PARADE!



  • @eViLegion said:

    Condition 3 should be: "Given a choice of goat doors, it doesn't matter what method is used to choose between them, so long as the player does not know that method".



    This is to ensure that if, for example,

    1. Monty always chooses the lowest numbered goat
    2. Player selects door #3
    3. Monty opens door #2

      If the player knows Monty's method he can deduce that the car MUST be behind door number 1.
      If he doesn't know the method, he cannot make that deduction.

    The question to be answered is not "does switching your choice exactly double your chance of winning?" but "Is it to your advantage to switch your choice?"

    In order for it not to be to the player's advantage to switch doors, there would need to be a scenario in which the player can use information about Monty's choice methods to raise the probability of the initial pick being correct to at least 1/2. PalmerEldritch appears to believe that the scenario where Monty always picks Door 3 when available is one such. So let's assume both that this is what Monty will do and that the player knows it and take it case by equally likely case.

    Player picks   Car is behind   Monty opens   Car deducible   Winning move
    ------------   -------------   -----------   -------------   ------------
    Door 1         Door 1          Door 3        No              Keep
    Door 1         Door 2          Door 3        No              Switch
    Door 1         Door 3          Door 2        Yes             Switch
    Door 2         Door 1          Door 3        No              Switch
    Door 2         Door 2          Door 3        No              Keep
    Door 2         Door 3          Door 1        Yes             Switch
    Door 3         Door 1          Door 2        No              Switch
    Door 3         Door 2          Door 1        No              Switch
    Door 3         Door 3          Door 1 or 2   No              Keep

    A few things become clear here: firstly, it's to the player's advantage not to pick Door 3, since deliberately depriving Monty of his preferred choice eliminates any chance of deducing the car's position from his failure to exercise it. Secondly, the only cases in which the car's position can be deduced are the switch-wins cases. Thirdly, there are still twice as many switch-wins cases as keep-wins cases, meaning that it is to the player's advantage to switch even if unable to deduce the position of the car.

    Unless I'm mistaken, the only fact that the player will ever be able to glean using knowledge of Monty's choice procedure is that the car is behind the door that the player didn't pick and Monty didn't open, leading to a win by switching. This follows from the fact that when Monty does open the door that the choice procedure says he should, there is no way for the player to find out whether or not that choice was forced. It's only the fact of Monty's having made a non-preferred choice that can reveal the car's location.

    Therefore, it's to the player's advantage to switch when given the opportunity even if Monty's choice procedure is known. The eViLegion version of condition 3 is superfluous as well.



  • @flabdablet said:

    stuff

    Yes, actually you're quite right... you can never be reduced below 1/2 chance by "always switching".



    With additional information you MAY get an opportunity to deduce that you mustn't switch, but it cannot reduce the expected win from "ALWAYS switching" below 50%.



  • @eViLegion said:

    With additional information you MAY get an opportunity to deduce that you mustn't switch
    Short of actual prior knowledge of the car's location, or being under strict instructions from your beloved to come home with a goat, can you give an example? Because it seems to me that all you could ever conceivably find out is that you really, really ought to switch, not that you mustn't.



  • Well you're changing your argument now. In response to an earlier post I made you replied as follows:

    @flabdablet said:

    @PalmerEldritch said:
    I'm just saying the problem as posed is ambiguous, and there are 3 conditions that need to be assumed (if they're not stated) to make the probability of winning by switching doors = 2/3.
    And I'm saying you're wrong, because as soon as you assume your first two conditions, which between them specify unambiguously that Monty must open a door to reveal a goat before giving the contestant the option of switching, then the probability of winning by switching becomes 2/3 without needing your third condition; that is, your third condition is superfluous.
     

    The highlighted part of your response is incorrect, the probability of winning by switching in the problem as stated only becomes 2/3 if Monty randomly picks a goat door when he has a choice of 2 goat doors

     



  • @eViLegion said:

    you can never be reduced below 1/2 chance by "always switching".
    Can you think of a scenario that includes PalmerEldritch's conditions 1 and 2 (mandatory reveal of a non-winning door, followed by an opportunity to switch) that can reduce your chance of winning by switching below 2/3? I haven't yet managed to, and although PalmerEldritch seems to be claiming that Bayesian Scripture says such a scenario must exist, he hasn't yet given a correct example.



  • @flabdablet said:

    @eViLegion said:
    With additional information you MAY get an opportunity to deduce that you mustn't switch
    Short of actual prior knowledge of the car's location, or being under strict instructions from your beloved to come home with a goat, can you give an example? Because it seems to me that all you could ever conceivably find out is that you really, really ought to switch, not that you mustn't.

    Actually, I have no example...

    I was imagining some unspecified hypothetical non-random process, which gives some you additional information.
    Regardless of that process, that additional information could only ever be usable in 1/3 of cases, and will not affect the 2/3 of cases which provide the main grounds for always switching.


    Further thought suggests that there is no non-random process that doesn't increase the players chances of deducing he should switch, nor is their any way for Monty to reliably communicate (via his choice of reveal, given their choice) that the player should stick (assuming he's trying to help the player). I certainly cannot contrive one.



