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Thread: Mythbusters- Monty Hall 3 Door Pick

  1. #31
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    Quote Originally Posted by SoFlaLaw View Post
    As for your second paragraph, that is my point. People are making subjective (not objective) value based decisions on how to reallocate the value of the original probability of the exposed door (having a 1/3 probability before being exposed by the game show host).
    well maybe, here is the thing though, people are at least doing something, at least really exerting some mental prowess in some regard over and on real phenomenon that exist, beyond just applying math with no justifiable reason.
    Okkkkkay?


    As for 1 above, how? How do we know we "screwed up" when we made our initial selection? We don't and can't unless we can read the future (See, Nicholas Cage as Chris Johnson in Next) or the answer to the problem was disclosed to us (or we stole the information).
    just playing the game is a screw up. gambling is a screw up. mindlessly following monty’s orders is a screw up. ok, it’s not a screw up because it’s a free roll, right? well, let’s imagine the monty devil thing again, where behind two of the doors is DEATH. is it a screw up to play the game then? hypothesis proved: it’s a screw up.
    So what, continuing to play the game might be a screw up. How do you know the devil (if such a thing/entity really exists) isn't just Monty Hall playing more mind games with you, and that there is really no deadly option behind 1 of the 2 remaining doors, or perhaps death lurks behind both doors. Booyah!


    As for 2 above, why? I can see where this would be true if our initial selection was made on a "known" bad choice. there yah go!!! our initial selection was made on a “known” bad choice! But since our initial selection was made in a truly random fashion, and not based on bad facts, it was done merely through gambling (cannot do more, but could choose not to play the game) on the odds 1:2 no matter which door we selected. So, how would our initial selection of Door 1, Door 2, or Door 3 merit "not deserv[ing] to have any more value added to it."?
    merely gambling!? SoFlaLaw you didn’t just say that did you and you a card counter?
    Okay, I was sloppy. You got me there. Yes I am a CC. I can appreciate your characterization of the decision to play the game in the first place as being a bad choice, thereby rendering the selection of Door 1 as a bad decision, assuming that you are placing your life, health, loved ones or significant financial interests at risk. But that doesn't change the bad status had we selected either of the other 2 doors. ALL ARE BAD CHOICES, SINCE THEY ARE SELECTED WITHOUT AN ADVANTAGE AGAINST THE HOUSE.


    As for 3 above, what "facts" do we "know" (or more appropriately, what reasonable assumptions can be made)?facts? reasonable assumptions? there yah go, now we are getting somewhere! and somewhere is where we want to get, not just spinning our wheels. hey it’s not a crystal clear perfect world out there in advantage play land but we do the best we can with what we’ve got, when we can. may seem like grasping at straws, but no out and out gambling, just gotta be thinking on your feet, savvy. i mean hey, you can go with your original losing gamble if you want, or you can use what information you have and refine your action, it’s up to you. science, my friend, but not rocket science.
    Science and math is what I have been trying to get to the underlying fundamentals of, and what I have been utilizing (proving or disproving a hypothesis). I have no way of prejudging whether my initial selection was "your original losing gamble" or if Door 3 would be a "losing gamble".


