Brainteaser

[i_am_the_swammi]

You are on a game show, and are given the choice of one of three doors (A, B or C). Behind two doors are goats, and the third door a new car.

 

You pick Door A, and the host opens and reveals a goat behind Door B. He then asks you if you want to change your pick.

 

Should you do so?

[skylive5]

No.

[driveby]

You are on a game show, and are given the choice of one of three doors (A, B or C). Behind two doors are goats, and the third door a new car.

 

You pick Door A, and the host opens and reveals a goat behind Door B. He then asks you if you want to change your pick.

 

Should you do so?

I just read the book. Yes you change.

[Scooby's Hubby]

you always knew you had picked a good door when they started out with free carpet and that carpet display.

[Savage Beatings]

I've heard about this before and the explanation never quite sat well with me.

[i_am_the_swammi]

I've heard about this before and the explanation never quite sat well with me.

+1

 

I was hoping someone here could explain it to me rationally

 

I think the basic answer is: at first, you have a 2/3 chance of being wrong and still have a 2/3 chance after a door is opened if you do NOT change. But if you change, you get a 50% chance and so your odds have improved by a 1/3. I think that's correct?

 

The bolded part is what I don't get.

 

How do you still have a 2/3 chance of getting it wrong if there are only two doors left?

 

Once you are down to 2 doors, the choice at that point is independent of whatever happened prior. You have B and C. You pick one. Done deal. How can your chance only be 1/3 if you stick with what happened to be your original choice?

 

i don't get it at all

[Hugh B Tool]

I'd trade my door for a Barkers Beauty

[wiegie]

the answer is yes

[TimC]

What if I want the goat? I would name him Billy.

[Hugh 0ne]

What if I want the goat? I would name him Billy.

 

Isn't part of your parole that you're not allowed near farm animals?

[Bier Meister]

What if I want the goat? I would name him Billy.

 

 

jeff is a much better name for him

[Big John]

Yes done before.

 

If the opened door was picked at random, it would have been a 1 in 3 chance that it revealed the prize. If the prize wasn't there, the other 2 doors would change from a 1 in 3 chance to a 1 in 2 chance. So no difference. Neither remaining door has better odds.

 

But they always pick a door to reveal that does not have the prize, thus not being random. After the rigged door is revealed. So the other unpicked door goes to a 1 in 2 shot, but the picked door remains at a 1 in 3 shot of the prize.

Edited December 9, 2010 by Big John

[gbpfan1231]

This is in the movie 21.

 

 

 

The answer never really made sense to me either.

[Big John]

jeff is a much better name for him

But he preferred sheep.

[Easy n Dirty]

Very good explanation of why switching is better right

 

Using the example from the link above - I show you 52 cards face down and ask you to pick one and put it aside, without looking at it. What are the chances that you pick the ace of spades? Clearly 1 out of 52. Now I take the remaining 51 cards and turn 50 of them face up, with the ace of spades not being one of them. Now, you and I each hold one card, and one of them is the ace of spades. Are the chances that your card is the ace of apdes suddenly 1 out of 2? Clearly not - in 51 out of 52 instances, my card will be the ace of spades. The key here is that when I turned up 50 cards, I did not choose them randomly - I looked at them and specifically chose 50 cards that were not the ace of spades, keeping one card aside. Same principle applies in the Monty Hall problem, where Monty knows which door the prize is behind, so his choice to reveal a door to you with a goat behind it is also not random. Video above does a very good job of explaining it.

Edited December 9, 2010 by Easy n Dirty

[Savage Beatings]

But they always pick a door to reveal that does not have the prize, thus not being random. After the rigged door is revealed. So the other unpicked door goes to a 1 in 2 shot, but the picked door remains at a 1 in 3 shot of the prize.

 

This is the part that I never quite get. Why once the rigged door is revealed don't both remaining doors go to a 1 in 2 shot?

[Savage Beatings]

Very good explanation of why switching is better right

 

Using the example from the link above - I show you 52 cards face down and ask you to pick one and put it aside, without looking at it. What are the chances that you pick the ace of spades? Clearly 1 out of 52. Now I take the remaining 51 cards and turn 50 of them face up, with the ace of spades not being one of them. Now, you and I each hold one card, and one of them is the ace of spades. Are the chances that your card is the ace of apdes suddenly 1 out of 2? Clearly not - in 51 out of 52 instances, my card will be the ace of spades. The key here is that when I turned up 50 cards, I did not choose them randomly - I looked at them and specifically chose 50 cards that were not the ace of spades, keeping one card aside. Same principle applies in the Monty Hall problem, where Monty knows which door the prize is behind, so his choice to reveal a door to you with a goat behind it is also not random. Video above does a very good job of explaining it.

 

For some reason this explanation makes more sense to me (the cards helps me visualize it I guess). Thanks!

[delusions of grandeur]

Yes done before.

 

If the opened door was picked at random, it would have been a 1 in 3 chance that it revealed the prize. If the prize wasn't there, the other 2 doors would change from a 1 in 3 chance to a 1 in 2 chance. So no difference. Neither remaining door has better odds.

