Monty Hall Problem

[Euclidean-Paradox]

Has anybody here heard of the Monty Hall problem? Basically, it goes like this: Suppose you are in a game show, and you have to choose between three doors. One of the doors contains a car, and the other two goats. Now say you pick door one, and the game host, who knows what is behind each of the doors, decides to open door three and reveals to the game player the goat. And now, you have two remaining doors left, and the game host asks if you want to switch your choice, now that you know that one of them has the goat. Would you switch?

Now, even though this is counterintuitive, you actually increase your chances of winning if you decide to switch. In this particular case, by switching the probability rises from 1/3 to 2/3 that you will win the car (Which can be proven mathematically BTW). Isn't that awesome? Given those odds under those circumstances I would switch. Now, of course, is there anyone who can try to prove otherwise?

[freeztar]

That doesn't make sense to me.
At first, your odds are 1/3 to pick the car. Once one of the goat doors is eliminated, your odds are 1/2 regardless if you switch or not.

But for fun, let's draw it out.
x=car o=goat

D1 x
D2 o
D3 o

If I choose D1 and it is revealed that D3 is a goat, I'm left with 2 doors with only one being a car, so 1/2 odds. If I stick with D1, I win, if I switch, I lose. It would be the same for D2.

So, I'm at a loss at to how you are getting 2/3. Can you explain?

[UncleAl]

It's on the Web explicitly proven on paper. You can run a Monte Carlo simulation and view the results each way, with or without switching, for tens of thousands of trials in a few seconds. Look it up or run it yourself.

Three three-foot lengths of adding machine tape with a thin line drawn down the long center of both sides:

1) Connect the first into a hatband with some tape and a lap join. Only cut along the line until you get back to the start. You obtain two unremarkable bands half the width of the original, same circumference.
2) Repeat (1), but before making the join rotate one end 180 degrees. Cut. You obtain one piece of half the width but twice the circumference.
3) Repeat (2), but instead of cutting on the line cut midway between the line and the long edge, all the way around an around, until you get back to the start.

The universe doesn't care what you expect to see.

[Erasmus00]

Originally Posted by freeztar

That doesn't make sense to me.
At first, your odds are 1/3 to pick the car. Once one of the goat doors is eliminated, your odds are 1/2 regardless if you switch or not.

But for fun, let's draw it out.
x=car o=goat

D1 x
D2 o
D3 o

If I choose D1 and it is revealed that D3 is a goat, I'm left with 2 doors with only one being a car, so 1/2 odds. If I stick with D1, I win, if I switch, I lose. It would be the same for D2.

So, I'm at a loss at to how you are getting 2/3. Can you explain?

The trick is after the door is open your odds aren't 50/50 because Monty knows where the car is, and Monty will NEVER open the door with the car, this adds information to the system. Hence, when Monty opens the door the probabilities change (its the simplest Bayes theorem problem).

Consider you guess first, and get it wrong (2/3 probability). Now if you choose to switch you get it right (100% probability).

Consider you guess first and get it right (1/3 probability). Now if you switch you get it wrong (100% probability).

Hence, 2/3 of the time switching wins the car.
-Will

[johnfp]

I disagree. If you always switch your choice after the first door is open, then essentially you have made your decision on a 1/3 probability and are stuck with it. No different then stayin with the first door you choose. It only goes to a 50/50 chance if you are allowed to actually choose between the two remaining doors.

Not to mention if you switch and you lose, you will hate yourself.

[modest]

Originally Posted by Erasmus00

The trick is after the door is open your odds aren't 50/50 because Monty knows where the car is, and Monty will NEVER open the door with the car, this adds information to the system.

I agree. The problem is better stated:

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? (Krauss and Wang 2003:10)

-modest

[freeztar]

Originally Posted by Erasmus00

Consider you guess first, and get it wrong (2/3 probability). Now if you choose to switch you get it right (100% probability).

Consider you guess first and get it right (1/3 probability). Now if you switch you get it wrong (100% probability).

Hence, 2/3 of the time switching wins the car.
-Will

Thanks Will, that makes sense.

[Euclidean-Paradox]

Wikipedia has a nice way you can visualize it:

Monty Hall problem - Wikipedia, the free encyclopedia

[ughaibu]

Originally Posted by Euclidean-Paradox

Given those odds under those circumstances I would switch

You might enjoy the two envelope problem: there are two envelopes, one of which contains twice as much money as the other does. A person who wants as much money as possible is offered the choice to take one of the envelopes and after choosing they're offered the opportunity to change envelopes. Should they change? And if they change, when offered the chance to change back, should they do so?

[HydrogenBond]

I think there are two affects going on. The contestant is using chance. But the game show host is using cause and affect since he knows what is behind each door with 100% certainty. The game show host is outside the land of chaos. The combined system is fuzzy dice plus reason. If we take the system average the odds improve. If the contestant watched the show previously, he may notice a logical pattern based on the necessity of shows ratings. He may also notice body language. He is going into the land of chaos with a flashlight allowing him to beat the odds. If Monty was also in the dark so there were two pairs of dice then the regular odds would apply.