    Fact 6 - We do not know if the game show host "knows" behind which door the prize has been placed (if relevant at all); incorrect, we know & it’s relevant but I contend it is not necessary info as long as the prize is not exposed by monty. but we know, if nothing other than this problem is on the internet, lol. and if it’s on the internet the switch solution must be true. Bonjur? 
    But he said he was a French model! Even if Monty Hall knows which door the prize is behind, provided that he exposes 1 of the 2 doors behind which the prize is not located, at best his knowledge is irrelevant. It could be behind the door I selected (Door 1) or the remaining door (Door 3 after he exposes Door 2).
    exactly !!!! the salient question becomes, how many times out of how many times is the prize gonna be the door you selected door 1?, for repeated renditions of this game? and how many times is the prize gonna be the door you refused to switch to from door 1 over those repeated renditions of this game ?
    for repeated renditions you expect your door #1 to have the prize one out three of those repetitions. same for each of the other two doors.
    now we introduce the show and switch offer
    you have door 1 and keep it, so door 2 is opened and no prize. so it’s just your door 1 and door 3 rendition 1
    you have door 1 and keep it, so door 3 is opened and no prize. so it’s just your door 1 and door 2 rendition 2
    for whatever door you keep, it’s always your door against two other possible doors as the renditions progress, no?
    we can do this over and over and over, but guess what the door you choose is never switched so it’s always the door you chose standing against one of either two different doors.
    as the renditions continue it’s a particular door, that door you chose against two other doors.
    over and over and over again…..
    your one against two
    your one against two
    your one against two
    …….
    does that translate to one chance against two chances?
    one chance of containing the prize against two chances of containing the prize.
    is that a total of three chances and is that 1/3 chance against 2/3 of being the prize?
    and what is the chance we theoretically say the door switched to has at the end stage of the game? 2/3 no?
    but you didn’t switch, so you just have a 1/3 chance
    this refusal to switch, the keeping of the original choice forces the non switcher to always be one against two, as renditions add up, over and over and over ad infinitium.
    but what if you switch?
    you have door 1 and door 2 is opened and no prize. so you switch to door 3, same as saying door 1 & door 3 versus door 2 rendition 1
    you have door 1 and door 3 is opened and no prize. so you switch to door 2, same as saying door 1 & door 2 versus door 3 rendition2
    so you could originally choose other doors but these renditions essentially can be repeated over and over and over
    and as renditions progress its your two doors against one door
    over and over and over again
    your two against one
    your two against one
    your two against one
    …..
    does that translate into two chances against one?
    two chances for the prize against one chance for the prize
    two out of three chances for the prize against one out three chances for the prize
    your 2/3 chances for the prize against 1/3 chance for the prize
    and what is the chance we say the door switched to has at the end stage of the game? 2/3 no?



    Fact 13 - The exposed door (Door 2) does not reveal the prize; no it doesn’t, but it reveals the fact that door 3 now has a 2/3 value since door 2 has a zero value far as the prize goes
    That is circular reasoning. Just because the prize is not behind Door 2 (now exposed), does not mean that its value automatically should be transferred to Door 3. That has been my point from the gitgo, and has not yet been disproven through objective method. Why can't its value be transferred to my initial selection of Door 1?

    Fact 15 - The odds/probability of Door 2 (1:2 or 33 1/3%) for success no longer applies to Door 2 (proven not to contain the prize); no, we now know how much door 2 contributes to the 2/3 sum of door 2 & door 3, it contributes 0
    Yes, it now contributes 0 to door 2, but likewise, before being exposed, it equally contributed to the 2/3 sum of Door 1 & Door 2.


    Fact 18 - If we nullify Door 2's chance/probability for success, Door 1 and Door 3 each have 1:1 odds or a 50% chance/probability for successfully selecting the prize; NO, NO, NO! door 1 has a 1/3 chance when we chose it, door 2 and door 3 have a 2/3 chance together. just because door 2 ends up having a goat doesn’t mean all of a sudden that door2 and door 3 chances don’t equal 2/3.
    You continue to refuse to acknowledge that although Door 2 & Door 3 had a 2/3 chance together, so did Door 1 & Door 2 OR Door 1 & Door 3. So, it means that there may be a goat (cow, sheep, lame horse, etc. or Monty Devil might kill you) behind either Door 1 or Door 3.


    Fact 19 - If we split Door 2's chance/probability for success, Door 1 and Door 3 each receive 50% of Door 2's 1:2 odds or 33 1/3% chance/probability for successfully selecting the prize, resulting in each now having 1:1 odds or a 50% chance/probability for successfully selecting the prize; and there is no justifiable reason for us to do that math in fact we have information that makes our brains unable to justify adding any value to door 1
    And you questioned my being a CC??? We always do the math! As for your conclusion, it is just that, a conclusion probably without any underlying solid foundation. The value from Door 2 can be added to either Door 1 or Door 3. You continue to refuse to recognize that possibility.