  • @PalmerEldritch said:

    The highlighted part of your response is incorrect, the probability of winning by switching in the problem as stated only becomes 2/3 if Monty randomly picks a goat door when he has a choice of 2 goat doors

    So you keep saying, but since you have neither given a coherent explanation of how the formula you keep mentioning actually applies, nor identified a step at which my own reasoning goes wrong, I remain quite convinced you're mistaken.

    You can keep quoting Scripture or you can reason your way to enlightenment. Totally your call.



  • @flabdablet said:

    @eViLegion said:
    you can never be reduced below 1/2 chance by "always switching".
    Can you think of a scenario that includes PalmerEldritch's conditions 1 and 2 (mandatory reveal of a non-winning door, followed by an opportunity to switch) that can reduce your chance of winning by switching below 2/3? I haven't yet managed to, and although PalmerEldritch seems to be claiming that Bayesian Scripture says such a scenario must exist, he hasn't yet given a correct example.
     

    I've already given an example (several times). In the problem as stated (you pick Door1, Monty opens Door3) if the probability that Monty opens Door3 if the car is behind Door1 = 1 then there is no benefit in switching to Door2 . I even reposted the Bayesian analysis from Wikipedia that shows that to be the case.



  • @PalmerEldritch said:

    @flabdablet said:
    then the probability of winning by switching becomes 2/3 without needing your third condition; that is, your third condition is superfluous.
     

    The highlighted part of your response is incorrect, the probability of winning by switching in the problem as stated only becomes 2/3 if Monty randomly picks a goat door when he has a choice of 2 goat doors

     

    The selection between 2 goat doors only occurs if you happen to choose the car first. So it only kicks in (1/3 of the time) at the exact point when you should stick.

    All other times (2/3) that decision doesn't occur, so can make no difference to the final outcome... so ALL those other times you should switch.



    So at the absolute minimum, your expected win if switching is 2/3.



    It might be possible to transform your expected win if switching to somewhere between 2/3 and 3/3, if you know something about the process to decide between two doors,

    but ONLY if you can deduce that process DIDN'T happen (e.g. you didn't pick the car first time, and should therefore switch).



  • @PalmerEldritch said:

    I've already given an example (several times). In the problem as stated (you pick Door1, Monty opens Door3) if the probability that Monty opens Door3 if the car is behind Door1 = 1 then there is no benefit in switching to Door2 . I even reposted the Bayesian analysis from Wikipedia that shows that to be the case.

    No... the problem as stated says you can pick ANY door, and monty will reveal one of the other doors with a goat.


    "Pick door 1, Monty reveals 3" is just given as one example of what MIGHT happen during a possible game. You need to consider all other possible combinations.



    If you contrive to say that the "player MUST pick door 1, and Monty MUST reveal door 3, should the player now switch?" then you'll get a very different result, but it stops being The Monty Hall Problem.



  • @eViLegion said:

    @PalmerEldritch said:
    @flabdablet said:
    then the probability of winning by switching becomes 2/3 without needing your third condition; that is, your third condition is superfluous.
     

     

    The highlighted part of your response is incorrect, the probability of winning by switching in the problem as stated only becomes 2/3 if Monty randomly picks a goat door when he has a choice of 2 goat doors

     

    The selection between 2 goat doors only occurs if you happen to choose the car first. So it only kicks in (1/3 of the time) at the exact point when you should stick.

    All other times (2/3) that decision doesn't occur, so can make no difference to the final outcome... so ALL those other times you should switch.



    So at the absolute minimum, your expected win if switching is 2/3.



    It might be possible to transform your expected win if switching to 3/3, if you know something about the process to decide between two doors,

    but ONLY if you can deduce that process DIDN'T happen (e.g. you didn't pick the car first time, and should therefore switch).

     

    Put it this way: if you picked Door1 and I (as Monty) opened Door3 to reveal a goat, would you accept a less than even money bet (say $100 gets you $70) that the car is behind Door2? You shouldn't if you don't know the method by which I chose to open Door3.

     



  • @eViLegion said:

    @PalmerEldritch said:
    I've already given an example (several times). In the problem as stated (you pick Door1, Monty opens Door3) if the probability that Monty opens Door3 if the car is behind Door1 = 1 then there is no benefit in switching to Door2 . I even reposted the Bayesian analysis from Wikipedia that shows that to be the case.

     

    No... the problem as stated says you can pick ANY door, and monty will reveal one of the other doors with a goat.

     

    No, the problem as stated quite clearly says ""Suppose you're on a game show, and you're given the choice of 3 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?"

    And that is invariably how the Monty Hall Problem is presented

     



  • @PalmerEldritch said:

    Put it this way: if you picked Door1 and I (as Monty) opened Door3 to reveal a goat, would you accept a less than even money bet (say $100 gets you $70) that the car is behind Door2? You shouldn't if you don't know the method by which I chose to open Door3.

    But thats NOT the monty hall problem, as YOU have just selected door 1 on my behalf, knowing where the car is already.

    Also, if the problem states "monty opens door 3", then there is no process to decide... its already decided for him.



    I can therefore assume that you have put the car behind door 1 in order to coax me into switching and getting a goat.





    [flippant]I actually refuse to pick door 1, I'm picking door 3 instead, now I know you were planning on revealing a goat to me... then regardless of whatever door you pick, I'm switching[/flippant]



  • How do I "pick a door" if you're saying "you picked #1"? If you're telling me I picked #1, then I didn't pick it... you picked it.



