The Monty Hall Paradox--eggheads explain?

[headbang8]

At work, I have an intelligent, youthful assistant. He maintains that the Monty Hall Paradox is true. That is, the counter-intuitive solution is correct. I don't see it. Can someone explain it/convince me? There must be either a simple explanation, or a refutation.

Maybe my brain is too old, and not flexible enough.

 

[dong20]

I hated probability, but it seems to me that the likelihood of picking the winning door is now 50% rather than 33.3%, the opening of the duff door being an attempt to sow doubt and confusion in the mind of the poor sap choosing.

If they do change door selection, the % of it being the right door doesn't change in either case, it's still 50% or 33.3%.

No, wait. If the sap changes his mind it's 50%, otherwise it remains 33.3%. I told you I hated probability!!!

My head hurts.

 

[ManlyBanisters]

The explanation is right there on the page you link to - what more is there to explain? Scroll down and look at the first big diagram. The explanation is right there.

There are 3 possible scenarios, right?

The player originally picked the door hiding the car.
The player originally picked the door hiding Goat A.
The player originally picked the door hiding Goat B.

From this the following happens:

The player originally picked the door hiding the car. The game host opens another door to show one of the two goats.
The player originally picked the door hiding Goat A. The game host opens another door to show the ONE other goat.
The player originally picked the door hiding Goat B. The game host opens another door to show the ONE other goat.

In the second TWO possibilities there the player switches doors and wins.

So in TWO out of the THREE possible scenarios switching doors is the right thing to do. So the ODDS of winning the car are higher if you switch. It is not guaranteed - just more likely.

Is that what you are asking - it seems fairly plain to me. Maybe I have misunderstood your misunderstanding :biggrin:

 

[Guy-jin]

It's because when the first choice is made, one is choosing one out of three doors.

Then the host removes an incorrect door.

The probability of you having chosen an incorrect door to begin with is 2/3.

After making the choice, you lose half of that probability because the host is guaranteed to remove a "goat" door.

So then you either stay with the one you picked, or the remaining one, and since there was an initial 2/3 chance that you picked the wrong door, you have a better probability of picking the winning door by switching.

 

[Pecker]

If I remember Marie Vos Savant's recommendation correctly, it's better to choose another door.

She said that, if you choose Door #1 and Monty shows you the goat behind Door #2, then it is to your advantage to change your choice to Door #3 because when you chose the first door you only had a 1 in 3 chance to win but, once Monty shows you what's behind another door, your second choice has a 2 in 3 chance of winning.

She got a lot of mail about this with smart people from all over the world disagreeing with her. But she's right.

 

[headbang8]

Um, er, it, uh, kinda makes sense. But I'm so dumb that I even misspelled eggheads.

My brain hurts.

 

[Guy-jin]

once Monty shows you what's behind another door, your second choice has a 2 in 3 chance of winning.

Your second choice has a 1 in 2 chance of winning, not 2 in 3. Given that there are always equal odds on the second choice, the odds have gone up from 1/3 on the first choice to 1/2 on the second, and that's why you ought to choose the other door.

At the end of the day, what it basically says is that your first choice doesn't matter because Monty will always open a door with a booby prize behind it, and then you'll always end up with a higher probability of choosing the grand prize at the second choice.

 

[2322]

I thought the same thing but, given the context of the original question, it really is 2/3!

In the first two cases above, the host reveals the car. What might happen in these cases is unknown — perhaps the contestant immediately wins or immediately loses. However, in the problem as stated, the host has revealed a goat, so only four of the six cases remain possible, and they are equally likely. In two of these four cases, switching results in a win, and in the other two, switching results in a goat. Staying with the original pick gives the same odds: a loss in two cases and a win in two others.

The player's probability of winning by switching increases to 2/3 in the problem as stated by Mueser and Granberg because in the two cases above where the host would reveal the car, he is forced to reveal the remaining goat instead. In the table below, the host's picks from the table above are highlighted. Because he cannot reveal the car, his behavior is altered in two cases. This change in the host's behavior causes the car to be twice as likely to be behind the "third door", and is what causes switching to be twice as likely to win in the "host knows" variation of the problem. -Wikipedia

Apparently it boils down to a single factor. If the host knows where the car is, switch. If the host doesn't know, it doesn't matter. The key is that the host will always reveal the goats before the car when the host knows where the car is.

Your second choice has a 1 in 2 chance of winning, not 2 in 3. Given that there are always equal odds on the second choice, the odds have gone up from 1/3 on the first choice to 1/2 on the second, and that's why you ought to choose the other door.

