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I Monty Hall Problem -- Questions

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  1. Jan 28, 2017 #1
    Monty hall theory(extra).



    Hi,

    Im not really a math guy here but more of a logical thinker and this question really confused me because of the way people explain it.
    Namely i do agree with the fact that taking a wrong door away after the contestant has chosen initially, rises the chance that the contestant will pick the right door by switching.
    However: people say there is a one third chance initially and a two third chance after. Maybe I am digging in to deep but let me explain with my next example.

    Me and two of my friends John and Mike used to play a lot of cards.
    We would play a simple game which only took 3 cards from the deck. The 1 clubs. the 2 clubs and the 3 clubs.
    We played as followed: John would shuffle the 3 cards and then would deal each of us a random card from the top of the deck. The one who got the highest card would win (It was a really exciting game).

    Now let me ask this: what are the odds each game Mike will get the highest card?
    Well if you ask me it isnt 1 in 3.

    Which one will be getting the highest card is really a matter of how the deck is getting shuffled. If John shuffles in a way which makes the 3 Clubs go on the top alot time and he deals to Mike first then it wouldn't be 1 in 3 would it?

    What im saying is that it really depends on the coincidence of the way the card is getting shuffled and therefore you can never predict (no matter the amount of games) how often (not even approximately) you will get the highest card.

    Now what if there would be 2 “3 clubs” and 1 “1 clubs” does that mean there now is a 2/3 of a chance of getting the high card? The way I look at it the chance has indeed increased but the ratio still isnt 2/3.

    The only way of saying what the ratio is of the highcard falling is is by testing each case uniquely. And then afterwards you could say what the ratio is of the amount of times you try d in that case. and not before hand where coincidence and random human decision making has way to much influence on the outcome.

    Its the same thing with coinflip.

    The ratio the coin is falling heads or tails depends on the way the person throws the coin.

    If the person throws the coin in a similar way every time, it will also fall on the same side everytime.

    In case of the Monty Hall scenario the the chance of knowing the ratio to the winning door is impossible to know because it depends on the random choice of the contestant and the random doors the gameshow put the goats at.

    Still i agree that if you switch you will win more often, but that is only because randomness will always try and go toward switching between 3 possible options(from which 2 in this case are the right ones).

    However there is no way to know for certain where the 2 goats will switch and which door the contestant will choose.

    Makes sense right?
     
  2. jcsd
  3. Jan 28, 2017 #2
    In the Monty Hall problem, it is assumed that everything is done completely randomly. Not 100% true to reality because as you said, the goats could be shuffled around in a non-random manner. However, a bias of the contestant cannot affect the game provided that the goats are placed in a completely random manner. By definition, a completely random manner is one such that the probability of choosing a goat will be 2/3, and the probability of getting a car will be a third. Unless the contestant has x-ray vision or something, they cannot have enough information to affect their choice in any significant way. Even if they have a bias, they can't actually
    be biased in such a way so as to significantly impact the game. Of course, when offered to swap their bias will affect their chances of winning, but there is a difference to be made. In the case of a bias, the contestant's chances of swapping doors is affected but the probability that a car lies behind the other door is something that depends only on how the goats are shuffled, which is random by assumption.

    The reason the car's winning chance doubles post-swap is because you initially have only a third chance to pick the car. So it is more likely that the door you chose has a goat behind. When Hall opens another door to reveal a goat, it becomes more likely that the other door contains the car. There still is a 2/3rds chance that the door you chose has a goat behind because once you choose, you can't go back in the past. But the chance of winning a car must have doubled by opening that door. Thus you get a 2/3rds chance to win the car.
     
  4. Jan 28, 2017 #3

    mfb

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    The car has equal chance to be placed behind each door. This is one of the requirements of the Monty Hall problem.

    It does not matter (=there is no good or bad strategy) which door you choose initially - the car will be behind this door with 1/3 probability.
     
  5. Jan 28, 2017 #4

    PeroK

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    No. The way your friend shuffles the cards has no bearing on the Monty Hall problem. Have you ever tried tossing a coin so that it lands Heads every time?
     