    The word "say" in "say No. 1" and "say No. 3" means "for example" not "this is the way it has to happen every time".



    There is no interesting maths/probability in the problem as you're contriving to state it.



    So, basically, you're just trolling now, but putting too much effort into doing it for it to score very highly.



  • @eViLegion said:

    @PalmerEldritch said:
    Put it this way: if you picked Door1 and I (as Monty) opened Door3 to reveal a goat, would you accept a less than even money bet (say $100 gets you $70) that the car is behind Door2? You shouldn't if you don't know the method by which I chose to open Door3.

    But thats NOT the monty hall problem, as YOU have just selected door 1 on my behalf, knowing where the car is already.



    I can therefore assume that you have put the car behind door 1 in order to coax me into switching and getting a goat.





    I actually refuse to pick door 1, I'm picking door 3 instead, now I know you were planning on revealing a goat to me... then regardless of whatever door you pick, I'm switching

     

    It doesn't matter, pick any door you like and I'll open 1 of the other 2 doors to reveal a goat. I'll still make you the same bet. So, if you pick Door1 I'll always open Door2 if it contains a goat (and offer the bet that the car is behind Door3), if you pick Door2 I'll always open Door3 if it contains a goat (and offer the bet the car is behind Door1), if you pick Door3 I'll always open Door1 if it contains a goat (and offer the bet the car is behind Door2). You'll lose money, I can guarantee it.



  • @PalmerEldritch said:

    It doesn't matter, pick any door you like and I'll open 1 of the other 2 doors to reveal a goat. I'll still make you the same bet. So, if you pick Door1 I'll always open Door2 if it contains a goat (and offer the bet that the car is behind Door3), if you pick Door2 I'll always open Door3 if it contains a goat (and offer the bet the car is behind Door1), if you pick Door3 I'll always open Door1 if it contains a goat (and offer the bet the car is behind Door2). You'll lose money, I can guarantee it.




    OK:



    My policy: I shall always switch, regardless of if I can deduce that I should stick

    -----------------------------------------------------------------------------------------------------------------------------------



    Case A:

    I pick door 1, car behind door 1, your policy reveals door 2, I switch to 3 and lose.



    Case B:

    I pick door 1, car behind door 2, you MUST reveal door 3, I switch to 2 and win.



    Case C:

    I pick door 1, car behind door 3, you MUST reveal door 2, I switch to 3 and win.



    Case D:

    I pick door 2, car behind door 1, you MUST reveal door 3, I switch to 1 and win.



    Case E:

    I pick door 2, car behind door 2, your policy reveals door 3, I switch to 1 and lose.



    Case F:

    I pick door 2, car behind door 3, you MUST reveal door 2, I switch to 3 and win.



    Case G:

    I pick door 3, car behind door 1, you MUST reveal door 2, I switch to 1 and win.



    Case H:

    I pick door 3, car behind door 2, you MUST reveal door 1, I switch to 2 and win.



    Case I:

    I pick door 3, car behind door 3, your policy reveals door 1, I switch to 2 and lose.



    --------------------



    3 cases where always switching loses, still at least 6 cases where always switching wins.





    2/3 * ($70 + $100) = $113.33

    1/3 * ($0) = $0



    Add those up, and the expected win is $13.33 more than my original stake.



    Unless of course, you're suggesting I give you $100, and EVEN IF I WIN you give me less back. No-one would agree to that.



  • @eViLegion said:

    The word "say" in "say No. 1" and "say No. 3" means "for example" not "this is the way it has to happen every time".
    ...and not even "this is the way it did happen this time". That might be an easy thing for somebody whose first language isn't English to miss; wonder if that's what's going on with PalmerEldritch.



  • @PalmerEldritch said:

    It doesn't matter, pick any door you like and I'll open 1 of the other 2 doors to reveal a goat. I'll still make you the same bet. So, if you pick Door1 I'll always open Door2 if it contains a goat (and offer the bet that the car is behind Door3), if you pick Door2 I'll always open Door3 if it contains a goat (and offer the bet the car is behind Door1), if you pick Door3 I'll always open Door1 if it contains a goat (and offer the bet the car is behind Door2). You'll lose money, I can guarantee it.
    The only way you can contort the thing into an even-money bet is by refusing to make a bet if you're forced to open a door that isn't the one you'd be opening anyway per policy. But doing that is tantamount to denying your stated Condition 1 (Monty must in all cases offer the contestant an opportunity to switch doors, emphasis mine).



  • @PalmerEldritch said:

    'flabdablet' repeatedly said condition 3 was superfluous. I pointed out that if condition 3 is NOT met then the probability of winning the car by switching to Door2 is NOT = 2/3, and that if in fact Monty has a preference for opening Door3 whenever possible then there is no advantage in switching to Door2.

     

    Condition 3 is entirely superflous.

    See, you can only have 2 states when Monty opens a door. Either (1/3 probability) you picked the car already and Monty is choosing from 2 goat doors. In this case, Monty's choice of door, whether random or non-random is completely irrelevant because it has no effect on whether the player will win or lose by switching. In the alternative, (2/3 probability) you chose a goat door and Monty only has 1 choice of door to pick from anyway. Which is no choice at all, and thus also irrelevant. The method that Monty uses to pick his door does not and CANNOT matter, as long as the other conditions are met.

