want to guess over about 3 or 6 sigma depending on if you like being 99% assured or if you like corporate buzzwords.

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    The explanation has always made sense to me, I suppose, but it seems more to be mathematical manipulation and probability semantics to me than anything 'real'. I guess I still don't get it.

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    it makes sense to me with the pickshers

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    Quote Originally Posted by lungs View Post
    it makes sense to me with the pickshers
    Same, I was really confused at first about how this would work. That was cool to think about though.

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    Quote Originally Posted by Shang Tsung View Post
    That's a great idea.

    Imagine there was a frequency to which these doors were revealed and a frequency to which one could change there guess and they may not ressonate. Modifying too much may disadvantage you, modifying at the right frequency would benefit you the most, and modifying at a slower frequency would benefit but diminish the further from ressonance. Thus one would want to guess over about 3 or 6 sigma depending on if you like being 99% assured or if you like corporate buzzwords.
    Here you go:


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    THE FOOL WHO DECIDES ONCE

    - has a 1 in three probability of choosing the prize
    - has a 1 in three probability of choosing one of the two decoys
    - has a 1 in three probability of choosing the other decoy

    Thus his chance of success is 1/3

    THE WISE MAN WHO MODIFIES HIS DECISION

    - has a one in three probability he chooses the correct door, upon switching this would mean failure
    - has a one in three probability of choosing the wrong door, upon switching this would mean success
    - has a one in three probability of choosing the wrong door, upon switching this would mean success

    thus he has 2/3 chance of success, which is greater than the 1/3 chance

    see its all about the narrowing of the options, whenever there is a smaller sample space decide, the odds favor you as the sample space approaches 1/inf.

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    Quote Originally Posted by InvisibleJim View Post
    Here you go:

    ...
    makes perfect sense, as you approach an infinite number of doors and modified decisions you always are correct.

    But this is assuming the decision is modified in perfect resonance with the doors opened, as would occur on a game show, but if instead the decision making were a separate mechanism from the game show host opening doors, then you could have someone modifying there decision something like twice for every time the game show host opens a door and narrows the sample space.

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    Quote Originally Posted by InvisibleJim View Post
    Opinions on this paradox? What does it tell you about the human psyche?
    It's not a paradox. The host introduces a subtle bias into the game. The human psyche... Maybe that true answers lie in the unknown areas.

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    No matter what I think, choose, say, or do; short of picking up the cars and goats myself and moving the fuckers myself, none of my thoughts are gonna change what actually exists behind those doors. My first choice was strong enough to send me through one round. I stand by my decisions.

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    Quote Originally Posted by woofwoofl View Post
    No matter what I think, choose, say, or do; short of picking up the cars and goats myself and moving the fuckers myself, none of my thoughts are gonna change what actually exists behind those doors. My first choice was strong enough to send me through one round. I stand by my decisions.
    You do know that choosing any door would get you past the first round, right?
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    Probability 101.

    Anyway:



    P.s. as InivisbleJim correctly argued and showed, the 2/3 frequency of winning (and consequently the value of 1 on the cdf which Jim shows) will only be approached when the game is played a decent number of times.

    none of my thoughts are gonna change what actually exists behind those doors. My first choice was strong enough to send me through one round. I stand by my decisions.
    But they will change the behavior of the host, since if you change and you had a goat behind the initially chosen door, he will be forced to show you the goat.

    makes perfect sense, as you approach an infinite number of doors and modified decisions you always are correct.
    Not exactly, he is showing the cumulative density function, so what it really means is that if an increasing number of people plays the game and switches, there will very likely be at least one winner after 5-6 games. If he's doing something else, I don't fully get it but the mathematical result is equivalent.
    Last edited by FDG; 02-05-2013 at 02:50 PM.
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    Quote Originally Posted by JWC3 View Post
    You do know that choosing any door would get you past the first round, right?
    and why on earth would that matter? I got a full head of steam here, and I'm gonna carry it through to the end. And after that first round, it's a one in two shot either way. Keeping the door or changing the door, at that stage it's a one in two shot. In the former case, I get to keep my convictions and dignity and go down with the ship with pride in the worst of worlds. In the latter case, I fucking squirmed and weaseled my way out of a car like some maggot. Much, much better to live with myself for keeping the decision and holding my ground.

    Quote Originally Posted by FDG View Post
    But they will change the behavior of the host, since if you change and you had a goat behind the initially chosen door, he will be forced to show you the goat.

