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Monty Hall Dilemma - Winning a GTI on a Game Show

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  • Swallowtail
    replied
    I am enjoying watching this....

    I agree with Dubya - and the Wikipedia page with its excellent explanation. The situation is not a 50/50, as elements of the original 67/33 apply, and the pre-knowledge of the host of which door holds the car changes the scenario. It is not a "fresh" situation once the first door is removed.

    Read through the Wikipedia page with its full explanation, not just the quick one.

    This is the math of the probability of winning by switching:

    Last edited by Swallowtail; 19-03-2010, 09:45 AM.

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  • team_v
    replied
    Originally posted by Dubya View Post
    But how do you know for sure something that is contentious if you have not tested the theory with a simple experiment, just as any researcher with a theory would do (if they possibly could).
    An application is too easy to modify to say what you want it to say.
    The only way you could test it is to do the card trick which i don't have handy.


    You simply cannot keep the probabilities from a 1/3 situation when applying it to only 2 doors, it just doesn't work like that.
    The situation has changed and so should the probabilities to reflect the change in variables.

    So if you flip a coin, do you have a 50/50 chance of getting heads or tails or a 33/67 chance?

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  • Rocket36
    replied
    Originally posted by team_v View Post
    And the reality is that it's 50:50.
    Originally posted by Dubya View Post
    But how do you know for sure something that is contentious if you have not tested the theory with a simple experiment, just as any researcher with a theory would do (if they possibly could).
    There is no need to test something that is mathematical FACT.

    ANY choice where there are ONLY 2 OPTIONS is a 50/50, or 1 in 2 chance... It's not a theory.

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  • Dubya
    replied
    Originally posted by team_v View Post
    And the reality is that it's 50:50.
    But how do you know for sure something that is contentious if you have not tested the theory with a simple experiment, just as any researcher with a theory would do (if they possibly could).

    They certainly wouldn't keep arguing tit for tat if the theory could so easily be disproved.

    So as it stands, the 67ers are asking (by now pleading with) the 50/50s to submit the theory to the test and the 50/50s, team v for one at least, appear to be reluctant to do so.

    Why the reluctance to prove the theory one way or the other by spending 10 minutes with some playing cards or an online simulator?

    PS - just so long as we do not get any "you didn't explain" / "I didn't realise" excuses when the penny does drop...
    Last edited by Dubya; 19-03-2010, 09:26 AM. Reason: Postscript

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  • team_v
    replied
    Originally posted by Dubya View Post
    So enough with theory guys, it's time to get real!
    And the reality is that it's 50:50.

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  • Dubya
    replied
    Originally posted by Buller_Scott View Post
    the funny thing is, and not to pick any bones at all with anyone (i dont take any disagreement etc as an affront- this is actually a pretty worthy discussion imo), but i have posted two scenarios, one theoretical, and one practical.

    and yet i dont see anyone disproving my conclusion RE the dub fest, the gti and the twenty raffle tickets reduced to ten- i dont see anyone proving that, should person A be allowed to switch until half the raffle tickets have been eliminated, that person A's chances of winning that GTI are now 55% or 11/20.

    i feel i was pretty comprehensive in my logic, and i would have welcomed rebuttal, but no one challenged my numbers in that scenario at all.

    then, i alluded to actually sitting at home and trying this 1000 times, but encountering the problem of KNOWING for a fact that the first card must fail, and thus my partner (the host) not even bothering to waste the energy to pick that first card up anymore- probability dealing with chance, i find it ironic that this first card is even relevant anymore as.... well..... there is no "chance" left- the rules dictate that it MUST NOT be the card i chose in the first place. it's no longer a "probable", and yet why do people insist that it must be counted?

    so, spilledprawn- whilst some of the 2/3 gang are quick to dismiss the 50/50 guys as ''a lost cause'', i'd like to invite YOU to address my question- im going to do this 1000 times. im not even gonna bother picking up 3 cards- just the TWO THAT MATTER, before i get offered the switch. disregarding the third card as a waste of time seeing as it must, by rule, fail- how are my chances 2/3?

    and please dont do what some of the others have done: "aww you dont get it, you're crap at math, you dont understand how it works, lol@your explanation that i will not acknowledge with any decent counter argument".

    p.s. not to come across as uppity or a sourpuss at all, i just reckon that the 50/50 argument is well established, but i cannot see too many people challenging our explanations directly (namely, my two explanations as mentioned above).

    i really really really really want someone to take my two scenarios and tear them apart- at this point of the thread, it'd be somewhat of a breath of fresh air.

    cheers,

    scotty
    I for one have not turned my mind to your purported analogy as we have a perfect analogy in the three playing cards and with the scenarios where there are more than three doors and only one GTI.

    Accordingly, if it is perfectly analogous it is otiose and if it is not perfectly analogous it is irrelevant. In any event, none of us wants a second debate about whether it is perfectly analogous or not. There is simply no need to consider it while there is no overwhelming consensus on the main issue.

