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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 04:19 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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    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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    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 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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    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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    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
    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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    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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    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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    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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    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 02:32 AM. Reason: correction

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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 03:06 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
    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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    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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    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
    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 01:50 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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    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 04:58 PM.

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    Oooooh, Gamma probabilities.

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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.

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    Also lets move on from the Monty Hall Problem to the Birthday Paradox. That'd be far more amusing.

    http://en.wikipedia.org/wiki/Birthday_problem
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    Quote Originally Posted by JWC3 View Post
    Also lets move on from the Monty Hall Problem to the Birthday Paradox. That'd be far more amusing.

    http://en.wikipedia.org/wiki/Birthday_problem
    I've met a few people w my bday, or people who know people w it.
    I had attributed that to having such an 'odd' bday that it gets remarked on out loud, and someone overhearing (or over reading) it and commenting on it. Same thing as my uncle's April Fools day bday, my friend's bday that sometimes shares easter day, my friend's thanksgiving bday, and my niece's christmas eve bday. These kinds of things get talked about more than an otherwise un'connected' bday. So I think that helps to contribute to the perception of how easy/hard it might be to find someone else w the same bday.
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    Quote Originally Posted by anndelise View Post
    I've met a few people w my bday, or people who know people w it.
    I had attributed that to having such an 'odd' bday that it gets remarked on out loud, and someone overhearing (or over reading) it and commenting on it. Same thing as my uncle's April Fools day bday, my friend's bday that sometimes shares easter day, my friend's thanksgiving bday, and my niece's christmas eve bday. These kinds of things get talked about more than an otherwise un'connected' bday. So I think that helps to contribute to the perception of how easy/hard it might be to find someone else w the same bday.
    The biggest perception error people make is generally assuming unequal likelihood for all possible birth days. Either way, the problem is more concerned with probability rather than an individual's awareness of others who share the same birthday. So perception aside, put any random 57 people in a room and it's 99% probable that a pair will share the same birthday.
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    Quote Originally Posted by JWC3 View Post
    The biggest perception error people make is generally assuming unequal likelihood for all possible birth days. Either way, the problem is more concerned with probability rather than an individual's awareness of others who share the same birthday. So perception aside, put any random 57 people in a room and it's 99% probable that a pair will share the same birthday.
    Heh, yeah, i'm obviously more interested in the perception side than the math, heh.
    I'll have to go back and reread the link, but i'd probably find it difficult to see feb 29 as having the same probability as the other days. But the link might have included that in their calculation.

    Edited to add: dur, it's in the first paragraph.
    That's an interesting probability to me. Cool.
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    Before you read my post further, please recognize that i already know my thoughts on this are mathematically wrong, I am merely describing how my mind thinks of the puzzle.

    I don't see the first choices as being meaningful choices. They are like a magician's slight of hand. Because no matter what door you choose, no matter what is behind the door you chose, the host will reveal a goat. In my mind, it's a fake choice that you've been given. The player is just being tricked into thinking that it's a meaningful choice.

    The meaningful choice comes when you are making a choice that ends in a result, one way or another (or another, etc). If the final result was revealed after making a choice between 3 doors, then you had a 1 in 3 chance of being correct. If it's between two doors, it's a 1 in 2 chance.

    I also recognize that I have problems w how it is mathematically approached. (remember, i also dont claim to be a mathematician.) All the scenarios given and the pictures given seem to me to cut out half of the equation. They start w the constraint that the car is behind door number 2, and the choice is door number 1. But I keep seeing that as a false constraint.
    What about if the car is behind door number 2, and the original choice was door number 2?
    Or to make the full possible scenarios easier to read:
    a) 1= car, 2= goat; choice= 1
    b) 1= car, 2= goat; choice= 2
    c) 1= goat, 2= car; choice= 1
    d) 1= goat, 2= car; choice= 2

    Which could be simplified to:
    a) meaningful choice = true
    b) meaningful choice = false

    ------
    Having said that, @Galen mentioned mythbusters did a show on it, perhaps I'll try to find that episode later so that I can see it actually in action and maybe finally get to see it as real, rather than a manipulation of numbers w false constraints.
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    Quote Originally Posted by anndelise View Post
    Before you read my post further, please recognize that i already know my thoughts on this are mathematically wrong, I am merely describing how my mind thinks of the puzzle.
    It might help to think of it this way:

    Each door has a 1/3 chance of having the car. Three doors, one car, that much is easy to see. When you choose a door, that door has a 1/3 chance of having the car. The other 2 doors together have a 2/3 chance of having the car. Two doors = 2/3 chances. Separate in your mind your door from the other two doors. See it as your door = 1/3, other 2 TOGETHER = 2/3.

    Keep that in mind. Your door = 1/3, the other 2 = 2/3. The host now opens one of the doors, and there's no car there. NOW, that one remaining door has a 2/3 chance of having the car, because the two doors together had a 2/3 chance, and you've just eliminated one of them, so the remaining one gets to hold the entire 2/3rds for itself. Your door was always one door, it always had a 1/3 chance.

    Think of it as my door vs not-my door. My door always = 1/3 chance. Not my door always = 2/3 chance.

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    Quote Originally Posted by squark View Post
    It might help to think of it this way:

    Each door has a 1/3 chance of having the car. Three doors, one car, that much is easy to see. When you choose a door, that door has a 1/3 chance of having the car. The other 2 doors together have a 2/3 chance of having the car. Two doors = 2/3 chances. Separate in your mind your door from the other two doors. See it as your door = 1/3, other 2 TOGETHER = 2/3.

    Keep that in mind. Your door = 1/3, the other 2 = 2/3. The host now opens one of the doors, and there's no car there. NOW, that one remaining door has a 2/3 chance of having the car, because the two doors together had a 2/3 chance, and you've just eliminated one of them, so the remaining one gets to hold the entire 2/3rds for itself. Your door was always one door, it always had a 1/3 chance.

    Think of it as my door vs not-my door. My door always = 1/3 chance. Not my door always = 2/3 chance.
    Ty, that helped me get a glimpse of what ya'll see. At least now I can move back and forth between the two views on it a bit easier, rather than an "I see" vs "They say" thing.

    Though if push came to shove, I'd still approach the puzzle as the last choice being the only meaningful choice, and thus the only one I'd pay attention to. (but then, it's a choice w/o info to base that choice off of, so it'd still be little more than a flip of the coin in my mind as i make that last choice, lol. Suffice to say, I'd likely not win a game show, )
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