  6. Jan 28, 2017 #5

    Thanks for the replies,

    I understand that the bias of the person doesnt affect the game itself, but it should affect the chance of picking the right door. assuming the choice of the contestant isnt completly random.
    So i think Monty hall is a bad example for the equation and might be why so many people get confused.
    Randomly means it has a chance to be behind everydoor but it does not say how often.
    If something is realisticly random, the car might as well be behind door number 3 90% of the time.
    But the more something gets randomized the bigger the chance the ratio will be more even. And in most cases it will be shuffled in such a way the outcome will be close to perfectly random. In most games of chance you would be more likely to end up with a ratio of for example: door1: 30% of the time a car, door2: 30% of the time a car, door3: 40% of the time a car.
    And that is why it seems 1/3 of the time.

    If the contestant now makes a random choice on top of the the random place the car is.
    Which could be depending on their current mood for example: 50% on door 3. It would be a random choice on top of a random winning door.
    In that case the contestants biased choice would have effect right?

    btw i do agree that your chances will get higher after he shows 1 goat.
    But i dont see how it would necessarily double the chance of 3 on 10 when it might as well be a chance of 4 on 10 created by the random biaz of the contestant to begin with. which then wouldnt be doubled but increased to 6 out of 10.
     
    Last edited: Jan 28, 2017
  7. Jan 28, 2017 #6

    PeroK

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    Monty Hall is a good example of probability theory because most people get confused!

    If you watched the show and came to the conclusion that the car was behind door #3 40% of the time, what would be your strategy?
     
  8. Jan 28, 2017 #7
    That is true, ill take that part back.
    But they way it is often explained is with an example of 1/3 which isnt realistic and can be confusing at first.
     
    Last edited: Jan 28, 2017
  9. Jan 28, 2017 #8

    mfb

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    It is realistic. Getting a uniform distribution for the car position is trivial. Making it non-uniform adds unnecessary complexity to th problem.
     
  10. Jan 28, 2017 #9

    pwsnafu

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    And then Michael Larson will play the game...
     
  11. Jan 28, 2017 #10

    Stephen Tashi

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    If you are saying that the probability of a repeated event is different from the actual frequency with which it happens, you are correct. Probability theory deals with probability. There is no mathematical connection between probability theory and observed frequency except that probability theory can tell you the probability of a given frequency actually being observed.

    However, the analysis of the Monty Hall problem is an analysis of probability. If people are confused by the fact that observed frequencies are different from probabilities, they are confused about any situation that involves analyzing probabilities.

    A connection between probabilities and actual frequencies is often made by people who apply probability to real world situations. They may choose to assume that actual frequencies will be equal to probabilities. However, that is not guaranteed by the mathematical theory of probability. It must be justified by the particulars of the real world problem and/or the belief that having "average luck" is guaranteed.
     
  12. Jan 29, 2017 #11
    Oh ok, well im glad we got that straight.
     
  13. Jan 29, 2017 #12

    FactChecker

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    We have been naturally casual in terminology. If we want to be formal, we should have said "uniformly random", but that has been assumed throughout. If you are being really formal, "randomly" without "uniformly" does not say there is a chance to be behind every door. It just says that there is a random process, which may always exclude a specific door, for picking the door for the car.
     
  14. Jun 14, 2017 #13
    Sorry to resurrect, but this is a pet peeve, and almost never addressed.

    The Monty Hall problem is indeed a wonderful example of how Conditional Probability confuses people. And the reason it continues to be controversial, is because the usual explanation of the correct answer, repeatable, doesn't use conditional probability. So although it gets the right answer of 2/3, it doesn't actually explain why the other answer is wrong, and as a solution is technically less correct than the solution that produces that answer.

    After the contestant picks a door - and without loss of generality we can call it #1 - there are four possible ensuing outcomes:
    1) The car is behind #3, and the host opens #2.
    2) The car is behind #2, and the host opens #3.
    3A) The car is behind #1, and the host opens #2.
    3B) The car is behind #1, and the host opens #3.

