tag:blogger.com,1999:blog-5944248932558389199.post585185924093554369..comments2024-02-15T19:40:29.872-08:00Comments on neopolitan's philosophical blog: Saving Monty Fallneopolitanhttp://www.blogger.com/profile/02501854905476808648noreply@blogger.comBlogger11125tag:blogger.com,1999:blog-5944248932558389199.post-16030226956009054432015-04-16T04:46:19.401-07:002015-04-16T04:46:19.401-07:00Hey this blog is great. Your analysis is right and...Hey this blog is great. Your analysis is right and I think the table you calculated is correct. We could call this variant of the game Fair Monty Hall. The solution is simply that there are two strategies, and both with a winning likehood of 1/6+1/6 = 1/3, so in this game you have the same winning probability as in the simple three door game (you pick one door and if it contains a car you win). This happens because the AutoLoss columns sum up to 1/6+1/6=1/3 as well. <br />I think this is the most exact way of saving Monty Fall, which is itself logically equivalent to Monty Hall (in which is stipulated that a car is never revealed). <br /><br /><br />FBAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-5944248932558389199.post-22454374752066611272015-03-13T23:31:22.773-07:002015-03-13T23:31:22.773-07:00Perhaps I wasn't clear earlier. My point is th...Perhaps I wasn't clear earlier. My point is that the entire difficulty of these kind of problems is translating the "word problem" into a mathematical model. Once you have a mathematical model, it's piece of cake. So the only thing people should be discussing is how to interpret the sentence.<br /><br />> This does make a difference if he intended to open one door but accidently opened another, because that means the other goat IS behind the door he intended to open in the first place.<br /><br />With this interpretation, your analysis is correct.<br /><br />But I did not interpret the sentence like that (and I still don't really understand how one can interpret it as you did). For me the sentence "accidentally pushes open another door" unambiguously means that a random event (the banana) makes him push open a door. I see the word "accidentally" as a proof that the banana peel was randomly placed in front of one of the doors, and hence it could have been any door (the door that the contestant selected or the door with the car). It turns out this door is "another door" (meaning a door other from the one the contestant already chose), and it also turns out that the door does not contain a car. <br /><br />Apparently, Pynes interprets the sentence differently. He says that somehow Monty is forced to push open a door that does not contain a car (for me, that's completely absurd). And you just interpreted the sentence in another way, saying that Monty wanted to open a goat door but accidentally open the other door because of the banana peel (that's not absurd, but it's really far-fetched). I would say that this is more an english problem than a mathematical problem ... <br /><br />Perhaps the fact that I'm used to read similar problems makes me interpret sentences in a very specific way. What I mean is the following :<br />I interpret Rosenthal's formulation to mean "we are in situation B", because I'm thinking that if the guy actually wanted to specify a situation that is not situation B, he would have formulated the problem very differently. <br />For example, if Rosenthal wanted to describe a situation where Monty was going to open a goat door but on his way he slips and open another door, then he would say something like :<br />"on his way to open a door as usual, Monty slips and accidentally pushes open THE OTHER door."<br />The fact that he did not mention that Monty had the intention to open a door, and the fact that he just says "another door", makes me believe that my interpretation is correct and that is what Rosenthal had in mind.Mathematiciannoreply@blogger.comtag:blogger.com,1999:blog-5944248932558389199.post-13128688491081685232015-03-13T15:32:32.186-07:002015-03-13T15:32:32.186-07:00I was not aware of this statement in Monty fall;
...I was not aware of this statement in Monty fall;<br /> "the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car"<br /><br />This does make a difference if he intended to open one door but accidently opened another, because that means the other goat IS behind the door he intended to open in the first place. This means he either opened the contestants door or the other door of the remaining 2 unpicked doors.In either case he has now revealed the location of the car in general terms on his side or Monty's side), if the contestant knows that Monty does a reveal in the game. If the contestant is unaware that Monty was going to perform a reveal on one of the two remain unpicked doors and reveal a goat, only then from not having knowledge are the choices 50/50. There is a significant difference from "falling on a door other than one of he goat doors" than a random pick. The math is entirely different as far as possibilities go, and far more simple. At this stage I am thinking Monty Fall is really a nothing when trying to understand it conceptually.KaySennoreply@blogger.comtag:blogger.com,1999:blog-5944248932558389199.post-18958516664143123802015-03-12T20:24:26.148-07:002015-03-12T20:24:26.148-07:00KaySen, I draw your attention to xor7486's com...KaySen, I draw your attention to <a href="http://www.reddit.com/r/math/comments/2ysbzq/saving_monty_fall_if_there_are_two_equally/cpcupkc" rel="nofollow">xor7486's comment</a>. The simulation has already been done.neopolitanhttps://www.blogger.com/profile/02501854905476808648noreply@blogger.comtag:blogger.com,1999:blog-5944248932558389199.post-35926637127441984682015-03-12T20:06:29.361-07:002015-03-12T20:06:29.361-07:00Apparently, you understand that there is a huge fu...Apparently, you understand that there is a huge fundamental difference between the two following scenarios :<br /><br />A - I consider one possible situation where Monty opens a goat door, in a Universe where the rules forces Monty to open a goat door.