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    Quote Originally Posted by woofwoofl View Post
    and why on earth would that matter? I got a full head of steam here, and I'm gonna carry it through to the end. And after that first round, it's a one in two shot either way. Keeping the door or changing the door, at that stage it's a one in two shot. In the former case, I get to keep my convictions and dignity and go down with the ship with pride in the worst of worlds. In the latter case, I fucking squirmed and weaseled my way out of a car like some maggot. Much, much better to live with myself for keeping the decision and holding my ground.



    Well you see that's the thing, the problem is designed to make you think the odds are 50/50 in the second round. It's actually 66.6% the door you can switch to, and only 33.3% the door you can keep. The odds actually aren't one in two, since the host informs the contestants second choice by altering the choices available. It would only be one in two if you had absolutely no information to go on, but that isn't the case here.
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    Quote Originally Posted by InvisibleJim View Post
    Consider the hypothesis of increasing the number of doors.

    Instead of 3 doors, consider 10,000 doors.

    You open 1 door, the host opens 9,998 doors and shows goats.

    Would you switch and why?
    After meditating a good deal on this, it is finally starting to make some sense. Although I still think the probability part is little more than mathematical manipulation, let me try to explain my current understanding:

    If you have 10000 doors, chances are very high that your guess is incorrect. If the host than opens 9998 doors... as far as I'm concerned, there still shouldn't be any reason the other door should have a higher 'probability' of not containing a goat, but it does, because the host let that door stay. He wouldn't throw out the door with the car behind it, and since your door very likely has a goat, it is therefore 'better' to switch. It seems to be a game of intentions rather than a game of math.

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    Oooooh, Gamma probabilities.

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    Quote Originally Posted by nil View Post
    After meditating a good deal on this, it is finally starting to make some sense. Although I still think the probability part is little more than mathematical manipulation, let me try to explain my current understanding:

    If you have 10000 doors, chances are very high that your guess is incorrect. If the host than opens 9998 doors... as far as I'm concerned, there still shouldn't be any reason the other door should have a higher 'probability' of not containing a goat, but it does, because the host let that door stay. He wouldn't throw out the door with the car behind it, and since your door very likely has a goat, it is therefore 'better' to switch. It seems to be a game of intentions rather than a game of math.
    It absolutely is.

    Interestingly a lot of scientific studies have been done that show the critical 'number of doors' when people start to want to switch rather than stick is 7. Perhaps they feel less attached to their original choice when it is made entirely clear that the doors are random and any choice leading to a door selection is nonsense?

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    Mythbusters did an episode on this principle, but wasted everynody's time by conducting a multi-manned operation and elaborate set of experiments on something that can be figured out using basic algebra skills.
    "And above all, watch with glittering eyes the whole world around you because the greatest secrets are always hidden in the most unlikely places. Those who don't believe in magic will never find it." -Roald Dahl

    http://forum.socionix.com/
    It's pretty cool

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    Quote Originally Posted by nil View Post
    After meditating a good deal on this, it is finally starting to make some sense. Although I still think the probability part is little more than mathematical manipulation, let me try to explain my current understanding:

    If you have 10000 doors, chances are very high that your guess is incorrect. If the host than opens 9998 doors... as far as I'm concerned, there still shouldn't be any reason the other door should have a higher 'probability' of not containing a goat, but it does, because the host let that door stay. He wouldn't throw out the door with the car behind it, and since your door very likely has a goat, it is therefore 'better' to switch. It seems to be a game of intentions rather than a game of math.
    Yeah, that's what I factored in too, but in the other way. That host is getting paid by the show, which loses money if the car goes. It's most likely to be in the host's best interests to try to steer you off course like that. If you got the goat, why would the host open up the other door in the first place, let alone ask the second question?

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    Not exactly, he is showing the cumulative density function, so what it really means is that if an increasing number of people plays the game and switches, there will very likely be at least one winner after 5-6 games. If he's doing something else, I don't fully get it but the mathematical result is equivalent.
    I thought he was showing the probability of winning the game versus the number of doors. What you are talking about involves comparing the probability of winning against the number of games.

    Consider the hypothesis of increasing the number of doors.

    Instead of 3 doors, consider 10,000 doors.

    You open 1 door, the host opens 9,998 doors and shows goats.

    Would you switch and why?
    That's what he is showing as I understood it, though maybe its a cdf, go ask jim.

