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#1 2006-10-19 22:49:07

Toast
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The Monty Hall Problem

Simple, yet twisted:

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. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Whaddaya reckon?

Last edited by Toast (2006-10-19 22:53:58)

 

#2 2006-10-19 23:36:16

mathsyperson
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Re: The Monty Hall Problem

Heh, I've lost count over how many heated arguments we've had about that on this forum in the past.

Yes, it is better to switch. When you picked the door, there were 3 to choose from and each had an equal chance. So the chance of you picking the right one was 1/3, meaning that the chance of you being wrong was 2/3.

Then the host removes an option, so now there are only 2 doors. We've established already that your door has a 1/3 chance of being right, so that means that switching must give you the 2/3 chance.


Why did the vector cross the road?
It wanted to be normal.
 

#3 2006-10-20 03:02:01

Ricky
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Re: The Monty Hall Problem

Perhaps someone should write and article explaining the problem?


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."
 

#4 2006-10-20 05:18:09

pi man
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Re: The Monty Hall Problem

Just thinking this through since it doesn't sound right.    Seems like it should be 50%-50%. 

We'll label the 3 prizes as G1 and G2 (goats) and C (car).   There are 6 different arrangements:

#      Door1                  Door2              Door3
1         C                        G1                   G2
2         C                        G2                   G1
3         G1                      C                     G2
4         G1                      G2                   C
5         G2                      C                     G1
6         G2                      G1                   C

Pick a door, any door.   Let's say Door 2.   In 2 of the cases (3 and 5), you don't want to switch.  In the other 4 case (1, 2, 4, and 6), you do want to switch.   So using this logic, you want to switch 2/3 of the time so it makes sense to switch.



Let's look at it from Monty Hall's point of view.   He knows where the car is.   Let's say the arrangement is #5 from above.   Now there 4 different possibilities:

1/3 of the time the contestant will pick door 1, Monty will show what's in Door3, and the contestant would want to switch.
1/6 of the time the contestant will pick Door 2 AND Monty will show what's in Door 1.  The contestent should keep the original pick.
1/6 of the time the contestent will pick Door 2 AND Monty will show what's in Door 3.  The contestent should keep the original pick.
1/3 of the time, the contestant will pick Door 3, Monty will show what's behind Door 1 and the contestent would want to switch.   

So, again, 2/3 of the time the contestent should switch.

I guess I convinced myself that it's better to switch.  I would prefer that someone find a hole in my logic!

 

#5 2006-10-20 08:26:06

mathsyperson
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Re: The Monty Hall Problem

Nope, no holes there. That's exactly right. Just because the solution is counter-intuitive doesn't mean it isn't correct. It's the counter-intuitiveness that sparks so much controversy.


Why did the vector cross the road?
It wanted to be normal.
 

#6 2006-10-21 00:39:10

Toast
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Re: The Monty Hall Problem

I think people think the solution is counter-intuitive mainly because people don't take into account the fact that because they picked door x, the host is unable to pick the same door, which affects the probabilities.
Had the guest not selected doors x, y or z, then after the host selected one of them, naturally the probability of the car being in one of the remaining doors would be 0.5.

 

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