    What MAY be useful, in a non-probalistic sense is that if Monty has a set pattern for picking doors (say he pick door 3 all the time, unless it is a car and then is forced to pick another one) then a player can rig the game by choosing door 2 first and only switching if Monty picked door 1 (or vice versa). But that relies on the player having external knowledge that he can't possibly have.And it not relevant at all to the problem because it alters the choice completely from a choice of 3 equally random doors to only 2 doors (the non-Monty pick).



  • OK. I used Excel (what else) to model the behavior where Monty always picks Door 3 if he can i.e. if neither the player picks it, nor the car is behind it.

    For 1000 runs I get the following values for "switch wins"....

    690
    637
    665
    662
    675
    664
    667



  • @RTapeLoadingError said:

    OK. I used Excel (what else) to model the behavior where Monty always picks Door 3 if he can i.e. if neither the player picks it, nor the car is behind it.

    So did I (though not using Excel). However, you won't convince PalmerEldritch with that. He does not dispute the outcome of 2/3 for this model, but he insists on changing the problem by excluding 1/3 of the cases, those in which Monty cannot pick Door 3, in order to match a situation of his own invention and (I think) calling this situation the real MHP.


  • Discourse touched me in a no-no place

    @eViLegion said:

    we DO need some kind of 3rd condition, and we DO need it to appear to the player as random
    Technically not. You can make do with just using the axiom of choice to make the player unaware of our labelling of the doors as “1”, “2” and “3”. What's more, we get that by simply saying that the contestant always picks “1” and the placement of the prize is random. From Monty's perspective, at the only point where he has a choice in his actions, that choice does not matter. (Well, I'd guess that it is actually the show's production staff who make the choice, but that's a technical detail that doesn't affect the problem statement.)

    The funny thing is, if Monty is non-random it doesn't actually matter. The case where you would gain information about the prize location leads you to taking exactly the same decision that you'd be taking with the fair probabilistic analysis.

    It's much more important to assume that the initial placement of cars and goats is random (with uniform distribution).



  • @HardwareGeek said:

    He does not dispute the outcome of 2/3 for this model, but he insists on changing the problem by excluding 1/3 of the cases, those in which Monty cannot pick Door 3, in order to match a situation of his own invention and (I think) calling this situation the real MHP.

    It seems likely to me that the PalmerEldritch reasoning process goes like this:

    1. The wording of the problem statement specifically mentions doors 1 (player's pick), 3 (which Monty opens) and 2 (for evaluation of switch advantage).

    2. Therefore, what is required is a calculation of the probability that door 2 conceals the car, given that the player has picked door 1 and Monty has opened door 3.

    3. If and only if we know the probability that Monty would choose door 3 rather than door 2 when given the choice, we can calculate the required probability using Bayes's Theorem.

    4. Forum members who refuse to accept either the premise of this calculation or its result have read something into the problem description that isn't there.

    If that is in fact where he's coming from, it seems to me that step 2 is where he steers off into the weeds. Understanding the MHP requires noticing that the use of "say" in the problem as presented makes the door numbering completely arbitrary, from which it follows that any correct evaluation of the player's door-switch advantage cannot possibly depend on any specific numbering, from which it follows that the result any calculation that does depend on a specific numbering is inapplicable to the problem as presented. You have to generalize, then calculate.



  • Also, Han shot first.



  •  

    flabdablet:

    It seems likely to me that the PalmerEldritch reasoning process goes like this:

    1. The wording of the problem statement specifically mentions doors 1 (player's pick), 3 (which Monty opens) and 2 (for evaluation of switch advantage).

    2. Therefore, what is required is a calculation of the probability that door 2 conceals the car, given that the player has picked door 1 and Monty has opened door 3.

    3. If and only if we know the probability that Monty would choose door 3 rather than door 2 when given the choice, we can calculate the required probability using Bayes's Theorem.

    4. Forum members who refuse to accept either the premise of this calculation or its result have read something into the problem description that isn't there.

    That would be a fairly accurate interpretation of my position. Let me just briefly address each point:

    1. True, though which door numbers are used is immaterial. The problem statement is just an example of a single instance of the game

    2. In order to evaluate whether it is advantageous to switch doors, you need to calculate the conditional  probabilites of the car being behind the door you picked to begin with and the door you're (potentially) switching to,

    3. You need to know, or be able to estimate, the prior probabilities of all the variables involved in the problem in order to calculate the conditional probabilities of 2) using Bayes Theorem

    4.  Forum members who don't acknowledge that Bayes is the standard method for calculating conditional probability values are ignoring or disagreeing with a vast body of accepted probability theory

    flabdablet:
    If that is in fact where he's coming from, it seems to me that step 2 is where he steers off into the weeds.

    I don't see where youn get that from, how else can you tell if it's advantageous to switch doors?

    flabdablet:
    Understanding the MHP requires noticing that the use of "say" in the problem as presented makes the door numbering completely arbitrary, from which it follows that any correct evaluation of the player's door-switch advantage cannot possibly depend on any specific numbering, from which it follows that the result any calculation that does depend on a specific numbering is inapplicable to the problem as presented.

    You can rephrase the problem statement to remove all reference to specific doors, and everything I've said to date still holds true. If you like phrase the problem as follows:

    "Suppose you're on a game show, and you're given the choice of 3 doors. Behind one door is a car; behind the others, goats. You pick a door and the host, who knows what's behind all the doors, opens one of the two doors which you didn't pick and reveals a goat. He then says to you, "Do you want to stay with your 1st pick or switch to the door I didn't open?" Is it to your advantage to switch your choice?"