<br />B - I consider one possible situation where Monty opens a goat door, in a Universe where the opening of the door was random. <br /><br /><br />As you said, in scenario A, the odds are 2/3 - 1/3. And in scenario B, the odds are 1/2 - 1/2.<br /><br /><br />The fact that we are in a situation where Monty did open a goat door, does not mean that in the Universe (in the mathematical sense) it is impossible for Monty to open the car door. In other words, just looking at the situation, we cannot deduce in what scenario we are. <br /><br />So to understand in which scenario (Universe) we are, we have to read carefully the formulation of the Monty Fall problem : <br /><br />> the host slips on a banana peel and accidentally pushes open another door, which just happens not to contain the car.<br /><br />So, for me, this is an unambiguous statement that the fact that there was a goat behind the door was the result of luck, and not of the rules of the game. It means that Monty could also have opened the car door accidentally. So, to me it's perfectly clear that Monty Fall corresponds to scenario B. I can see no reasonable interpretation of the word "accidentally" that would somehow indicate that the "choice" of which door is open is not uniformly random.<br /><br /><br />Apparently you (and also Pynes) disagree with this analysis of the sentence. But I don't understand why ... What makes you think "Monty always falls on a goat door" in the Monty Fall scenario ? I don't understand how the phrasing of the problem could make you (and Pynes) think that.<br />Mathematiciannoreply@blogger.comtag:blogger.com,1999:blog-5944248932558389199.post-10794960564104547842015-03-12T18:28:19.878-07:002015-03-12T18:28:19.878-07:00If you ran an actual simulation on Monty fall, it ...If you ran an actual simulation on Monty fall, it will show the exact same results as Monty Hall. Because Monty always falls on a goat door. If in some simulations he can fall on a car door, then it will be 50/50. Run the simulation yourself and see the results where monty falls on a goat door every time.KaySennoreply@blogger.comtag:blogger.com,1999:blog-5944248932558389199.post-37458980798239638102015-03-12T15:38:42.909-07:002015-03-12T15:38:42.909-07:00With respect to your first paragraph, I draw your ...With respect to your first paragraph, I draw your attention to <a href="http://www.reddit.com/r/math/comments/2ysbzq/saving_monty_fall_if_there_are_two_equally/cpcrysd" rel="nofollow">Vietoris' comment</a> - he argues that Pynes simply got it wrong.neopolitanhttps://www.blogger.com/profile/02501854905476808648noreply@blogger.comtag:blogger.com,1999:blog-5944248932558389199.post-27313744089971019512015-03-12T15:34:31.719-07:002015-03-12T15:34:31.719-07:00In a sense, the perspectives are sort of right tho...In a sense, the perspectives are sort of right though - even if the final conclusion of each is wrong. Winning as a consequence of both switching and staying <strong>is</strong> more likely, but only to the extent that both options move from 1/3 to 1/2 in likelihood. I think we tend to fill in the gaps with what makes more sense to us, making a statement more precise even if by doing so we make it less accurate. In this case we turn what I wrote:<br /><br />"it’s more likely that the car is behind the remaining unselected and unopened door"<br /><br />into<br /><br />"there’s a greater likelihood of the car being behind the remaining unselected and unopened door than of it being behind the selected door (2/3 vs 1/3)"<br /><br />rather than<br /><br />"the likelihood of the car being behind the remaining unselected and unopened door in now greater than it was originally (1/2 rather than 1/3)"<br /><br />Perhaps this is where the gut feeling comes from, especially when we are familiar with Monty Hall.<br /><br />I know that you are a mathematician and you have worked through this dispassionately, however, are you really entirely free of gut-feelings? Do you not have any intuition as to what would be the best option, despite knowing what probabilities tell you?<br /><br />Finally, I think that the sense that there are two equal but incompatible likelihoods is a very good hint that something has been left out in one's considerations. It's reassuring that, now at least, I've got it right with my computations.neopolitanhttps://www.blogger.com/profile/02501854905476808648noreply@blogger.comtag:blogger.com,1999:blog-5944248932558389199.post-83584294813585925322015-03-12T15:07:43.233-07:002015-03-12T15:07:43.233-07:00Sorry can't edit where I accidently pasted the...Sorry can't edit where I accidently pasted the first paragraph to the middle :(KaySennoreply@blogger.comtag:blogger.com,1999:blog-5944248932558389199.post-72712887167895998972015-03-12T14:28:26.257-07:002015-03-12T14:28:26.257-07:00In a game where it is possible for Monty to reveal...In a game where it is possible for Monty to reveal a car then conditional probability applies