    The probabilities of these are 1/3, 1/3, 1/6, and 1/6, respectively.

    Most people actually use a naive application of conditional probability. If door #3 is opened, they know case #1 is ruled out, Reasoning that cases #2 and #3 (combined) must still be equally likely, they update the 1/3 probabilities to 1/2. This is a correct approach, with one incorrect interpretation that I'll get to soon.

    The above solution does not use conditional probability at all; it says the probability of case #3 can't possibly change, which is untrue (again, I'll get there soon), so whatever is left must be 2/3.

    The error of interpretation is how to apply the information we have as a conditional probability. If door #3 is opened, case #3A is also ruled out. Case #2 is now twice as likely as what is left of case #3. The same approach taken by the naive solution now gives the right answer.

    The solution I quoted is technically incorrect, because it implies that the probability of case #3 is fixed, regardless of any information gained. This is true only if cases #3A and #3B are the same. That turns out to be so, but the implication is incorrect.
     
  15. Jun 14, 2017 #14

    mfb

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    It is, if the information which door is opened is uncorrelated with the event "it is behind door 1". And that is indeed the case.
     
  16. Jun 14, 2017 #15
    And I said as much. But the solution is still incorrect if the assumption of this state of no correlation is left out. Nor is there an explanation of why it affects the answer.

    The reason people are almost never convinced by explanations like that, is because they know that their solution is closer to the technically correct approach. It just makes a simple mistake, and that explanation does nothing that would point it out.

    Think about it: say you had two students who gave different answers to the same problem. Would you just say "Susan is right," or explain why John is wrong?
     
  17. Jun 14, 2017 #16

    mfb

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    Both, and most explanations I see do that properly.
     
  18. Jun 14, 2017 #17

    WWGD

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    I don't see why conditional probability is needed: If the person chose a door randomly, 2/3 of the time s/he will have selected a goat, and only one third of the time s/he would have selected the car.
     
  19. Jun 14, 2017 #18

    FactChecker

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    It's an easy application of Bayes' Rule, which is all about conditional probabilities. So the proof is easy. The thing that trips people up is the intuitive feeling that the probability of both unopened doors are equally affected by opening the third door.
     
  20. Jun 15, 2017 #19

    Stephen Tashi

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    In this portion of the thread, we haven't heard any Susan who gave a complete and coherent answer. Susan should state the probability space, state the conditional probability to be computed and demonstrate how Bayes rule is applied. Until that is done, we'll have to be content with John-type answers.
     
  21. Jun 15, 2017 #20
    Since the two options were mutually exclusive, you can't do both. And not to pick on WWGD, but the following is a perfect example of what I meant.

    This is a form of the typical answer. If the host always opens the highest-numbered door he can after the contestant chooses door #1, it is true that s/he will win by switching 2/3 of the time. But that will be all of the time when the host opens door #2, which comprises 1/3 of all games, and half of the time when he opens door #2, which comprises 2/3. Since s/he knows which case she is in, the 2/3 result WWGD arrived at is meaningless.

    Now, I'm not suggesting this bias exists, I'm saying that you can point out the error in the 1/2 answer only by addressing it with conditional probability, like that solution does incorrectly.

    I'm going to assume you reversed the names, so John gave the correct answer.

    And I'm saying that (my) Susan's is less complete, and technically less coherent, than John's. It's only virtue is that it gets the right answer, which turns out to be the average of the two conditional cases.

    But considering a probability space is a good idea. Neither does this explicitly. Both imply one where the initial probability the car is behind door #1 is 1/3.

    John's incorrect solution goes on to imply that it is 1/3 for each door, that one door is eliminated, so the conditional probability for the remaining two is found by dividing by sum of the remaining initial probabilities. THIS IS CORRECT TECHNIQUE. It just missed one random variable.

    Susan's correct answer ignores any other part of the probability space, implying, without explicitly saying, that it doesn't matter. My point is that John can see that he used correct technique, even if it was incomplete. And that Susan didn't. So why should John accept her answer?
     
    Last edited: Jun 15, 2017
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