Total bullshit

[TheMajor]

http://www.grand-illusions.com/monty.htm

 

[Dave]

Winning cars?

Dave

 

[TheMajor]

No, I just got it from a website. I thought it would be untrue. But at this moment I am in doubt. I will see if they are right with my TI-83 later this afternoon.

 

[Chankama]

I read the question. Yeah, it sounds right TheMajor. Perhaps, it might be easier to visualize with the 100 door example.

100 doors, one car, 99 goats.

You pick one, your probability of winning is 1/100.
The host opens 98 doors (and he ALWAYS has 98 doors to open since there are always "at least" 98 goats left). So you don't learn anything new. When you picked your first door, you KNEW that he'd have 98 goat doors to open. Due to this certainty, and only due to this certainty, your initial probability of 1/100 doesn't change.

So, this "fact" of opening 98 goat doors is irrelevant WITH RESPECT TO THE INITIAL PICK (not the remaining door). It will ALWAYS happen successfully. So, you don't learn anything new about your initial pick.

Well.... What's the probability of the "remaining other door" containing the car?.. 1 - 1/100 = 99/100 .. b/c one of them has the car..

Now try it on the 3 door problem.

 

[Apokalipse]

I see how it works, looking at only the first way of explaining

 

[idiotec]

The 100 door example does help... makes sense to me...

 

[joshd]

hmmm.... surely it should be a 50:50 chance? once there are two doors, you know the car is in one, and the goat is in the other. surely you should treat the whole thing as only being two doors, as you know one of them will get discounted anyway.

 

[Snump]

In college I studied this exact situation, we prived again and again it is quite true.

 

[Apokalipse]

okay, there are 3 possible solutions:

car | goat | goat
goat | car | goat
goat | goat | car

if you picked one door, you have a 1 in 3 chance of getting the car. say you picked door 3.
only the last one is a possible solution.
if the person opened a door, which wasn't door 3, that had a goat, then the possible solutions now looks like this:

car | removed | goat
removed | car | goat
removed | goat | car - it doesn't matter which goat he eliminates in this solution

remove the ones that say removed, and you get this:

car | goat
car | goat
goat | car

and the third door becomes the second door if you look at it this way. now, you see that 2/3 of the solutions end up giving you a goat if you stick with your door. but if you switch to the first of those solutions, 2/3 of the solutions give you a car.

 

[Chankama]

hmmm.... surely it should be a 50:50 chance? once there are two doors, you know the car is in one, and the goat is in the other. surely you should treat the whole thing as only being two doors, as you know one of them will get discounted anyway.

I believe I clarified this in 100 door example more clearly. The reason "switching" is better is b/c the probability of the first door does not change from 1/3. The opening of the door by the host does not change this probability. If you buy a lottery ticket and I tell you, "what did you have for lunch? I had an apple.", you probability of winning big money does not change. Similarly, the opening of the door doesn't mean anything with respect to the initial probability.

Therefore, the other door = 1 - 1/3 = 2/3.