and the chances are 50/50. Where there is no possibility he can reveal a car then the 2/3 applies on a switch. In Monty Fall he always falls on a goat, so it is the same as classic Monty. However your new game is 50/50 because he can reveal a car. It does not matter what strategy the contestant takes, switching or staying yields a 50/50 chance. Not being a math person, this is how I work out the probability. first move is to list he possible outcomes;<br /><br />Where the contestant always picks door 1. ( ) represents which door Monty opens being 2 or 3 and ** is the location of the car<br /><br />a: *1* (2) 3<br />b. *1* 2 (3)<br />c: 1 (*2*) 3<br />d: 1 *2* (3) <br />e: 1 (2) *3* <br />f: 1 2 (*3*)<br /> <br />A. and B. become compressed due to Monty being able o pick either one and only reveal a goat. <br /><br />In a game where it is possible for Monty to reveal a car then conditional probability applies and the chances are 50/50. Where there is no possibility he can reveal a car then the 2/3 applies on a switch. In Monty Fall he always falls on a goat, so it is the same as classic Monty. However your new game is 50/50 because he can reveal a car. It does not matter what strategy the contestant takes, switching or staying yields a 50/50 chance. Not being a math person, this is how I work out the probability. first move is to list he possible outcomes;<br /><br />Where the contestant always picks door 1. ( ) represents which door Monty opens being 2 or 3 and ** is the location of the car<br /><br />a: *1* (2) (3)<br />c: 1 (*2*) 3<br />d: 1 *2* (3) <br />e: 1 (2) *3* <br />f: 1 2 (*3*)<br /><br />So now to play it out. Monty randomly picks door 2 and there is a goat. So eliminate all possibilities where he picks door 3 and also the possibility there is a car behind door 2. his is what is left.<br /><br />In a game where it is possible for Monty to reveal a car then conditional probability applies and the chances are 50/50. Where there is no possibility he can reveal a car then the 2/3 applies on a switch. In Monty Fall he always falls on a goat, so it is the same as classic Monty. However your new game is 50/50 because he can reveal a car. It does not matter what strategy the contestant takes, switching or staying yields a 50/50 chance. Not being a math person, this is how I work out the probability. first move is to list he possible outcomes;<br /><br />Where the contestant always picks door 1. ( ) represents which door Monty opens being 2 or 3 and ** is the location of the car<br /><br />a: *1* (2) (3)<br />e: 1 (2) *3* <br /><br />You have a 1 in 2 chance of having the car and Monty has a 1 in 2 chance of having he car.<br /><br />Just thought I would list how I work it out, it may give it another conceptual outlook.KaySennoreply@blogger.comtag:blogger.com,1999:blog-5944248932558389199.post-77349443501757859462015-03-12T08:48:45.190-07:002015-03-12T08:48:45.190-07:00As far as I can tell, the computations in your tab...As far as I can tell, the computations in your table are perfectly correct. Which means that the question :<br /><br />> what is the best strategy for the contestant?<br /><br />should be easy to answer . You just have to look at the table :<br /><br />Strategy 1, not taking the "risk" : 1/3 chances of winning.<br /><br />Strategy 2, taking the risk and switching if possible : 1/3 chances of winning<br /><br />Strategy 3, taking the risk and staying if possible : 1/3 chances of winning<br /><br />All three strategies are completely equivalent, probabilitiy-wise. <br /><br />> This means that the likelihood of benefitting as a result of switching, after surviving the coin toss, would be equal to the likelihood of benefitting as a result of not switching. In which case, there was no point in going through the charade, was there?<br /><br />Exactly.<br /><br /><br /><br />Now concerning Perspective one and two, they are in a nutshell exactly the same kind of mistake you did before. Namely that you take into account only part of the situation, and you forget to take into account that probabilities change when something is revealed (that's the point of conditional probabilities).<br /><br />In perspective one, you only look at the choice of the contestant. You say that apriori there is a 1/3 chance that the car is behind the selected door. That's true. But once a door is opened revealing a goat, this probability goes to 1/2. This is just a simple computation of conditional probabilities :<br /><br />P(selected door contains car | open door contains a goat) =(1/3)/(2/3) = 1/2.<br /><br />In perspective two, you only look at the placement of the car. You say that it's twice more likely to reveal a goat if the car is behind the selected door. That's true. But it's twice more likely to have a goat behind the selected door, so the two things cancel out. And you can use Bayes theorem to compute the same probability :<br /><br /><br />P(selected door contains car | open door contains a goat) = P(open door contains a goat | selected door contains car) * P(selected door contains car) / P(open door contains a goat ) = 1*(1/3) / (2/3) = 1/2<br /><br /><br />> I do recognise that the probability calculations don’t seem to support this gut feeling but perhaps someone has some better probability calculations.<br /><br />As your calculations are right, it means that your gut feeling is wrong. I don't have better probability calculations to offer, as your computations are already correct and complete. Mathematiciannoreply@blogger.com