    You now have: 'The Door you Picked', 'The Door Monty Opened', and 'The Door Monty didn't Open' . In order to evaluate whether it's advantageous to switch doors you need to calculate the conditional probability of both 'The Door you Picked' and ''The Door Monty didn't Open''. To do this you use Bayes Theory which isn't dependent on any specific door numbering.

     

    flabdablet:
    You have to generalize, then calculate

    I'm not sure what you mean by this sentence..



  • @PalmerEldritch said:

    If you like
    phrase the problem as follows:

    "Suppose you're on a game show,
    and you're given the choice of 3 doors.
    Behind one door is a car; behind the others, goats. You pick a door and
    the host, who knows what's behind all the doors, opens one of the two
    doors which you didn't pick and reveals a goat. He then says to you, "Do
    you
    want to stay with your 1st pick or switch to the door I didn't open?" Is
    it to your advantage to switch your choice?"

    You now have:
    'The Door you Picked', 'The Door Monty Opened', and 'The Door Monty
    didn't Open' . In order to evaluate whether it's advantageous to switch
    doors you need to calculate the conditional probability of both 'The
    Door you Picked' and ''The Door Monty didn't Open''. To do this you use
    Bayes Theory which isn't dependent on any specific door numbering.

    I'd like to see you work that calculation step by step, if you don't mind.



  • @dkf said:

    It's much more important to assume that the initial placement of cars and goats is random (with uniform distribution).

    I just imagined Monty Hall distributing some kind of uniform to the goats for them to wear. This would immeasurably improve the game.



  • @flabdablet said:

    ]The only way you can contort the thing into an even-money bet is by refusing to make a bet if you're forced to open a door that isn't the one you'd be opening anyway per policy. But doing that is tantamount to denying your stated Condition 1 (Monty must in all cases offer the contestant an opportunity to switch doors, emphasis mine).
     

    You got that right. Why would I offer a bet in a situation where you are guaranteed (100%) to win and I'm guaranteed (100%) to lose. If I'm forced to open a door that isn't the one I'd open according to my policy I'll still open it (so conditions 1 and 2 are still satisfied) I'm just not prepared to bet on the outcome.

    So you seem to acknowledge that in the 1/3rd of all cases where I refuse to make a bet ,switching wins 100% of the time. Since the average chance of winning over multiple trials by always switching  is 66.67% (which we both agree on) then in the 2/3rd of all cases where I do make a bet switching must win only 50% of the time? (Since [100% * 1/3] + [50% * 2/3] = 66.67%)



  • @flabdablet said:

    @PalmerEldritch said:

    If you like phrase the problem as follows:

    "Suppose you're on a game show, and you're given the choice of 3 doors. Behind one door is a car; behind the others, goats. You pick a door and the host, who knows what's behind all the doors, opens one of the two doors which you didn't pick and reveals a goat. He then says to you, "Do you want to stay with your 1st pick or switch to the door I didn't open?" Is it to your advantage to switch your choice?"

    You now have: 'The Door you Picked', 'The Door Monty Opened', and 'The Door Monty didn't Open' . In order to evaluate whether it's advantageous to switch doors you need to calculate the conditional probability of both 'The Door you Picked' and ''The Door Monty didn't Open''. To do this you use Bayes Theory which isn't dependent on any specific door numbering.

    I'd like to see you work that calculation step by step, if you don't mind.
     

    Mmmm...... I thought you might ask that :)  I'll do it but it'll take me a bit of time as it's a bit more complicated than the Bayes solution presented earlier from the Wikipedia website.

    BTW, if I came across yesterday as rude or arrogant I apologise. I'm enjoying this discussion, and as I work through my arguments I understand more about the intracies and variations inherent in what appears, at first at least, to be a simple probability problem .

     

     


  • ♿ (Parody)

    @PalmerEldritch said:

    "Suppose you're on a game show,
    and you're given the choice of 3 doors.
    Behind one door is a car; behind the others, goats. You pick a door and
    the host, who knows what's behind all the doors, opens one of the two
    doors which you didn't pick and reveals a goat. He then says to you, "Do
    you
    want to stay with your 1st pick or switch to the door I didn't open?" Is
    it to your advantage to switch your choice?"

    You now have:
    'The Door you Picked', 'The Door Monty Opened', and 'The Door Monty
    didn't Open' . In order to evaluate whether it's advantageous to switch
    doors you need to calculate the conditional probability of both 'The
    Door you Picked' and ''The Door Monty didn't Open''. To do this you use
    Bayes Theory which isn't dependent on any specific door numbering.

    The great thing about doing things with math is that there is often more than one way to do it, and it can be interesting to consider them to get a better understanding of what's going on.

    Before Monty opened a door, I knew there was a 2/3 chance it was behind one of the two doors I didn't pick. That's still true after Monty reveals the goat. It doesn't really require any explicit calculations to note that and correctly state the value of switching. You could represent all of that using things like Bayes Theorem, I guess, but it's not worth it for this relatively simple problem.



  • @boomzilla said:

    The great thing about doing things with math is that there is often more than one way to do it, and it can be interesting to consider them to get a better understanding of what's going on.

