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Why the Monty Hall puzzle is categorically 1/2 and not 2/3rds

  1. Oct 20, 2012 #1
    So I was browsing the Wikipedia article concerning The Monty Hall Paradox, and I seem to take great issue with the assumption that switching results in a 2/3 probability of winning a car.
    (hold on pressed enter by mistake... editing now watch this space)

    My reasons are as follows (and I don't believe this has anything to do with intuition bias)

    Conditions and assumptions:

    * This puzzle uses the word "say" before No.1 and No3... Therefore, it could be equally valid that contestant choose 2, or 3 initially
    * Monty Hall must always choose a goat door. If guessed correctly, this would either be 2 of the goat doors (irrespective of number) decided by an a-priori coin toss before the show and not chosen out of the whim of the presenter. Alternatively, the presenter knows what door has the goat and always opens the door with the goat. Therefore the switch door and the door Monty opens can be in variable order.
    * Ive never watched this show, so I'm going to ignore all subjective behavioural characteristics (e.g. no right/left door bias)
    * the presenter must always choose a door with a goat, and always offer a choice to switch or not.
    * The contestant has no idea what's behind the door.

    The original puzzle:
    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?

    It is claimed that switching results in a probability of 2/3rds, but it has to be 1/2, factoring all those conditions outlined above.

    I didn't study Maths or Physics, but this makes perfect logical sense to me.
    Last edited: Oct 20, 2012
  2. jcsd
  3. Oct 20, 2012 #2


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  4. Oct 20, 2012 #3


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    After Monty opens the door, he gives you the option to stay or switch. The probability of the stay strategy being correct is equal to the probability you were correct to begin with. That probability is of course 1/3, because you chose that door when there were three options. So the switch strategy is 2/3. QED.
  5. Oct 20, 2012 #4
    Thanks Borek for that link I'm reading those threads to get my head round this...

    This is where I'm having trouble grasping. It is true at the start your odds are always 1/3 of winning (before the door is opened and selected at random).

    I accept that the odds change when the door is opened...but

    (this is how my mind is processing the problem)
    1) If Monty has apriori knowledge and always chooses one of the remaining goat doors (100% behaviour). In my head, this leaves 2 doors... with 1 goat, 1 car... Always

    Door 1: It could be a goat, it could be a car.
    Door 2: It could be a goat, it could be a car.

    It no longer becomes 2 goats and 1 car, but 1 car and 1 goat.
    I'm going to substitute goat and car with head and tail.

    You toss two coins: it could be heads, it could be tails
    coin 2: it could be heads, it could be a tails.

    Each time you toss a coin, the odds are always 50:50 for the subsequent coin toss to match the correct guess irrespective if it is heads/head, tails/tail, heads/tail, tail/heads etc.

    The decision to switch from the playing perspective is like saying "heads or tail?" before the coin is actually tossed.

    Am I misunderstanding the problem?
  6. Oct 20, 2012 #5
    I found a comment from DaveC426913 which matches my conclusion... I have also included another quote from another poster which concludes with 2/3rds probability and I've highlighting the part that I'm having trouble following.

    DaveC426913 wrote:
    Mister X: 2/3 hypothesis

    (I'm not a maths wizard, but doesn't the terms of the original puzzle change to match the new circumstances?)
  7. Oct 20, 2012 #6
    Consider this: you have 1000 doors, one of which has a car. After you pick a door, the quizmaster opens 998 doors that do not contain the car. You then get the chance of switching. Do you switch in this scenario?
  8. Oct 20, 2012 #7
    Well it's a different puzzle, but.

    1000-998 = 2

    2 doors.

    1 car for 1 door, 1 goat for another.

    For the remaining 998 doors if posed this question than this 50% chance assumption is incorrect. When it gets to two possible doors left than it is squarely 50%.

    Is this a problem of linguistic semantics and maths (akin to a riddle?). I just don't know why 2/3's assumption of one of the two doors come in... this is like saying the mean average number of children born per couple in the world is 2.33, but this is does not equate to 2 babies, and a baby who is only a 1/3 born?

    2 doors left... 50% chosen door is car
    3 doors left... 33.33%... chosen door is car
    4 doors left... 25% chosen door is car
    5 doors left 20% chosen door is car

    in all 2+ door instances, surely switching presents the same percentage odds? if I switch door if there were 5 left, then there is still 20% chance that the door I switched too is correct. No change?

    Bit confused.
    Last edited: Oct 20, 2012
  9. Oct 20, 2012 #8
    Ahh...I sort of get it now but i'm debating whether it will work in a social science context as opposed to a mathematical concept.

    I realised if the rules are extended to multiple doors, and the doors chosen by Monty are duds than this can lead a 2/3 interpretation.


    I'm wondering (purely from a Pyschologist perspective) If the Monty problem would be congruent with a maths (theory) interpretation.

    Assuming 2 rooms...Each room had one participant. Both are blind to each other and no information can be leaked to another (sound proof).

    The psychologist is the arbiter... Participant A Selects one of these doors. There are two left.

    Participant B is aware of Participant's choice, and must either a) choose the remaining goat or B) select from a possible door to challenge Person A.

    Person A is than relayed this information of Person's B choice through a binary receipt inputted by the psychologist. Person A has to make a choice to stay or switch.

    I seriously wonder, given the fact that confounding variables are removed and this set up is blind and cannot be leaked. Whether this does actually does lead to the same conclusion, with 100 randomly assigned participants running 50 different tests.

