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Monty Hall Problem

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StevenCaton

Mon Aug 17 2009, 09:28PM

Registered Member #1845 Joined: Fri Dec 05 2008, 05:38AM
Location: California
Posts: 211

I have been fascinated by this problem for the last few days. I just finished a book on it. Here is the problem.

Suppose you're on a game show and you're given the choice of three doors. Behind one door is a car; behind the others, goats. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?

ANSWER: You should always switch doors. If you switch doors, then there is a 66.6% chance that you will get the car.

Hon1nbo

Mon Aug 17 2009, 09:46PM

Registered Member #902 Joined: Sun Jul 15 2007, 08:17PM
Location: Pacific Northwest USA
Posts: 1042

this problem, and the explanation, was actually featured in the movie "21" - it took me a few days to understand how the math worked, but like the Columbus egg once you see how it's done it seems obvious

Frosty90

Tue Aug 18 2009, 10:26AM

Registered Member #1617 Joined: Fri Aug 01 2008, 07:31AM
Location: Adelaide, South Australia
Posts: 139

I have a book with mathematical puzzles and problems like this, it is very fascinting. DaJJHman, exactly, once you get the maths it seems perfectly obvious and trivial! From what I've seen, the trick in the Monty hall problem is the actual definition of probability; we all have a sort of 'intuition' about what the answer should be, and a sort of intuitive feeling for what probablility actually is, but when you look at just the definition and mathematics of probability, it doesn't always match up to your 'gut-feeling'!

Steve Conner

Tue Aug 18 2009, 11:42AM

Registered Member #30 Joined: Fri Feb 03 2006, 10:52AM
Location: Glasgow, Scotland
Posts: 6706

There are a few different versions of this problem, with different named game show hosts and different prizes.

I think ultimately the crux is that, when the host opens one of the doors he's showing you one bit more information than you had. If you act on that information, then you improve your chances of winning, compared to your original random choice that was made without the benefit of the information.

The wording of the problems of course goes to great lengths to disguise this.

HV Enthusiast

Tue Aug 18 2009, 11:43AM

Registered Member #15 Joined: Thu Feb 02 2006, 01:11PM
Location:
Posts: 3068

Google is your friend . . .

Renesis

Tue Aug 18 2009, 01:46PM

Registered Member #2028 Joined: Mon Mar 16 2009, 08:13PM
Location: Norway
Posts: 319

Iv'e been twisting my head to try to make some sence out of this stupid paradox, until i read wikipedia's great explanation. I've got it now, and it is so simple! Its silly really, but i can see why people insist this is wrong.

IamSmooth

Wed Aug 19 2009, 12:48AM

Registered Member #190 Joined: Fri Feb 17 2006, 12:00AM
Location:
Posts: 1567

This website explains the puzzle quite well and offers many others.

Chris Russell

Sat Aug 29 2009, 04:34AM

... not Russel!
Registered Member #1 Joined: Thu Jan 26 2006, 12:18AM
Location: Tempe, Arizona
Posts: 1052

Without peeking at the explanations, my logic goes something like this:

First choice: odds of winning are one in three if I decide to stick with it.

Second choice: there are two possible states here. My door has a car (one in three), or my door has a goat (two in three). If I have the car, one goat is eliminated, I change to the other goat, and I lose. If I have a goat, the other goat is eliminated, I change to the car, and I win. Thus, so long as I didn't pick the car to start with, I will win. That's two out of three times. Not bad!

StevenCaton

Sun Aug 30 2009, 04:19AM

Registered Member #1845 Joined: Fri Dec 05 2008, 05:38AM
Location: California
Posts: 211

If I have the car, one goat is eliminated, I change to the other goat, and I lose. If I have a goat, the other goat is eliminated, I change to the car, and I win. Thus, so long as I didn't pick the car to start with, I will win. That's two out of three times. Not bad!

Chris, thats possibly the most straightforward explanation to this problem, and the same explanation I've given to other people. As long as one uses the switching strategy and he initially picks a goat, then he will win. Its as simple as that.

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