    Before Monty opened a door, I knew there was a 2/3 chance it was behind one of the two doors I didn't pick. That's still true after Monty reveals the goat. It doesn't really require any explicit calculations to note that and correctly state the value of switching. You could represent all of that using things like Bayes Theorem, I guess, but it's not worth it for this relatively simple problem.

     

    It depends on what you mean by "relatively simple". Consider the Ignorant Monty variation of the problem..Before Monty opens a door at random you know there's a 1/3 chance that the car is behind the door you picked first. In 2/3rd of all the cases (and that's the vast majority of cases) where Monty randomly picks a door and reveals it's a goat, the chance that the car is behind the door you picked to begin with has increased from 1/3 to 1/2.  Bayes Theory easily explains why that is, although "intuitively" it doesn't make sense.

     

     

     



  • @PalmerEldritch said:

    the average chance of winning over multiple trials by always switching  is 66.67% (which we both agree on)

    Do we also now agree that the always-switch strategy is guaranteed a better win rate over multiple trials than any strategy involving keeping, regardless of Monty's method for choosing which non-winning door to open in the 1/3 of all cases where such a choice is needed, and regardless of the player's knowledge about any such method?



  • @PalmerEldritch said:

    In 2/3rd of all the cases (and that's the vast majority of cases) where Monty randomly picks a door and reveals it's a goat, the chance that the car is behind the door you picked to begin with has increased from 1/3 to 1/2.

    No. In 2/3 of all the cases, Monty doesn't get an opportunity to pick a door at random, because that 2/3 of all the cases are the ones where the contestant's initial pick has already used up one of the goat doors and Monty has no option but to reveal the other. Those and only those are the cases where the contestant will win by switching.

    You are of course quite free to mis-apply conditional probability formulae to convince yourself that some other result matters, if that pleases you.



  • @flabdablet said:

    @PalmerEldritch said:
    the average chance of winning over multiple trials by always switching  is 66.67% (which we both agree on)

    Do we also now agree that the always-switch strategy is guaranteed a better win rate over multiple trials than any strategy involving keeping, regardless of Monty's method for choosing which non-winning door to open in the 1/3 of all cases where such a choice is needed, and regardless of the player's knowledge about any such method?

     

    I''d already agreed that was the case many posts ago. I also said (many posts ago) that the question "Is it to your advantage to switch to Door2" is  different to the question "(over multiple trials) what is the proportion of games that you win if you always switch?". On average you'll win 2/3rd of the time. The thing is you don't get to play multiple times. The way the MHP is always phrased is "you're on a gameshow..." and not "you on a game show 3,000 times...". In the Ignorant Monty version you'll win 1/3 of the time on average by always sticking, but if Monty reveals a goat door you've got a 50% chance of wining by sticking.

    Do you agree, that in scenario I posted where I make a bet (or not) that in 2/3rd of all cases you only win 50% of the time by switching?

     

     



  • @flabdablet said:

    @PalmerEldritch said:
    In 2/3rd of all the cases (and that's the vast majority of cases) where Monty randomly picks a door and reveals it's a goat, the chance that the car is behind the door you picked to begin with has increased from 1/3 to 1/2.

    No. In 2/3 of all the cases, Monty doesn't get an opportunity to pick a door at random, because that 2/3 of all the cases are the ones where the contestant's initial pick has already used up one of the goat doors and Monty has no option but to reveal the other. Those and only those are the cases where the contestant will win by switching.

    You are of course quite free to mis-apply conditional probability formulae to convince yourself that some other result matters, if that pleases you.

     

    I was talking, in that particular post, about the Ignorant Monty variation, where Monty doesn't know where the car is and opens a door at random. If you're going to edit my posts don't leave out the important bits.

     



  • @flabdablet said:

    @PalmerEldritch said:
    In 2/3rd of all the cases (and that's the vast majority of cases) where Monty randomly picks a door and reveals it's a goat, the chance that the car is behind the door you picked to begin with has increased from 1/3 to 1/2.

    No. In 2/3 of all the cases, Monty doesn't get an opportunity to pick a door at random, because that 2/3 of all the cases are the ones where the contestant's initial pick has already used up one of the goat doors and Monty has no option but to reveal the other. Those and only those are the cases where the contestant will win by switching.

    You are of course quite free to mis-apply conditional probability formulae to convince yourself that some other result matters, if that pleases you.

    Kindly disregard. I had not noticed the fact that you were talking about the "ignorant Monty" variant.

    Has to be said, though, that people who use the phrase "vast majority" in a discussion about conditional probability risk damaging the credibility of their arguments.



  •  Well ignore that phrase if you wish. I'd have thought that saying 66.67% represented a "vast majority" wasn't an exagerration, but ....


  • ♿ (Parody)

    @PalmerEldritch said:

    It depends on what you mean by "relatively simple". Consider the Ignorant Monty variation of the problem..Before Monty opens a door at random you know there's a 1/3 chance that the car is behind the door you picked first. In 2/3rd of all the cases (and that's the vast majority of cases) where Monty randomly picks a door and reveals it's a goat, the chance that the car is behind the door you picked to begin with has increased from 1/3 to 1/2.  Bayes Theory easily explains why that is, although "intuitively" it doesn't make sense.