    A number of people, most lately Hawk (a sceptic intitially, if I may say) with gleaming logic - but with incorrect link to my post, have explained how the higher odds from switching are as certain as a 50/50 long-run average from flipping a coin.

    The 67ers are asking the 50/50s to simply apply the highly-testable theory. We're not talking about life on Mars here!

    If the doubters do, they will see that if they never switch they will "win" ~33% of the time and if they always switch they will win ~67% of the time.

    So instead of arguing until you're blue in the face, test the theory.

    All of the 67ers have satisfied themselves in one or more ways such as:

    - the ever more compelling logic as the number of doors is increased above 3 (with still only one GTI);

    - used online simulators - the randomness of which is irrelevant as the only randomness is the order in which the cards are laid - you determine which card you will pick initially after the computer has determined what each card hides (or at least a proper simulator will);

    - using three playing cards (one Ace and two Jokers, say) for 10 minutes with another person to test the theory.

    By now it should only be a question of which of the 50/50s is going to take 10 minutes to test the theory and be first to share their enlightenment.

    As opposed to arguing ad nauseum against a proposition that is so easily verifiable as being true in theory and reality!

    So enough with theory guys, it's time to get real!

    Leave a comment:


  • team_v
    replied
    Originally posted by The_Hawk View Post
    I'm not trying to offend, just want to try and lay out the logic in this scenario, to illustrate mote clearly, lets start with 100 doors, 99 goats and one GTI.

    Your chances of choosing the GTI are 1/100 (and of getting a goat are 99/100).

    The host then opens the doors one at a time, finally getting down to the final two doors then gives you the choice to switch.

    When we started you had a 1% chance of being right and a 99% chance of being wrong. Now that the host has dragged things out to this point, slowly showing you all the goats you original odds haven't changed, there is still a 1% chance your first choice was right and a 99% chance you chose wrong.

    So now, even though you have only two doors left, there is a 99% chance your door is still the wrong one, it's in your best interests to switch.

    The thing is though, you have changed the probablitlities as the other doors are open and thus, their probablility shouldn't be included since they are no longer a probablility but a certainty.

    As said earlier, it is a dynamic situation and the probabilities change to reflect that.
    Each time a choice has to be made, the probabilities are reset and adjusted to suit the remaining choices.
    It is simply not reasonable to say out of 2 choices, you have a 1/3 chance of getting a goat as there are only 2 options remaining (due to 1 door being opened and revealing the goat)

    Just my view point though to back the 50/50.
    Computer applications will never be truly random.



    Originally posted by The_Hawk View Post
    So you see, it can be two things at the same time.
    *Insert schroedinger's cat*
    Last edited by team_v; 19-03-2010, 08:11 AM.

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  • The_Hawk
    replied
    I think Dubya said it best here: http://www.bigpond.com/internet/plan...d=bph-access-3

    In short it's the culmative effect that make the probability of winning higher if you switch.

    I'm not trying to offend, just want to try and lay out the logic in this scenario, to illustrate mote clearly, lets start with 100 doors, 99 goats and one GTI.

    Your chances of choosing the GTI are 1/100 (and of getting a goat are 99/100).

    The host then opens the doors one at a time, finally getting down to the final two doors then gives you the choice to switch.

    When we started you had a 1% chance of being right and a 99% chance of being wrong. Now that the host has dragged things out to this point, slowly showing you all the goats you original odds haven't changed, there is still a 1% chance your first choice was right and a 99% chance you chose wrong.

    So now, even though you have only two doors left, there is a 99% chance your door is still the wrong one, it's in your best interests to switch.

    Lets say at this point they drag in a bystander off the street and say he gets what veers behind the other door. He has a 50/50 chance of getting a GTI since by the time he comes in there is only two options left.

    So you see, it can be two things at the same time.

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  • Buller_Scott
    replied
    Originally posted by Spilledprawn View Post
    Dubya, perfect description. I'm sorry to say that if the 50/50s haven't got it yet, it is a lost cause.
    the funny thing is, and not to pick any bones at all with anyone (i dont take any disagreement etc as an affront- this is actually a pretty worthy discussion imo), but i have posted two scenarios, one theoretical, and one practical.

    and yet i dont see anyone disproving my conclusion RE the dub fest, the gti and the twenty raffle tickets reduced to ten- i dont see anyone proving that, should person A be allowed to switch until half the raffle tickets have been eliminated, that person A's chances of winning that GTI are now 55% or 11/20.

    i feel i was pretty comprehensive in my logic, and i would have welcomed rebuttal, but no one challenged my numbers in that scenario at all.

    then, i alluded to actually sitting at home and trying this 1000 times, but encountering the problem of KNOWING for a fact that the first card must fail, and thus my partner (the host) not even bothering to waste the energy to pick that first card up anymore- probability dealing with chance, i find it ironic that this first card is even relevant anymore as.... well..... there is no "chance" left- the rules dictate that it MUST NOT be the card i chose in the first place. it's no longer a "probable", and yet why do people insist that it must be counted?