    As for the fourth paragraph, how do we know our initial selection was "unwise"? How was that selection being taken from a "mathematically disadvantageously large set" different from whether we initially selected either of the other two doors? Or all you saying selecting any of the 3 doors initially was unwise? YES! or no, or maybe lol, but yes If so, then why would you reward any selection with Door 2's odds/probability for success? because one is either an advantage player to begin with and knows the score about monty, or one is trapped like a rat in a stupid move, but still is a smart enough rat to figure out how to maybe escape the trap. the trapped rat realizes his mistake and fixes it as best he can.
    But the rat might have unknowingly fixed "it as best he can" when he made his initial selection of Door 1, and not switching to the possibly deadly decision awaiting behind Door 3. One thing you repeatedly disregarded in replying to my questions is that I assumed from the start that the game show host did not know the identity of the door behind which the prize (or life and $1 million) rested. Alternatively, I posited that it is irrelevant whether Monty knew where the prize rested, provided that he exposed 1 of the 2 doors behind which the prize was not located, and left my initial selection (Door 1) and the other door. Based upon that hypothetical that I posited, the rat cannot know behind which door the prize is located, and cannot know that switching would improve his odds (as his odds would grow to 1:1 or 50%) when the value of Door 2 is split between Door 1 and Door 3.


    Even if you expand the set of options from 3 to 1000 doors, and expose 998 doors without showing the prize, each of those options initially were from (your words) a "mathematically disadvantageously large set" of options. But once the door(s) is(are) exposed, we have a relatively small set of options (Door 1 and Door 3, or if a 1000 Doors, Door 1 and Door 537), just 2. true, the example just dramaticizes the error inherent in picking from a mathematically disadvantageously large set.
    I will leave this one alone, other than to state my objection to your characterization of the "mathematically disadvantageously large set" on the record based upon relevancy grounds.


    I understand your points, but disagree that you are arguing from a point of relevant facts. So, why does our "unwise" initial choice from a ""mathematically disadvantageously large set" deserve to be recognized and be taken into account (punished, definitely not an objective process)? this is why i suggested the devil and his goat angels scenario. such a scenario proves the error of initially playing the game. that error may not be 100% fixable but one can improve one’s odds by using the information monty provided, and switching doors. but what? come on, have you never made a mistake, realized it was a mistake and went back and corrected it? that’s the idea of not putting any more value onto door 1, we realize it was a mistake in the first place. and on top of that we realize that there is no compelling reason to remove any value from the combined values of door 2 and door 3, just because door 2 ends up having zero value doesn’t change the combined value of door 2 and door 3.
    Once again, you keep assuming that switching will fix the "mistake", but have failed to prove that the initial selection was indeed a mistake, or that it needed to be corrected. The assignment of Door 2's value solely to Door 3 is an assumption, without proof that it cannot be assigned to Door 1. Also, "just because door 2 ends up having zero value doesn't change the combined value of door 2 and door [1]". Come on, we're playing for my life!


    I truly would like to understand your reasoning to attempt to internalize your position and see if I change my approach (an analytical process akin to selecting Door 1 versus Door 3, albeit there does not appear to have been a Door 2 in our discussion and analysis).
    ok, tell yah what, i’ll switch, you are correct. 
    Au contraire. Maybe I'll switch, because your arguments appear on the internet! Any chance you are a french model?

    Thanks for trying, but I am still looking for proof or disproof (is that a legitimate word?) of my understanding of the Let's Make a Deal hypothetical.

    PS - Obviously, yes I am a lawyer, and I represent working people, their unions, pension plans, health and welfare plans and apprenticeship training programs. It's just the David in me that wants to defeat Goliath, just like when I take money away from the thieves, oops, meant casinos.
    uhhmm, i dunno if i got the additional reply above in off-red correct or not. it makes sense to me now, but my head is spinning, just thought i'd run it past yah, lol.
    edit: don't like my attempt above? try reading this:
    Solution:
    Assume that you always start by picking Door #1, and the host then always shows you some other door which does not contain the car, and you then always switch to the remaining door.