    See if there are 10,000 doors there is only 1/10,000 chance of choosing correctly the first time and 9,999/10,000 chance you choose wrong. If you choose wrong, when the host opens up 9,998 door, switching will win the game. Thus the equation is n-1/n for success and 1/n for failure as the number of doors n approaches infinity... then the chance of failure approaches 0. It's completely correct. However in reality you'd never have such a case of an infinite number of doors. Whatever the doors represent can at best be considered to be infinite in comparison to something else by the law of large numbers.

    Basically I'm not wrong and think your criticism is unfounded. Cdf's are relationships of probability against a random variable, the number of doors isn't a random variable in the game but rather a different configuration to the game all together.

    Also you can write

    n-1/n

    as

    1/n+(n-1)/n * n-2/n-1 = 1/n+n-2/n = n-1/n

    so analysis in which you use a calculation such as 1/3+(2/3)(1/2) or 1/4+(3/4)(2/3) are just as valid logically --> which is strangely how intuitively I would calculate it out, although its less straightforward admittedly than the n-1/n formula.

    I don't really see where I have made an error.
    Last edited by male; 02-05-2013 at 05:58 PM.

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    I wonder what would happen if instead of the host revealing a goat, if the host was choosing randomly, there were a large number of doors and the guest could choose to change after the host revealed a door.

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    Quote Originally Posted by Vois View Post
    Well, I already have a car, so I'd pick the goat. Problem solved.

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    Quote Originally Posted by Shang Tsung View Post
    I wonder what would happen if instead of the host revealing a goat, if the host was choosing randomly, there were a large number of doors and the guest could choose to change after the host revealed a door.
    I imagine (without a full analysis) that there would be no need to change because you would be as likely to have the door with the car regardless of how often you switched as would the host.

    Say 10 doors
    Pick door 1, 1/10 chance of car (at random) Host opens 10, 1/10 chance of car (at random)
    Game effectively resets because the host had no insider knowledge of what was behind the doors
    Pick/Hold door 1, 1/9 chance of car (at random) Host opens 9, 1/9 chance of car (at random)
    Game effectively resets because the host had no insider knowledge of what was behind the doors
    Pick/Hold door 1, 1/8 chance of car (at random) Host opens 8, 1/8 chance of car (at random)

    Either way, I'd be very annoyed in a game where someone opens all the other doors at random and you never get to see whats behind the door you pick unless they open it. In the case above they have a 90% probability of success, whereas you have only 10% regardless of how often you switch doors.

    This is most often how people envision the monty hall problem unless they realise the whole problem is rigged such that the host won't open the door with the car. They envision that the problem resets and there is a 50/50 chance which is a base rate fallacy. This is of course driven by decision inertia, if you make the decision once you 'own' the decision and don't want to appear wrong, even when the reason for making the initial decision are just random anyway and actually have no bearing on anyones capability to make it, whereas the reason for switching doors is an actual mathematical one.

  39. #39
    Creepy-male

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    Yea I understand that... the host actually has insider knowledge, and if you choose wrong (a goat) the first time he will always open up the door with the goat rather than the car. Therefore assuming you were wrong the first time (a goat), you are now presented with a smaller sample set of options (doors) from which only a single has the car. In this case there is only one remaining option, so you are always correct. But even if there were more options, say n, the host opening up a door with goat has narrowed down the available options to n-1, and n-i on a subsequent number of decisions i. Since n-i is always less than n, you are always favored to switch if the host is using there insider knowledge to tip the odds in your favor (by showing you i wrong answers). But say the host wasn't a host, but an opponent, also trying to get the car? How then could you tip the odds in your favor. That's essentially what I'm asking =p.

    http://en.wikipedia.org/wiki/Monty_H...host_behaviors

    "Monty from Hell": The host offers the option to switch only when the player's initial choice is the winning door. (Tierney 1991) Switching always yields a goat.
    http://en.wikipedia.org/wiki/Prisoner%27s_dilemma >> ARE THE POLICE EVER "ANGELIC MONTY"?
    Last edited by male; 02-05-2013 at 07:17 PM.

  40. #40

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    Quote Originally Posted by woofwoofl View Post
    Yeah, that's what I factored in too, but in the other way. That host is getting paid by the show, which loses money if the car goes. It's most likely to be in the host's best interests to try to steer you off course like that. If you got the goat, why would the host open up the other door in the first place, let alone ask the second question?
    It seems to me that any bias the host may have is irrelevant, as it doesn't seem that he is trying to steer the contestant off course, but instead merely offer him the choice to switch. I think it shouldn't make any difference if the host asks you if you would like to switch whether you originally chose the door with the car or not.

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