    so, spilledprawn- whilst some of the 2/3 gang are quick to dismiss the 50/50 guys as ''a lost cause'', i'd like to invite YOU to address my question- im going to do this 1000 times. im not even gonna bother picking up 3 cards- just the TWO THAT MATTER, before i get offered the switch. disregarding the third card as a waste of time seeing as it must, by rule, fail- how are my chances 2/3?

    and please dont do what some of the others have done: "aww you dont get it, you're crap at math, you dont understand how it works, lol@your explanation that i will not acknowledge with any decent counter argument".

    p.s. not to come across as uppity or a sourpuss at all, i just reckon that the 50/50 argument is well established, but i cannot see too many people challenging our explanations directly (namely, my two explanations as mentioned above).

    i really really really really want someone to take my two scenarios and tear them apart- at this point of the thread, it'd be somewhat of a breath of fresh air.

    cheers,

    scotty

    Leave a comment:


  • gareth_oau
    replied
    try the Poll

    Ive just created a poll to see what the population thinks.

    Please cast your switch/not switch vote!!

    Leave a comment:


  • Spilledprawn
    replied
    Dubya, perfect description. I'm sorry to say that if the 50/50s haven't got it yet, it is a lost cause.

    Leave a comment:


  • Dubya
    replied
    Originally posted by Buller_Scott View Post
    sorry dubya, i simply cant.

    this is going to sound pretty claimish on my part, but no one has answered my question- if im really going to do this a thousand times with some cards, but knowing that the first one, as a rule, must be a failure, why would i bother even picking it up? i'll just deal with switching between the two remaining cards, where my chances are 1 in 2.

    the 50/50ers were accused of not having the courage, bravado, nor the confidence blah blah to try this at home. i will. but i would like to know first, when im only going to be dealing with cards 2 and 3, not bothering to pick up card 1 because everyone knows it must fail- how will this prove in real life that the chances are 2/3?
    Try it. But you need two people. Or try one of the online simulators.

    Either way you will see a trend within 20 - 30 attempts:

    If you never switch you will see that you win around a third of the time, as you would expect.

    If you always switch, you will see that you win around two-thirds of the time, as you might not expect.

    Leave a comment:


  • Buller_Scott
    replied
    sorry dubya, i simply cant.

    this is going to sound pretty claimish on my part, but no one has answered my question- if im really going to do this a thousand times with some cards, but knowing that the first one, as a rule, must be a failure, why would i bother even picking it up? i'll just deal with switching between the two remaining cards, where my chances are 1 in 2.

    the 50/50ers were accused of not having the courage, bravado, nor the confidence blah blah to try this at home. i will. but i would like to know first, when im only going to be dealing with cards 2 and 3, not bothering to pick up card 1 because everyone knows it must fail- how will this prove in real life that the chances are 2/3?

    Leave a comment:


  • Dubya
    replied
    Originally posted by Rocket36 View Post
    Regardless of what ANYONE says, the second choice is ALWAYS going to be 50/50. That is mathematical FACT and anyone who disagrees with that is simply wrong. When choosing between two things, whatever they are, it's ALWAYS 50/50, or 1 in 2.
    No more conversions I see . . . does this mean the 50/50ers are set in their thinking?

    If so, one last try:

    If you had the choice of sticking with the first choice or switching to both of the other doors, would you switch?

    I'm presuming the reader would rather have two chances of winning (67%) instead of just one (33%) and so switch from their original selection to the two other doors. . .

    Now consider:

    What is the difference between:

    - switching to the two closed doors; and

    - switching from one door to one of the two other doors when you know for certain which of the two hides a goat?

    Does revealing which of the two other doors conceals a goat alter the probability that one of the two door doors will conceal the GTI?

    So after the goat is revealed behind one door, there is still a 67% chance that one of these two doors conceals the GTI. And a 33% chance the door we first chose conceals the GTI.

    So we get to choose between:

    A door with 0% chance of concealing the GTI (the one the host has already opened);

    A door with 67% chance (the door we did not first choose);

    A door with a 33% chance (the door we chose first).

    We already knew one of the doors we did not choose would contain a goat.

    Switching to the only other closed door after the first goat is revealed is the same as being allowed to open two doors.

    Whether you open both doors at once or Monty helps by first opening one for you makes no difference to the 67% probability that one of these two doors conceals a GTI.

    The only question is, are you going to act on this information and have the benefit of effectively opening two doors instead of just one?

    Or do you just go with your instinct and stick with your gut feel despite it offering only half the probability if you switched?

    You'd kick yourself if you switched and got the goat . . . but twice as often you'd be glad you switched.

    Anyone changing their position?
    Last edited by Dubya; 18-03-2010, 11:28 PM.

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  • Buller_Scott
    replied
    what rocket said!

    and the graph is in the link:

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