But they always pick a door to reveal that does not have the prize, thus not being random. After the rigged door is revealed. So the other unpicked door goes to a 1 in 2 shot, but the picked door remains at a 1 in 3 shot of the prize.

 

That does sound like some flawless logic, and according to the law of probabilities may be true. But the reason why the 2 bolded statements are opposite conclusions is, I think, because of the law of independent trials.

 

I understand that the first opened door "not being random" is the crux of the argument, but the subsequent choice is still random for the person choosing, which should be the only person that matters in the "should you change" question...

 

As such, I don't think there is a correct answer. There were only 2 "real" choices in the first place (the other was predetermined to be shown, and it was determined there'd be a second trial where it'd actually be revealed). So now that we are onto the second trial, what happened in the first trial is irrelevant...

 

In other words, "the picked door remains at a 1 in 3 shot of the prize" only if we're talking about the first "trial" (if you can even call it that, since they had planned to only show the wrong choice)... On the second trial, which is the only one that is now of any concern to the contestant, "Neither remaining door has better odds".

[Savage Beatings]

I think Easy n Dirty's post above yours actually provides the nicest explanation.

 

Yep I totally agree... the additional number of 52 cards makes it easier for me to see now than the original example of just 3 doors.

[Big Country]

It is very simple. The one assumption is that the location of the prize does not change.

 

 

You pick door A. The other side is door B and C. Think of B and C as now being a collective.

 

So, Door A has a 33% chance of being right

 

Door B and C have a 67% chance of being right.

 

They reveal that Door C has nothing behind it.

 

This does not change the fact that Door A has a 33% chance of being right, and the collective Door B/C side has a 67% chance of being right, as nothing has changed except that now, instead of that 67% chance being split between two options (B and C), it is now solely on one option (.

 

 

Thus, for once, I can state that Big John's statement above about the unpicked door going to a 1 in 2 shot of being right is actually a wrong statement, as it goes to a 2 in 3 shot of being right.

[i_am_the_swammi]

In other words, "the picked door remains at a 1 in 3 shot of the prize" only if we're talking about the first "trial" (if you can even call it that, since they had planned to only show the wrong choice)... On the second trial, which is the only one that is now of any concern to the contestant, "Neither remaining door has better odds".

 

Thats pretty much my issue with this perplexing problem

 

Once the first "1 out of 3" trial is over, it becomes a pure "1 out of 2". On the surface, it would seem you have an equal chance, at that point, of being right with your original choice. The math (and Easy's 52-card explanation) prove it otherwise.

[i_am_the_swammi]

It is very simple. The one assumption is that the location of the prize does not change.

 

 

You pick door A. The other side is door B and C. Think of B and C as now being a collective.

 

So, Door A has a 33% chance of being right

 

Door B and C have a 67% chance of being right.

 

So door B & C, collectively, have a 67%...meaning they each have a 33.333% chance. So when one is eliminated, the remaining door is left with its independent 33% chance.

 

So A has a 33% chance, and B has a 33% chance. Looks pretty equal to me

 

I know I am wrong, I just can't come to terms with it

[Big Country]

Thats pretty much my issue with this perplexing problem

 

Once the first "1 out of 3" trial is over, it becomes a pure "1 out of 2". On the surface, it would seem you have an equal chance, at that point, of being right with your original choice. The math (and Easy's 52-card explanation) prove it otherwise.

 

 

It is only a pure 1 out of 2 situation if they are again reshuffling where the prize is located. As they are not, the odds for your door vs. the collective odds of the other 2 remains unchanged. The fact they have shown you which of the other 2 has no prize does nothing to help the odds that you initially picked correctly.

[Big Country]

So door B & C, collectively, have a 67%...meaning they each have a 33.333% chance. So when one is eliminated, the remaining door is left with its independent 33% chance.

 

So A has a 33% chance, and B has a 33% chance. Looks pretty equal to me

 

I know I am wrong, I just can't come to terms with it

 

 

Once C is revealed, we now know that the full 67% chance that the prize is behind door B or C resides with door B.

A is 33%

B is 67%

C is 0%

[delusions of grandeur]

It is very simple. The one assumption is that the location of the prize does not change.

 

 

You pick door A. The other side is door B and C. Think of B and C as now being a collective.

 

So, Door A has a 33% chance of being right

 

Door B and C have a 67% chance of being right.

 

They reveal that Door C has nothing behind it.

 

This does not change the fact that Door A has a 33% chance of being right, and the collective Door B/C side has a 67% chance of being right, as nothing has changed except that now, instead of that 67% chance being split between two options (B and C), it is now solely on one option (.

 

 

Thus, for once, I can state that Big John's statement above about the unpicked door going to a 1 in 2 shot of being right is actually a wrong statement, as it goes to a 2 in 3 shot of being right.

 

So without any more math talk, that means you'd change the door from the one you picked, even though you can now see with your own eyes that it is just as likely to be wrong as it is right?

 

I don't really see how what happened in the past now gives you a statistical advantage to change. Regardless of whether you were right in the first place, they open one of the wrong doors, so you gain no additional insight at this stage of the game, where your odds are 50%.

 

Aside from being technically correct, I don't see how it gives you any advantage either way, now that you have a second choice to make between two doors.

Edited December 9, 2010 by delusions of granduer