    Surely that would be the decider to determine if the internal maths system theory is congruent with happenings observed in the external world.
  10. Oct 20, 2012 #9


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    I'm a bit confused as to why you did not ask why the answer is 2/3, originally, rather than asserting that it is 1/2.
  11. Oct 20, 2012 #10
    The answer really is 2/3 instead of 1/2. There are multiple ways to see this fact. If the math doesn't convince you, then maybe a practical experiment will. Do the experiment a large number of time and see how many times you get the right answer by switching the doors.
  12. Oct 20, 2012 #11
    I'm not quite 100% convinced it is 2/3rds. In my head, still debating whether it is a logic trap, akin to St Anselm "Ontological Proof of God" which 'logically' holds but empirically is shaky and can be challenged (from an atheistic/scientific/realist perspective from acquiring evidence outside of the original argument). I know that's a slight jump from the original math puzzle posed but I am now going to look online for evidence from psychology a-posteriori tests with real participants to rest my mind.

    Maybe I'm experiencing cognitive dissonance but I'm questioning whether it is a systemic flaw in mathematical modelling or a flaw in my understanding of the original problem. Even a weather man, with a team of meteorologists behind him could be wrong sometimes, and not all models of ecology are predictive of what actually occurs (e.g. humans are not on top of the food chain as they still can be eaten by sharks etc)
  13. Oct 20, 2012 #12


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    Dearly Missed

    Remember that MONTY's opening of doors is basically non-random, due to his knowledge of where the car is.
    Allowing for his ignorance as well, not just yours (which would add the possibility that he opened the door with the car, giving you a guaranteed loss), switching from your part would never increase your odds.
  14. Oct 20, 2012 #13


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    I think you are getting sidetracked by talking about modelling here. There is nothing to model here - or, if you prefer, Monty's model of the situation is perfect, because he knows all the relevant facts. And your model of the situation needs to include the fact that Monty is omniscient. The apparent paradox only arises if you ignore Monty's omniscience, and the game situation is set up to disguise that fact - especially if you only play once.
  15. Oct 20, 2012 #14
    Think of a stack of all the tickets for a lottery - one is the winner - you choose one.
    Then, except for one all the remaining non-winning tickets are destroyed, so either yours or the one remaining is a winner.
    Which ticket do you think is the winner?

    Go back to the 1000 doors scenario...

    1000 doors... you choose one... what are your chances of picking the one with the car?

    998 non-car doors are opened.

    There are now just two doors, your original choice, and the other one.

    The chance that you originally selected the car with your first choice was a thousand to one.
    Now, you know for certain that the car is behind either your first choice or the remaining door... remind yourself how little chance your first choice was of being the car. Think about it... how incredibly lucky you would have had to have been to have chosen correctly the first time.
    Now you have a choice between two doors, the one you chose out of a thousand, an the other one...

    Substitute any large number for the original number of doors until you see it...
    A million doors, a billion doors, a googleplex of doors... your chance of selecting the right one on your first choice approaches infinitesimal.
  16. Oct 20, 2012 #15


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    Fwiw, here's where DaveC's logic goes wrong.
    He lists four cases and assumes they're equally likely, but they're not. There are three equally likely possibilities: you chose A, B or C. The C case he then splits into two possibilities, but the probabilities are now:
    A 1/3
    B 1/3
    C1 1/6
    C2 1/6
    So the odds of winning by switching are 2/3 (A+B).
  17. Oct 20, 2012 #16


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    Why are you thinking about psych? You should be looking for computer simulations of the problem.

    It can't be case one because the problem is purely mathematical to begin with. It is its own mathematical model.

    But he is statistically correct, which is what you want.

    You are thinking of a food web. A linear sequence along the web from basal to consumer is a food chain. Ecologists are not interested in food chains per se, they are interested in the average over all possible chains in a web. This is called the food chain length.
  18. Oct 20, 2012 #17


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    That doesn't respond to my question. If you are NOT "100% convinced it is 2/3rds" why did you start by asserting that it is 1/2? Those are quite different statements.
  19. Oct 20, 2012 #18
    This is how I think about it:

    The only outcome where it would be beneficial to you to not switch is when you guess correctly with your initial choice. As you know this happens 1/3 of the time. This would mean that the other 2/3 of the time you will get it correct if you switch.

    In other words, every time you do not choose the correct door on your first try, switching is guaranteed to give you the correct door since you are currently on one wrong door and the other wrong door has been eliminated. You do not choose the correct door on your first try 2/3 of the time, therefor 2/3 of the time when you switch you will end up on the correct door.
  20. Oct 20, 2012 #19


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    This is incorrect.
    If you happen to be on a show with an IGNORANT Monty Hall, who happens to open a door with a goat, then it is irrelevant if you switch or not.
    The critical issue is that a rule of the game is that you can rest assured that Monty Hall WILL open a door for the second chance (rather than revealing your first picked goat), and that he ALWAYS will reveal a goat after your first pick.
    Last edited: Oct 20, 2012
  21. Oct 20, 2012 #20


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    You choose 1 door, so have a 1/3 chance of making the right pick. That means the odds of the car being behind one of the remaining 2 doors is 2/3. That should be clear and obvious, right?

    Why does the host opening one of those doors change the odds?

    The host shows you which one is NOT the car, there is a 2/3 chance of winning the car if you switch, 1/3 if you don't.
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