    If the car is behind Door #1, then after you pick Door #1, the host will open another door (either #2 or #3), and you will then switch to the remaining door (either #3 or #2), thus LOSING.

    If the car is behind Door #2, then after you pick Door #1, the host will be forced to open Door #3, and you will then switch to Door #2, thus WINNING.

    If the car is behind Door #3, then after you pick Door #1, the host will be forced to open Door #2, and you will then switch to Door #3, thus WINNING.

    Hence, in 2 of the 3 (equally-likely) possibilities, you will win. Ergo, the probability of winning by switching is 2/3

    source: http://www.probability.ca/jeff/writing/montysimple.html
    Last edited by sagefr0g; January 7th, 2015 at 12:49 AM.
    best regards,
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    that's my take on it your mileage may vary.
    for senior citizen fuzzy count click link:
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  2. #32
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    Quote Originally Posted by sagefr0g View Post
    uhhmm, i dunno if i got the additional reply above in off-red correct or not. it makes sense to me now, but my head is spinning, just thought i'd run it past yah, lol.
    edit: don't like my attempt above? try reading this:
    Solution:
    Assume that you always start by picking Door #1, and the host then always shows you some other door which does not contain the car, and you then always switch to the remaining door.

    If the car is behind Door #1, then after you pick Door #1, the host will open another door (either #2 or #3), and you will then switch to the remaining door (either #3 or #2), thus LOSING.

    If the car is behind Door #2, then after you pick Door #1, the host will be forced to open Door #3, and you will then switch to Door #2, thus WINNING.

    If the car is behind Door #3, then after you pick Door #1, the host will be forced to open Door #2, and you will then switch to Door #3, thus WINNING.

    Hence, in 2 of the 3 (equally-likely) possibilities, you will win. Ergo, the probability of winning by switching is 2/3

    source: http://www.probability.ca/jeff/writing/montysimple.html
    Your outline of the 3 options does not provide for any difference based upon exposure of Door 2 or 3, than the game player's initial selection decision 1/3. Behind Door 1, once. Behind Door 2, once. Behind Door 3, once. Remember, the player can switch to Door 2 or 3, and LOSE. So, it is not 2/3 if the player switches. It is only so, if the car is behind Door 2 or Door 3, and it was not behind Door 1 the whole time.

  3. #33
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    Default Let's Wager

    Quote Originally Posted by SoFlaLaw View Post
    Your outline of the 3 options does not provide for any difference based upon exposure of Door 2 or 3, than the game player's initial selection decision 1/3. Behind Door 1, once. Behind Door 2, once. Behind Door 3, once. Remember, the player can switch to Door 2 or 3, and LOSE. So, it is not 2/3 if the player switches. It is only so, if the car is behind Door 2 or Door 3, and it was not behind Door 1 the whole time.
    SoFlaLaw: I will give you 250 ZenZone Kudos for every win ("games stayed and won") less 100 ZenZone Kudos for every play ("games stayed") - minimum 30 plays: http://www.shodor.org/interactivate/...mpleMontyHall/

    Thirty plays ought to take you three minutes or less (but you might want to play more than 30 times). And since you expect to win 50% of the time, 30 plays (on average) should net you 15 x 250 - 30 x 100 = 750 ZZ Kudos.

    If you end up with less than zero Kudos, you can deliver the Kudos you owe via my thread: http://www.zenzoneforum.com/threads/...n-of-Ken-Uston

    I will accept the equivalent to ZenZone Kudos in US$ when we meet in person, if you rather.
    Last edited by Dodo; January 7th, 2015 at 12:25 PM.

  4. #34
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    Default Let's Make a Deal Without Collusion!