    "Intuitively," it does to me! Most hated words in any math class: "It is therefore obvious that..." Opening that door doesn't reduce the 2/3 probability for the two doors.


  • ♿ (Parody)

    @PalmerEldritch said:

    I''d already agreed that was the case many posts ago. I also said (many posts ago) that the question "Is it to your advantage to switch to Door2" is  different to the question "(over multiple trials) what is the proportion of games that you win if you always switch?". On average you'll win 2/3rd of the time. The thing is you don't get to play multiple times. The way the MHP is always phrased is "you're on a gameshow..." and not "you on a game show 3,000 times...". In the Ignorant Monty version you'll win 1/3 of the time on average by always sticking, but if Monty reveals a goat door you've got a 50% chance of wining by sticking.

    It hurts to see someone so happy to tout Bayes making fallacious frequentist arguments. Or did we acquire our second long term MHP troll in the thread?



  • @PalmerEldritch said:

    I also said (many posts ago) that the question "Is it to your advantage to switch to Door2" is  different to the question "(over multiple trials) what is the proportion of games that you win if you always switch?".

    It is not, however, different to the question "(over multiple trials) can you reasonably expect switching doors to win more often than not?"

    @PalmerEldritch said:

    On average you'll win 2/3rd of the time. The thing is you don't get to play multiple times.

    For somebody so willing to pontificate about the vast body of probability theory, your grasp of the idea of the expected return seems surprisingly poor.

    Here's a very simple non-MHP-like first-principles game for you to consider. You call "high" or "low", then I roll a fair die. You win a prize only if you called "low" and the die shows 1 or 2, or you called "high" and the die shows 3, 4, 5 or 6. Is it to your advantage to call "high"? Why or why not?

    @PalmerEldritch said:

    Do you agree, that in scenario I posted where I make a bet (or not) that in 2/3rd of all cases you only win 50% of the time by switching?

    Sure. And in the other 1/3 of cases I win 100% of the time by switching. So what? That scenario is not the same as the MHP setup, where Monty doesn't get to choose whether or not to "take the bet".

    @PalmerEldritch said:

    In the Ignorant Monty version you'll win 1/3 of the time on average by always sticking, but if Monty reveals a goat door you've got a 50% chance of wining by sticking.

    Again: so what? This is a fact about some game other than the one specified in the MHP; a game, furthermore, that needs a new rule to deal with what happens when Monty opens the door with the car. Is it an instant win? An instant loss? Do the doors get reshuffled and the game restarted? Does the audience vote on calling in an air strike? Inquiring minds must know.



  • @PalmerEldritch said:

    You now have:
    'The Door you Picked', 'The Door Monty Opened', and 'The Door Monty didn't Open' .
    In order to evaluate whether it's advantageous to switch doors you need to calculate the conditional probability of both 'The Door you Picked' and ''The Door Monty didn't Open''.

    To make that work surely you also need to calculate the probability that Monty would open the door he opened, given you picked the door you picked? Since Monty can't open the door you picked....



  • @PalmerEldritch said:

    I'd have thought that saying 66.67% represented a "vast majority" wasn't an exagerration
    I don't know what "vast majority" means to you; to me it suggests that the person using the phrase believes that minority cases occur negligibly often. In a discussion about probability, cases with p < 1/2 still count.


  • Discourse touched me in a no-no place

    @boomzilla said:

    "Intuitively," it does to me! Most hated words in any math class: "It is therefore obvious that..."
    Mathematics uses a non-standard definition of “intuitive” and “obvious”, as reflected in the old story about the math professor who said in a lecture “It is therefore obvious that…” before trailing off, picking up a chalk, writing on the blackboard very rapidly and densely for about half an hour, and then resuming “Yes. It is indeed obvious that…”


  • Discourse touched me in a no-no place

    @flabdablet said:

    Has to be said, though, that people who use the phrase "vast majority" in a discussion about conditional probability risk damaging the credibility of their arguments.
    That depends on whether you're talking about inhabitants of trailer parks in Alabama.


  • ♿ (Parody)

    @dkf said:

    @boomzilla said:
    "Intuitively," it does to me! Most hated words in any math class: "It is therefore obvious that..."

    Mathematics uses a non-standard definition of “intuitive” and “obvious”, as reflected in the old story about the math professor who said in a lecture “It is therefore obvious that…” before trailing off, picking up a chalk, writing on the blackboard very rapidly and densely for about half an hour, and then resuming “Yes. It is indeed obvious that…”

    Yep. In this case, however, it's pretty simple, even though many people have a difficult time seeing it. It can be explained in plain language and easily grasped by the mathematically inept. The fact that people can distract themselves with irrelevancies doesn't change any of that.



  • @PalmerEldritch said:

    @flabdablet said:

    @PalmerEldritch said:

    If you like
    phrase the problem as follows:

    "Suppose you're on a game show,
    and you're given the choice of 3 doors.
    Behind one door is a car; behind the others, goats. You pick a door and
    the host, who knows what's behind all the doors, opens one of the two
    doors which you didn't pick and reveals a goat. He then says to you, "Do
    you
    want to stay with your 1st pick or switch to the door I didn't open?" Is
    it to your advantage to switch your choice?"

    You now have:
    'The Door you Picked', 'The Door Monty Opened', and 'The Door Monty
    didn't Open' . In order to evaluate whether it's advantageous to switch
    doors you need to calculate the conditional probability of both 'The
    Door you Picked' and ''The Door Monty didn't Open''. To do this you use
    Bayes Theory which isn't dependent on any specific door numbering.

