Flash

15226. Wed Feb 09, 2005 11:35 am 


Is this trick question very old hat? I hadn't encountered it until today. It's apparently called the Monty Hall Dilemma:
There are three doors, and behind one of them is a nice surprise. The contestant nominates one door, and the host then opens one of the others, revealing that there is nothing behind it. The question is, which door should the contestant open if he wants to find the nice surprise? The one he chose originally? The third door? Or is it a 50/50 call?
The intuitive response is that it's a 50/50 call, but that's wrong. The right strategy is to choose whichever door you didn't choose the first time, and to see why you scale the question up. Say there are 100 doors. The chance that you nominate the right one the first time is 1/100. In 99 cases out of 100 you'll choose the wrong one and the host will open up the 98 other wrong ones and leave the right one shut  in other words, 99 times out of 100 he'll tell you the right answer and 1 time out of 100 he'll tell you the wrong answer  so you should go with the clue he's giving you, not your original (oddsagainst) guess.
For n doors the chance of winning when you switch doors is (n1)/n, ie 2/3 if there are 3 doors. 




eggshaped

15228. Wed Feb 09, 2005 11:55 am 


I have encountered this before, when I was checking out Uni's. I was going to study maths, and this was one of the puzzles they gave us. (I forget the university)
I didn't understand it then, and I don't understand it now. The difference that a University degree makes is now I'm happy to admit it. 




Flash

15229. Wed Feb 09, 2005 12:01 pm 


I think the nonmathematical way to understand it is this: when you choose right, the host's actions tell you nothing. When you choose wrong (which is most of the time) the host's actions give you the answer. So onethird of the time it makes no difference whether you change or not, but twothirds of the time changing will give you the right answer. So if you always change you'll be no worse off in the first case and much better off in the second. 




Gaazy

15230. Wed Feb 09, 2005 12:28 pm 


I read about this last year, until my brain said, "that's enough; I'm out of here", packed an overnight bag and went on a long and welldeserved holiday.
There's a multitude of websites which seek to explain this paradox, and the easiest form of explanation for my nonmathematical mind is the flowchart model. However, when I used such a flowchart to explain it to my family, all I got were pitying looks.
The paradox received public attention when it was published as a puzzle in the Parade magazine by Marilyn vos Savant in her "Ask Marilyn" column.
After her answer was published she received tens of thousands of letters, the great majority disagreeing with her, and many  from respected mathematicians  couched in quite aggressive terms.
However, the proofs she supplied were incontestable, and the mathematicians were forced to accept the wholly counterintuitive conclusion. 




Icarus

15231. Wed Feb 09, 2005 12:29 pm 


Flash,
This is a very well known conundrum, with repeated (and very snippy) letters to the NY Times over the years about the decay of mathematics in the US as evidenced by the number of supposedly bright people who can't see the answer. This conditional probability question has caused at least two of my evenings to be given over to debates (because once the intuition sinks in, both the correct and the incorrect one, neither side seems to want to let go) which started at drinks before dinner and extended well beyond drinks after pudding. 




Icarus

15232. Wed Feb 09, 2005 12:31 pm 


I had to build a spreadsheet to prove that it's twothirds, not having Marilyn vos Savant's persuasive proof to hand. 




Flash

15233. Wed Feb 09, 2005 12:36 pm 


Blimey, what a terrible evening that must have been. Still, it implies that we might be able to use this. For those of you who have wrestled with the question before, is my attempt at a simplified explanation (ie "when you choose right, the host's actions tell you nothing. When you choose wrong (which is most of the time) the host's actions give you the answer") reasonable, would you say? If we're going to use this subject we'll need a oneline, nonmathematical bit of glibberypokery for Stephen to wrap it up with. 




Gaazy

15235. Wed Feb 09, 2005 12:50 pm 


I don't think it's possible to boil the actual question down to any reasonably concise length  for example, it's necessary to include the host's invitation to switch choices, otherwise there's a flaw in the logic (do a Find for "ambiguity" in this extract). 




Gray

15239. Wed Feb 09, 2005 2:17 pm 


Flash, your explanation works for me  in fact it's the only explanation I remember from a previous encounter with this conundrum.
You could make it look like Stephen was doing a silly magic trick  that should get some laughs. Of course, Alan will get the correct door first time!
Humans are famously poor at intuiting the correct answers to probability questions, which is exploited to the full by casinos. Psychology textbooks are littered with experiments that show the common mistakes that we make  generally listed in the 'Decisionmaking' chapter. 




Flash

15242. Wed Feb 09, 2005 4:27 pm 


I've thought about this a bit more and there's another approach: if you always change your first call, the result will always be that if you guessed right first time (which you will one time in three) then you'll get it wrong the second time, whereas if you guessed wrong the first time (which you you will two times in three) then you'll get it right the second time. So you'll get it right 2/3 of the time.
IE: if the right answer is A, and you guess A, the host will reveal one of the remaining two and you will choose the other one: wrong.
If the right answer is A and you guess B the host will reveal C and you will change your guess to A (the only one left): right
If the right answer is A and you guess C the host will reveal B and you will change your guess to A (the only one left again): right.
This pleases me. 




eggshaped

15254. Thu Feb 10, 2005 4:38 am 


I worked this out on the drive home last night, in exactly the way described above. If you choose the wrong door (2 in 3 chance) then you are guaranteed to win by switching.
What it got me thinking though was:
You're stood in front of 2 doors and you pick one, then the gameshow host says "actually, there's another door over here that you didn't know about  but don't worry it's empty". Do your chances after a switch suddenly improve, just cos he's told you about this door?
What if there's 10 hidden doors? Or 100? What if the doors are there, but he doesn't tell you, cos they're empty and he doesn't want to confuse you?
Oh, and my girlfriend had heard of this paradox, she said she'd seen it in some teambuilding kind of meeting. 




Flash

15256. Thu Feb 10, 2005 5:03 am 


I guess if there are hidden doors the proposition ceases to work because you can't choose the hidden door in round 1  ie you get only the first two possibilities in the list of three that I posted. 




Icarus

15333. Tue Feb 15, 2005 7:53 am 


Flash,
The key to this is, as you pointed out, the information. Monty always revealed a door that didn't have the big prize. Thus by switching you are making two guesses for the price of one (and thus it's 2/3rds, not 50/50). 




Jenny

15344. Tue Feb 15, 2005 10:55 am 


We could have Stephen doing a kind of 'Take your Pick' thing here  'Open the box!' 'Take the money!'
You could have a £10 behind the door that is opened by Stephen, and a £100 note behind one of the other two doors. Encourage audience participation! 




Flash

15679. Sat Feb 26, 2005 6:55 am 


Under conundra: I think we have discussed elsewhere the business about how mirrors reverse leftright but not updown. Anyone got anything insightful to say about this?
Here's a start: your image in the bowl of a spoon is reversed updown but not leftright. 