    Quote Originally Posted by sagefr0g View Post
    Dodo is right there is an element of collusion going on in this problem, due to the inside knowledge that monty has. thing is i don't believe the collusion factor or inside knowledge factor is absolutely needed for the problem to exist in a still fundamentally valid way. you just need for the prize to not end up being revealed before the game is over, is all, then the problem can be played out with or without collusion.
    Quote Originally Posted by sagefr0g View Post
    Fact 6 - We do not know if the game show host "knows" behind which door the prize has been placed (if relevant at all); incorrect, we know & it’s relevant but I contend it is not necessary info as long as the prize is not exposed by monty. but we know, if nothing other than this problem is on the internet, lol. and if it’s on the internet the switch solution must be true. Bonjur?
    sagefr0g: Try http://www.math.ucsd.edu/~crypto/Mon...esnotknow.html

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    Quote Originally Posted by Dodo View Post
    SoFlaLaw: I will give you 250 ZenZone Kudos for every win ("games stayed and won") less 100 ZenZone Kudos for every play ("games stayed") - minimum 30 plays: http://www.shodor.org/interactivate/...mpleMontyHall/

    Thirty plays ought to take you three minutes or less (but you might want to play more than 30 times). And since you expect to win 50% of the time, 30 plays (on average) should net you 15 x 250 - 30 x 100 = 750 ZZ Kudos.

    If you end up with less than zero Kudos, you can deliver the Kudos you owe via my thread: http://www.zenzoneforum.com/threads/...n-of-Ken-Uston

    I will accept the equivalent to ZenZone Kudos in US$ when we meet in person, if you rather.
    Since I did not develop the program that was written to generate that application, I cannot know whether there was a bias toward assuming that the probability value of Door 2 vsn only be reallocated to Door 3, and not to Door 1. Consequently, my acceptance or rejection of your challenge has no bearing on resolving this discussion, but rather, would only prove/disprove that based upon the parameters of the computer program, a certain result can and should be expected.

    For instance, I could design a shoe for blackjack which would be comprised of decks from which you have removed 4 ten value cards from each deck of cards in that shoe (think Spanish 21 or Fun 21). With that shoe of cards, I suspect that you would agree that it is less likely that the player and the dealer will draw as many natural blackjacks as they might ordinarily draw, and that it is also less likely that they will bust before drawing to a pat hand, as opposed to decks of cards that contain all of the ten valued cards. Conversely, if I remove all of the 2s and 3s from the decks, would the odds increase for natural blackjacks being dealt? More hands busting?

  6. #36
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    Quote Originally Posted by Dodo View Post
    wowser, i see that i may well be in error regarding the need for monty's collusion for the problem to work as it does or not. gotta look into that and try and understand it. thank you Dodo.
    edit: my guess is that since my stipulation that the car never comes up if monty is not colluding is unrealistic, then by golly monty needs to be colluding for the gig to work.
    edit: perhaps this subtle aspect is what is throwing SoFlaLaw, the article states that it may be the reason that intuitively so many people get the wrong answer.
    Last edited by sagefr0g; January 7th, 2015 at 03:44 PM.
    best regards,
    mr fr0g MMOA honorary predator
    STRENGTH - HONOR - HEART
    that's my take on it your mileage may vary.
    for senior citizen fuzzy count click link:
    http://www.youtube.com/watch?v=DrTiP4ZIUfI

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    Quote Originally Posted by sagefr0g View Post
    wowser, i see that i may well be in error regarding the need for monty's collusion for the problem to work as it does or not. gotta look into that and try and understand it. thank you Dodo.
    edit: my guess is that since my stipulation that the car never comes up if monty is not colluding is unrealistic, then by golly monty needs to be colluding for the gig to work.
    edit: perhaps this subtle aspect is what is throwing SoFlaLaw, the article states that it may be the reason that intuitively so many people get the wrong answer.
    sagefr0g, thanks for the link:

    From that article, the following information really shines a light on why I have looked at this from a different perspective than you, zg and dodo:

    This question seems to have a non-intuitive answer. Why were so many convinced that Marilyn Vos Savant was wrong? They had all decided that it did not matter if the contestant switched or did not switch. There may be a reason so many disagreed with her. Omitting one phrase in the statement of this problem changes the answer completely and this might explain why many people have the wrong intuition about the solution. If the host (Monty Hall) does not know where the car is behind the other two doors, then the answer to the question is "IT DOESN'T MATTER IF THE CONTESTANT SWITCHES." The change in the statement of the problem is so slight that this might be the reason this problem is such a "paradox."