    I'd like to see you work that calculation step by step, if you don't mind.
     

    Mmmm...... I thought you might ask that :)  I'll do it but it'll take me a bit of time as it's a bit more complicated than the Bayes solution presented earlier from the Wikipedia website.

    I don't believe it's complicated at all.

    The Bayes rule is P(B given A) = P(A given B) × P(B) ÷ P(A); this follows directly from the conditional probability formula, P(A given B) × P(B) = P(A).

    If we're determined to apply the Bayes rule to the MHP, and we're convinced that we know P(monty-opens-some-given-door given monty-has-a-choice-of-doors), then it's only natural to assume that using Bayes to calculate P(monty-has-a-choice-of-doors given monty-opens-some-given-door) will help us work out P(monty-has-a-choice-of-doors), which will give us the chance that our initial pick has deprived Monty of a choice by using up one of the two available goats, which will tell us whether we improve our expected return by switching vs. staying.

    In order to do that, as well as P(monty-opens-some-given-door given monty-has-a-choice-of-doors) which we're convinced we have, we're going to need P(monty-opens-some-given-door) and P(monty-has-a-choice-of-doors). Well, we don't have the first one yet so maybe we can work it out from the other two? Ah! The conditional probability formula tells us that P(monty-opens-some-given-door) = P(monty-opens-some-given-door given monty-has-a-choice-of-doors) × P(monty-has-a-choice-of-doors). Good. Looks like we're getting somewhere. And we're now in a position to apply Bayes:

    P(monty-has-a-choice-of-doors given monty-opens-some-given-door) = P(monty-opens-some-given-door given monty-has-a-choice-of-doors) × P(monty-has-a-choice-of-doors) ÷ P(monty-opens-some-given-door)
    = P(monty-opens-some-given-door given monty-has-a-choice-of-doors) × P(monty-has-a-choice-of-doors) ÷ [P(monty-opens-some-given-door given monty-has-a-choice-of-doors) × P(monty-has-a-choice-of-doors)]
    = 1
    WTF?

    And it's at this point that we're pretty much forced to step back from Bayes, having realized that (a) all it's telling us is that Monty's having a choice of doors means that he can open one of them and (b) we only wanted P(monty-has-a-choice-of-doors given monty-opens-some-given-door) in order to work out P(monty-has-a-choice-of-doors), which we can just do directly anyway.

    @PalmerEldritch said:

    BTW, if I came across yesterday as rude or arrogant I apologise. I'm enjoying this discussion, and as I work through my arguments I understand more about the intracies and variations inherent in what appears, at first at least, to be a simple probability problem.

    Don't you worry about that. "Rude and arrogant" is the cultural norm here.



  • @flabdablet said:

    I don't believe it's complicated at all.

     <long complicated probability discussion>

     

    ;)

    Amusingly, the real mistake is thinking that you need to calculate any probability [i]after you've chosen the door[/i]. You don't. It stops being a probability problem the minute you make your choice, because at that point the outcome is fixed: at that point you've either picked a goat or a car, so the probability of switching wins is already either 0 or 1. Everything that happens after you pick a door is - as far as probabilityis concerned - irrelevant.

     And that's why the problem isn't complicated: you only need to calculate one probability.


  • ♿ (Parody)

    @JimM said:

    Amusingly, the real mistake is thinking that you need to calculate any probability after you've chosen the door. You don't. It stops being a probability problem the minute you make your choice, because at that point the outcome is fixed: at that point you've either picked a goat or a car, so the probability of switching wins is already either 0 or 1. Everything that happens after you pick a door is - as far as probabilityis concerned - irrelevant.

    And that's why the problem isn't complicated: you only need to calculate one probability.

    False. It's true that you either picked a car or you didn't, but you still don't know which, so non-trivial probabilities are still involved. All probabilities are conditional, and you don't get to use the knowledge of whether the car is there or not until you actually know that.

    Can I offer you a ride to my next poker game?



  • @boomzilla said:

    False. It's true that you either picked a car or you didn't, but you still don't know which, so non-trivial probabilities are still involved. All probabilities are conditional, and you don't get to use the knowledge of whether the car is there or not until you actually know that.

    Not really. The probability is 1/3 that you picked a car, 2/3 that you didn't. The opening of a door doesn't change that. There's still a 1/3 probability that you picked a car, and a 2/3 probability that you didn't. The only thing that changes is that you know that there is 0 probability that it's behind the open door. It changes the distribution of the probability, but not the probability itself. Perhaps I should have phrased that "you don't need to calculate [i]the probability of anything that happens[/i] after you've chosen the door", but I thought that was kind of obvious (the real obvious, not the mathematical one ;) ). The only probability that matters is the one of whether you originally chose the car or not.

    @boomzilla said:

     Can I offer you a ride to my next poker game?

     

    No, but not for the reason you think.

     


  • ♿ (Parody)

    @JimM said:

    Perhaps I should have phrased that "you don't need to calculate the probability of anything that happens after you've chosen the door", but I thought that was kind of obvious (the real obvious, not the mathematical one ;) ). The only probability that matters is the one of whether you originally chose the car or not.

    Ah. Yes, this is correct.


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