Brainteaser

[Rovers]

One one of the doors is elimated, the odds are 50-50.

 

Easy's example is rigged. He removed all of the cards except the ace unless in the unlikely event that the ace was pulled. out of the full deck.

 

Assuming the prize door does not get changed, it's no a 50% probablilty. For Easy's example to work, the unknown must be removed. take the two last reamianing cards and shuffle them. The odds go back to 50%. Look at it this way...

 

If it were known that one door would be opened, being one of the two goat doors, the player actually had a 2 in 3 chance of not picking the goat door Monty opened.

[redrumjuice]

The odds do not change. Once they reveal a door, the odds you are playing with are the same.

 

If you play with 2 doors, it's 50/50.

 

If you play with 3 doors, and they reveal one, well 50/50 again.

 

You can write it down and proof the odds are in your favor, but you will just be wasting time and not changing anything.

[Rovers]

The only way to prove or disprove this is to do it 1000 times. Then and only then can I buy into it.

[Big Country]

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.

 

 

The thing is you are not making a new, independent choice. If you were, and the were shuffling the prize, then yes, it is 50/50, but that is not the case.

 

Look at it this way.

 

Are you more likely to win with Door A or with both Door B and C?

 

I'll take the two doors every day of the week and twice on Sunday. I know one of the doors on my side is a blank, but I still have a 2 in 3 chance of being right, even after they have shown me which one of the doors is blank.

[delusions of grandeur]

The thing is you are not making a new, independent choice. If you were, and the were shuffling the prize, then yes, it is 50/50, but that is not the case.

 

Look at it this way.

 

Are you more likely to win with Door A or with both Door B and C?

 

I'll take the two doors every day of the week and twice on Sunday. I know one of the doors on my side is a blank, but I still have a 2 in 3 chance of being right, even after they have shown me which one of the doors is blank.

 

I really don't think that's applicable for hokey game-show game of "You choose, I'll tell you which one it ain't, and we'll go from there"....

 

Although at first there may appear to be a 33% you could be correct, that's not really the case... One of those doors is determined to be eliminated from the get-go... So you're really only choosing between 1 non-donkey door, and the 1 donkey door that they don't arbitrarily choose to eliminate based on your previous decision.

 

There never is a "real" first round where it's 33%, because all that round is, is an elimination round of the wrong choices, with no chance of being right. This decision begins in this case, as is shown by the picture in Opies' link, when the only two legitimate options are left...

 

I completely disagree that 1 door gets upgraded to 66%, just because in a hokey game another became 0%.

Edited December 10, 2010 by delusions of granduer

[Chief Dick]

I feel like I'm sitting on a bar stool next to Cliff and Norm.

 

[i_am_the_swammi]

I feel like I'm sitting on a bar stool next to Cliff and Norm.

 

 

I feel like Mr. Pitt from Seinfeld trying to see the hidden image in the picture

[Chief Dick]

I feel like Mr. Pitt from Seinfeld trying to see the hidden image in the picture

 

[Easy n Dirty]

One one of the doors is elimated, the odds are 50-50.

 

Easy's example is rigged. He removed all of the cards except the ace unless in the unlikely event that the ace was pulled. out of the full deck.

 

Assuming the prize door does not get changed, it's no a 50% probablilty. For Easy's example to work, the unknown must be removed. take the two last reamianing cards and shuffle them. The odds go back to 50%. Look at it this way...

 

If it were known that one door would be opened, being one of the two goat doors, the player actually had a 2 in 3 chance of not picking the goat door Monty opened.

 

My example is "rigged", because I get to look at the cards when I remove 50 of them. But recognize that the game show is "rigged" also, because Monty Hall knows where the car is too.

 

You posted later that you'd need to see this run a thousand times to be convinced. Try this out (it's from the end of the video that I posted above) - when you initially picked a door, there were three possible guesses. And the car is in one of three possible places, which yields 9 possible scenarios as follows:

 

You Pick Door # - Car is Behind Door #

 

1 - 1

1 - 2

1 - 3

2 - 1

2 - 2

2 - 3

3 - 1

3 - 2

3 - 3

 

In every one of these scenarios, Monty opens a door and shows you a goat after your first guess, and now you have the option to stay put or to switch - in scenarios 1, 5 and 9 above, you're better off staying put. In the other 6 instances, you are beter off switching.

