Monty Hall Problem

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Monty Hall Problem

Post by Edwin Ashworth » Tue Jul 04, 2006 11:12 pm

There may be some members who haven't been exercised by this famous mindbender. The problem is loosely based on a US game-show hosted by the said gentleman, though as usually stated the format differs from that actually used in the show, and is as follows:

You, the contestant, are in a room with three identical-looking doors (assume you've teleported there). Behind two of the doors are booby-prizes (traditionally goats - but quiet and odourless!) Behind the other door is the star prize (traditionally a luxury car - or more likely, the keys to one).
The host, who is in the room with you and knows which door the car lies behind, asks you to provisionally select one door, when he will open one of the remaining two doors, but not the door with the prize behind it.
You may then stick with the door you originally nominated, or swap to the other unopened door.
The question is:
(a) Are you more likely to win if you stick with your original nomination?
or (b) Are you more likely to win if you swap doors?
or (c) Doesn't it make a scrap of difference?
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Monty Hall Problem

Post by paulwiggins » Wed Jul 05, 2006 1:08 am

Binomial distribution springs to mind here.
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Monty Hall Problem

Post by Shelley » Wed Jul 05, 2006 12:11 pm

I will guess that it's (c) -- no difference. When Monty reveals one of the "no prize" doors, one's chances of having picked the door with the prize become 50/50. However, I gave up on logic and math in 10th grade. Is this a trick question?
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Monty Hall Problem

Post by gdwdwrkr » Wed Jul 05, 2006 12:19 pm

I will guess (b)--swap doors, because these are always trick questions, and it may expedite door opening. One good thing...no screaming audience!
Also, see
but not the door with the prize behind it.
Though, Shelly, I still secretly agree with you.
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Monty Hall Problem

Post by Edwin Ashworth » Thu Jul 06, 2006 9:26 pm

gdwdwrkr's cynicism would stand him in good stead (in this case at least).
An educationalist giving an INSET lecture at the school my wife used to teach at asked for a show of hands from the staff about their opinions on this question. Only three gave the correct answer ( (b) ). He was using it to illustrate how fallible even mighty teachers are. (In fact, there have been heated arguments over which is the correct answer, even amongst mathematics professionals - the issue is well documented on the Internet.)
The intuitive answer is, as Shelley says, (c), and it can be hard to accept that this is based on false logic (I know from personal experience).
There are various ways of showing that one should swap doors to maximise the chances of a win, but none seems intuitively very convincing. Perhaps a start can be made by altering the situation to, say, 102 doors, one prize and 101 boobies, and the host opens 100 losing doors when you have made your initial selection. Would you swap then?
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Monty Hall Problem

Post by Wizard of Oz » Fri Jul 07, 2006 1:13 am

.. one of the best ways to convince yourself using a concrete method .. and provided you have a few hours to spare .. is to use 3 playing cards to represent the two goats and the car .. then settle back and make and record your choices and the results .. of course that is provided you give two hoots about this or any similar problem in probability ..

WoZ of Aus 07/07/06
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Monty Hall Problem

Post by gdwdwrkr » Fri Jul 07, 2006 1:20 am

hoot
hoot

BUT WAIT! THERE'S MORE!!
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Monty Hall Problem

Post by kagriffy » Fri Jul 07, 2006 12:26 pm

Why, I'd pay upwards of FOUR hoots for that . . . don't answer yet . . . if you call in the next 20 seconds, we'll also throw in a set of ginsu knives . . . NOW, how much would you pay??????
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K. Allen Griffy
Springfield, Illinois (USA)

Monty Hall Problem

Post by Debz » Fri Jul 07, 2006 6:52 pm

I think I would forget logic and try echo-location.
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Monty Hall Problem

Post by Edwin Ashworth » Fri Jul 07, 2006 7:37 pm

It's not the Batmobile.
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Monty Hall Problem

Post by Ken Greenwald » Fri Jul 07, 2006 10:13 pm

Edwin, I think that this is a far trickier problem than it appears to be and, in fact, I got the wrong answer when I tried it (I had worked this years ago, but couldn’t remember a thing other than it paid to switch), but for the life of me couldn’t figure out what I was doing wrong after attacking it in the simplest way I knew how (there are other surer but more complicated approaches). I finally had to e-mail my number one son, the real mathematician in the family, to tell me where I had gone wrong.

I had counted 4 possible outcomes: 1) The original pick being the car and the game host picks goat 1. 2) The original pick is the car and the game host picks goat 2. 3) The original pick is goat 1 and the game host shows goat 2. 4) The original pick is goat 2 and the game host shows goat 1.

I took the above to be 4 equally probable outcomes, two of which (1 & 2) would cause a loss if the player switched and two of which (3 & 4) would cause a win if the player didn’t switch. So the probability was 2/4 in each case, so there was no advantage to switching.

Why the above is wrong is very subtle (at least to me whose long suit in mathematics was never probability, although I’m not exactly a total slouch). I was counting the number of possible outcomes, after both actions ((i) initial choice, (ii) revealing a door) have taken place. And of these there are correctly 4. But the error here is that the probability of winning-after-switching should be calculated based on how many initial choices the player has, not how many distinct ways the game can end. And I don’t think that this is all that obvious to see, but I do now. (see Wikipedia’s first solution at http://en.wikipedia.org/wiki/Monty_Hall_problem).

Well, I felt like a jerk after all my years of having studied and taught this stuff, but I did get a warm feeling knowing that my son has surpassed me by light years. And the boy did console me with the following words after pointing out my mistake:
<“It's not an easy problem though, don't think it was obvious for me.”>
But was he just being kind to an old guy on the skids? (&lt)

________________

Ken – July 7, 2006
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Monty Hall Problem

Post by Edwin Ashworth » Mon Jul 10, 2006 6:33 pm

Don't ask me, Ken. I'm trying to get rid of a flock of goats.
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Monty Hall Problem

Post by mike2005 » Mon Jul 10, 2006 6:59 pm

This mindbender has got my goat. I think it has been milked for all it's worth.
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