At the end of the day, what it basically says is that your first choice doesn't matter because Monty will always open a door with a booby prize behind it, and then you'll always end up with a higher probability of choosing the grand prize at the second choice.

 

[Mem]

My thought is that you probably were gonna pick the wrong one if you had a 1/3 shot. Now that one is removed you have a 1/2 shot at it. Since you probably would have chosen the wrong one, your best bet is to switch. Sometimes you will be wrong to switch, but your odds are better if you do.

 

[Guy-jin]

I thought the same thing but, given the context of the original question, it really is 2/3!

Ah, read what I wrote again, Jason.

"Your second choice has a 1 in 2 chance of winning."

If you always choose to switch, there is no second choice at all, and the probability defaults to initial probability on the initial choice, which was 2/3s.

I should just give an example:

If you wrote a computer program to choose randomly from three options, throw one incorrect one away, then switch to the other and pick it, you'd win 2/3s of the time.

If you wrote a computer program to choose randomly from three options, throw one incorrect one away, then choose from the remaining two randomly, you'd win 1/2 of the time.

That ought to make it clearer, I hope?

 

[Calcium]

What's always blown my mind about this is that if you picked completely randomly on the second go around, like, flipping a coin, you still have a better chance if the result of your coin-flippery is to switch doors.

 

[dong20]

What's always blown my mind about this is that if you picked completely randomly on the second go around, like, flipping a coin, you still have a better chance if the result of your coin-flippery is to switch doors.

How so? The likelihood of a head or tail is the same each time you toss a coin regardless of how many times you do so. It's probably I may be misunderstanding what you're saying - probability always made my eyes glaze over.

 

[dong20]

...I even misspelled eggheads.

It's OK, Chaucer may have spelled it the same way.:biggrin1:

My brain hurts.

I'm not looking anymore so mine has stopped (hurting that is).

 

[2322]

Ah, read what I wrote again, Jason.

"Your second choice has a 1 in 2 chance of winning."

If you always choose to switch, there is no second choice at all, and the probability defaults to initial probability on the initial choice, which was 2/3s.

I should just give an example:

If you wrote a computer program to choose randomly from three options, throw one incorrect one away, then switch to the other and pick it, you'd win 2/3s of the time.

If you wrote a computer program to choose randomly from three options, throw one incorrect one away, then choose from the remaining two randomly, you'd win 1/2 of the time.

That ought to make it clearer, I hope?

Yes, in that scenario it's true. It's not in the Monty Hall scenario because the emcee (the computer) can only throw away two of the three options (the goats). The emcee cannot throw away the car. I believe in your scenario the computer can discard any of the three. Is that correct?

 

[Guy-jin]

Yes, in that scenario it's true. It's not in the Monty Hall scenario because the emcee (the computer) can only throw away two of the three options (the goats). The emcee cannot throw away the car. I believe in your scenario the computer can discard any of the three. Is that correct?

No, it can only throw away the incorrect ones (the booby prizes), just like Monty.

 

[Quite Irate]

Shell game, anyone?

 

[2322]

Not to dispute your genius here... but I'm missing something then. Your conclusion appears to dispute people far brainier than I am.

No, it can only throw away the incorrect ones (the booby prizes), just like Monty.

 

[Guy-jin]

Not to dispute your genius here... but I'm missing something then. Your conclusion appears to dispute people far brainier than I am.

Not at all. That's precisely the computer program that people have created to prove that it's 2/3s.

 

[rob_just_rob]

It makes it easier to conceptualize if you don't think about your odds of being right, but rather, your odds of being wrong. It also helps if you imagine 10 doors, and Monty opening 8 of them.

If you pick with all the doors closed, you have a 1 in 10 chance of being correct. More importantly, there's a 9 in 10 chance you picked wrong.

Now Monty reveals 8 goats, skipping over your door and one other door. You were 90% likely to have been wrong in your first choice. If you were 90% likely to have been wrong, of course you should change your answer.

In the classic paradox you are 66% likely to be wrong in your first choice. Again, it makes sense to change it.

 

[2322]

Does that actually work in lab testing with humans? Does changing actually result in more wins?

It makes it easier to conceptualize if you don't think about your odds of being right, but rather, your odds of being wrong. It also helps if you imagine 10 doors, and Monty opening 8 of them.

If you pick with all the doors closed, you have a 1 in 10 chance of being correct. More importantly, there's a 9 in 10 chance you picked wrong.

Now Monty reveals 8 goats, skipping over your door and one other door. You were 90% likely to have been wrong in your first choice. If you were 90% likely to have been wrong, of course you should change your answer.

In the classic paradox you are 66% likely to be wrong in your first choice. Again, it makes sense to change it.