    My underlying assumption all along was that the game show host did not know nor have an ability to impact the decision to switch from Door 1 to Door 3. So, assuming no knowledge on the game show host's part, "IT DOESN'T MATTER IF THE CONTESTANT SWITCHES."

    Thank you, thank you, thank you.

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    Hmmmmmmm....ok I almost have my head wrapped around this........
    But where does Vanna White fit into this whole thing!!

    Mac
    So look buddy it's nothing personal, but either your the unluckiest person in the world or your stealing from me. Either way............ I WANT NO PART OF YOU!!!!!! NEVER!!!!

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    Quote Originally Posted by Machinist View Post
    Hmmmmmmm....ok I almost have my head wrapped around this........
    But where does Vanna White fit into this whole thing!!

    Mac
    Pat, I'll buy a pair (well, Vanna's pair) of UU's (used to be OO's) for .....

    Is that what you were thinking Machinist?

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    Here is the stumbling block for SoLaw:
    When the curtain#3 is first chosen we know that the probability of success is 1/3 -
    - that 1/3 likelihood NEVER CHANGES, how could it? (that was my error as well in my initial answer)

    End of story...

    "The dogs bark but the caravan moves on."
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    Quote Originally Posted by zengrifter View Post
    Here is the stumbling block for SoLaw:
    When the curtain#3 is first chosen we know that the probability of success is 1/3 -
    - that 1/3 likelihood NEVER CHANGES, how could it? (that was my error as well in my initial answer)

    End of story...

    zg,

    Really quite simple. When you started you had 3 options. Each had 1/3 value/odds.

    With 1 option removed, it no longer retains its value, its value/odds is transferred equally between 2 remaing options, so Door 1 = 1/3 + 1/6 or 3/6 (50%), and Door 3 = 1/2 + 1/6 or 3/6 (50%). Otherwise you are preferentially saying that Door 3 is now de facto superior to Door 1 (remember they were and I argue remain equal) and that only Door 3 can increase from 1/3, from 1/3 to 2/3.

    Any other reallocation of Door 2's odds/value is not an objective exercise in mathematics, but rather an expression of subjective desire for 1 door to be superior to the other.

    Does that clarify my reasoning for you?
    Last edited by SoFlaLaw; January 7th, 2015 at 11:04 PM.

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    Quote Originally Posted by SoFlaLaw View Post
    With 1 option removed, it no longer retains its value, its value/odds is transferred equally between 2 remaing options,,,
    Does that clarify my reasoning for you?
    Yes, but the fallacy of your reasoning is revealed more accutely when we consider the higher# of curtains scenario:
    - There are 20 curtains with one prize only.
    - We pick one curtain, a 1/20 chance.
    - Monty eliminates all but one closed curtain, inviting us to switch.
    - Did our 1/20 curtain call suddenly become 1/2?

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    Quote Originally Posted by DDutton View Post
    Yes, but the fallacy of your reasoning is revealed more accutely when we consider the higher# of curtains scenario:
    - There are 20 curtains with one prize only.
    - We pick one curtain, a 1/20 chance.
    - Monty eliminates all but one closed curtain, inviting us to switch.
    - Did our 1/20 curtain call suddenly become 1/2?
    Not clear on your premise. If I can assume that you mean that Monty Hall eliminated all but one closed curtain, AS WELL AS THE CURTAIN I INITIALLY SELECTED, then there are now 2 curtain options. I can stay with the curtain I initially selected, or I can switch to other remaining curtain.