 

You can also think through these same 9 scenarios to validate what Big Country is saying, i.e., that the door you could switch to now has a 2/3 probability of being the right door (for all but the goat loving TimC anyway).

Edited December 10, 2010 by Easy n Dirty

[delusions of grandeur]

Completely understand where you're coming from with the math, which is hard to argue with, but I look at it like this:

 

Say I go to my new friends' restaurant, and have narrowed down my choices to 3 dishes... I'm leaning towards A, but I ask my friend what he suggests, to which he replies: "Can't really say, but dish B has spinach, so I know you won't like that".

 

Okay, so now we've narrowed it down to 2 choices, before I make my final choice. I'm none the wiser on which one will be best, but I now know choice B isn't what I'm looking for... So regardless of the 33% chance I was taking before on choosing Dish A, I now have a second opportunity to look at it again, now that there only 2 dishes I'm deciding between, with a current 50% chance I'll like one more than the other....

 

So should I go for dish C simply because I had chosen dish A at a time when I had 3 choices rather than 2? Or further, say I was looking at everything on the menu when I originally picked A. Does that matter now, that I had in the past made a low % choice, when now there is equal probability that I will like one more than the other?

 

I think it's mistaken to place any importance on the first round, when it is GUARANTEED that your actual final choice will only happen once 1 of the options is eliminated... And in my mind, this final question is all that matters, in "should you change doors"... What's in the past is in the past, without having any necessary bearing on the present choice.

 

 

(ETA: I think the sticking point is that people view it as choosing to "stay" with the same door, which allows for "staying" with the 33% probability; Whereas I see this choice as a totally independent decision from the first one, and this difference in view alone allows for drastically different interpretation.)

Edited December 10, 2010 by delusions of granduer

[gbpfan1231]

Completely understand where you're coming from with the math, which is hard to argue with, but I look at it like this:

 

Say I go to my new friends' restaurant, and have narrowed down my choices to 3 dishes... I'm leaning towards A, but I ask my friend what he suggests, to which he replies: "Can't really say, but dish B has spinach, so I know you won't like that".

 

Okay, so now we've narrowed it down to 2 choices, before I make my final choice. I'm none the wiser on which one will be best, but I now know choice B isn't what I'm looking for... So regardless of the 33% chance I was taking before on choosing Dish A, I now have a second opportunity to look at it again, now that there only 2 dishes I'm deciding between, with a current 50% chance I'll like one more than the other....

 

So should I go for dish C simply because I had chosen dish A at a time when I had 3 choices rather than 2? Or further, say I was looking at everything on the menu when I originally picked A. Does that matter now, that I had in the past made a low % choice, when now there is equal probability that I will like one more than the other?

 

I think it's mistaken to place any importance on the first round, when it is GUARANTEED that your actual final choice will only happen once 1 of the options is eliminated... And in my mind, this final question is all that matters, in "should you change doors"... What's in the past is in the past, without having any necessary bearing on the present choice.

 

 

(ETA: I think the sticking point is that people view it as choosing to "stay" with the same door, which allows for "staying" with the 33% probability; Whereas I see this choice as a totally independent decision from the first one, and this difference in view alone allows for drastically different interpretation.)

This makes more sense to me than all the other responses that are trying to explain the other side.

 

Maybe it is because I also don't like spinach??

[T_bone65]

My head hurts

[Seahawks21]

I can't solve this one. I've been trying for a good hour now. This kind of problem was my forte when I had a much younger and faster mind.

 

You've essentially got 2 unopened doors, one has a goat, one has a car. I just cannot fathom how your odds are anything more or less than 50/50 at that point. Switching seems so arbitrary. Uggh. This one is gonna cost me sleep.