    The value of the other 18 curtains, each having a 1/20 (5%) value/odds (collectively 18/20 or 90% value/odds), is now split evenly amongst my initial curtain and the other remaining curtain, so we are back to 1:1 or a 50% chance. 45% of their odds/value is added to my selection, and the other 45% of their odds/value is added to the other curtain.

    So, if I do understand your premise that there are now 2 curtains, my initial selection and one other remaining curtain, then YES, my 1/20 curtain (the number I could select at the beginning - 1, versus the total population of curtains to choose from - 20), has now become 1 out of 2, or 1:1 (50%). The magnitude of the population of options is irrelevant to the analysis. We are still left with options having equal value / odds.

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    Hey who cares....change the station and let's see if Vanna has a wardrobe malfunction!

    Mac

    So look buddy it's nothing personal, but either your the unluckiest person in the world or your stealing from me. Either way............ I WANT NO PART OF YOU!!!!!! NEVER!!!!

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    Quote Originally Posted by aslan View Post
    On the Monty Hall game show there is a three door pick-a-door for a prize segment; it goes like this: The person picks a door. Immediately another door is opened which does not have a prize behind it. Now you are asked it you want to change your choice of door, with only the two remaining doors unopened. Part of the Mythbusters test was to see if people would stick with their original choice, or change. The claim is that most people will stick to their original choice. In fact, in the test, 100% decided to stick with their first choice, confirming that part of the test.

    The part that confuses me is the second part. Mythbusters states that it is statistically wiser to switch doors. How can that be? What have I missed? It seems like with two doors remaining, you have a 50/50 chance of winning no matter which door you pick. Why is it that they state that statistically you should always switch doors for a better chance?

    Are they saying that there is a 1/3 chance in the initial selection and that once one door is opened, it makes leaves the remaining door with a 2/3 chance and the original door still has a 1/3 chance? This is very difficult to see.

    Besides simulation, is there a good way to prove it? It would be useful in creating gambling propositions, if it holds true.
    Surprisingly, the picture is still NOT clear to a large percentage of people! You didn’t really see Parpaluck’s page on solving the Monty Paradox? It was published more than a decade ago! It has offered also specialized combinatorial software! Just generate the sets in lexicographic order, Asslan/Scobe!

    There are 6 permutations of 3 elements (e.g. doors): 1 x 2 x 3 = 6.
    1 2 3
    1 3 2
    2 1 3
    2 3 1
    3 1 2
    3 2 1

    Only one door is a winner; e.g. door #1; doors #2 and #3 are therefore losers.

    Let's remember also that the 1st position in the 6-element set represents the very 1st choice of the game: The Player chooses a doorWin or Lose. That's a given. Monty cannot erase the door in that position.

    Next, the host eliminates one losing door from the remaining 2 doors (position 2 or 3).

    Let's update the 6-element set following the Host's action. Monte Hall cannot eliminate door 1 — because we designated it as the winning choice; Monty may only eliminate a losing choice in step 2; the losing choices (doors) are #2 and #3 in the 6-element set.

    Code:
    The original game: N = 3 (three doors or choices)
     
                     Stay      Change
                     ----      ------
     1  x  3          W          L
     1  x  2          W          L
     2  1  x          L          W
     2  x  1          L          W
     3  1  x          L          W
     3  x  1          L          W
    No matter what losing door Monte Hall “erases” (the x in the graphic), the Player will always have 2 of 6 winning choices (1/3) on Stand and 4 of 6 winning choices on Change (2/3). You can see that there are 4 W remaining for Change as opposed to 2 for Stay.

    Read it carefully and study the specialized software as well. But in order to be successful, YOU MUST PULL YOUR HEAD OUT OF THAT BARREL OF WHISKEY!

    Mathematics of the Monty Paradox, The Classical Occupancy Problem, Ion Saliu's Paradox


    Ion Saliu,
    Founder of Probability Theory of Life

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