[i_am_the_swammi]

I think I get it

 

Here's how I reason it: there are, basically, three scenarios that can play out:

 

1. You pick Door A (car), leaving B & C both with goats. Host takes away one goat, leaving another. You switch You lose.

 

2. You pick Door B (goat), leaving A & C, one car, one goat. Host takes away goat, leaving car. You switch, You win.

 

3. You pick Door C (goat), leaving A & B,one car, one goat. Host takes away goat., leaving car. You switch, you win.

 

Knowing these are the only three outcomes, you have a 2 in 3 chance of winning if you switch.

 

Conversely, if you stay put and dont switch:

 

1. You pick Door A (car), leaving B & C both with goats. Host takes away one goat, leaving another. You stay put. You win.

 

2. You pick Door B (goat), leaving A & C, one car, one goat. Host takes away goat, leaving car. You stay put, you lose.

 

3. You pick Door C (goat), leaving A & B,one car, one goat. Host takes away goat., leaving car. You stay put, you lose.

 

Knowing these are the only three outcomes if you don't switch, you only win 1 in 3 times.

 

Make sense??

[rhippens]

i think the point is that just by switching, you aren't going to win every time. but by switching you have increased your odds. at first you had a 33% chance of hitting the car and 66% chances of hitting the goat. by switching, you are upping your chance of car to 66% and putting the goat at 33%.

[redrumjuice]

but by switching you have increased your odds.

 

No, you do not at all.

[redrumjuice]

You can try and create any scenario you want, word play doesn't make it right.

 

The Monty Hall problem, in its usual interpretation, is mathematically equivalent to the earlier Three Prisoners problem, and both bear some similarity to the much older Bertrand's box paradox. These and other problems involving unequal distributions of probability are notoriously difficult for people to solve correctly; when the Monty Hall problem appeared in Parade, approximately 10,000 readers, including nearly 1,000 with PhDs, wrote to the magazine claiming the published solution ("switch!") was wrong.

 

They claim different, but I'll never believe it.

 

More info than you want: http://en.wikipedia.org/wiki/Monty_Hall_problem

[matt770]

This is why I always say math is a bunch of horsesh*t.

[Seahawks21]

We had a going-away party for somebody at work last night, and I took this problem around to about 20 educated people, hoping for any help whatsoever. Everybody came up with 50/50. I had forgotten all about this for a few hours, now I wake up today and check the huddle and get sucked right back in. Swammi, I appreciate the math, and I can almost touch the answer, then I lose it again.

[tazinib1]

Knock Knock

[chester]

I think I get it

 

Here's how I reason it: there are, basically, three scenarios that can play out:

 

1. You pick Door A (car), leaving B & C both with goats. Host takes away one goat, leaving another. You switch You lose.

 

2. You pick Door B (goat), leaving A & C, one car, one goat. Host takes away goat, leaving car. You switch, You win.

 

3. You pick Door C (goat), leaving A & B,one car, one goat. Host takes away goat., leaving car. You switch, you win.

 

Knowing these are the only three outcomes, you have a 2 in 3 chance of winning if you switch.

 

Conversely, if you stay put and dont switch:

 

1. You pick Door A (car), leaving B & C both with goats. Host takes away one goat, leaving another. You stay put. You win.

 

2. You pick Door B (goat), leaving A & C, one car, one goat. Host takes away goat, leaving car. You stay put, you lose.

 

3. You pick Door C (goat), leaving A & B,one car, one goat. Host takes away goat., leaving car. You stay put, you lose.

 

Knowing these are the only three outcomes if you don't switch, you only win 1 in 3 times.

 

Make sense??

 

Best explaination yet

[westvirginia]

I haven't looked at any answers, but the exact wording was "behind the 3rd door is a new car"? Or am I over thinking?

 

ETA: of course...

Edited December 14, 2010 by westvirginia

[The Irish Doggy]

If the car was a Honda Odyssey, I would switch doors. If it was a cheap American peener-compensating car, I'd stay put and hope for the goat.

[TimC]

If the car was a Honda Odyssey, I would switch doors. If it was a cheap American peener-compensating car, I'd stay put and hope for the goat.

 

I would rather ride a goat with bushwacked pulling the harness with both of us dressed in Ancient Roman robes than a minivan.

 

ETA: I believe that is Door D.

Edited December 14, 2010 by TimC