Statistical Mindfuck

[BinoAl]

So, a buddy at work told me about an interesting statistical problem. Let's say there are 3 doors. Behind one door is 1 million dollars. Behind the other 2 doors, there is absolutely nothing. So, you choose a door, let's say door 1. Now, door 3 is opened to reveal that it is empty. Do you stay, or switch? (post in spoilers before reading the answer :p)

ANSWER:

Spoiler

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It is a 33% chance of your door being right, and a 66% chance of the other being right. When you choose a door, it is a 33% chance of being correct. So, a door is eliminated, and it is only ever a door with nothing behind it. If the prize was behind door 2, door 3 opens, and vice versa. That means that as long as you were originally wrong (a 2 in 3 chance), then the other door is correct.

It took me a few to understand originally, but I got it :)

[PatrickSJ]

I would switch.

[Awfulcopter]

Think about it in a different order of operations.

First, one empty door is opened. Then you choose one of the two remaining doors. Do you still think there is a 66% probability your first instinct is wrong?

[Itamarcu]

Yeah, it's a known riddle. I never thought anyone did not know it :P

[Graphite]

If you're wondering what bit ends up skewing that choice, then it's the fact that, to remove that third door, someone had to have knowledge of what door contained the prize. If you picked the door with the prize, they'd be free to choose, but if you didn't, then they are forced to pick the empty door of the remaining two.

[jorgebonafe]

Think about it in a different order of operations.

First, one empty door is opened. Then you choose one of the two remaining doors. Do you still think there is a 66% probability your first instinct is wrong?

You can't assume the first door is empty. Its all a matter of probability. If you don't know and pick at random, and then, after you pick, someone who knows open one empty door, then switching is gonna give you a higher probability of success. Assuming you know which door has the prize invalidates the test.

[jorgebonafe]

I find its a bit easier to understand this test like this:

Step one: Pick one door.
Step two: Someone opens an empty door (assume this person knows what doors are empty)
Step tree: switch.

On step 1, lets say you choose door #2. When you do that, assume that doing that means you didn't actually pick door #2, you actually picked both doors #1 and #3. Thats what swiching later means in the end, means you chose two doors instead of one, because you know that one of those doors is empty and that empty door wil be opened in step 2. That gives you double the chance of getting the right door when you switch.

If instead I ask you directly, "what gives you more chance to win, chosing one door, or choosing two doors?" the answer is evident.

[Catox]

It's funny how I find most of the explanations given here to be quite unclear while I understood it on the spot when I watched this episode of Numb3rs.

Anyway, my way of seeing this is simply that you have an opportunity to literally invert the result of your first choice.
Since this first choice is statistically the wrong one 2 times out of 3, inverting its result means you're right 2 times out of 3. So YES you pick the other door, no question asked.

[The great randomo]

This is the Monty Hall Problem, which originally came from the show 'Monty Hall'. Contestants were asked to pick from 3 doors, 1 with a car behind, and two with goats. The contestant would pick a door, and the host would open a door with a goat. The contestant then had to stick or swap.
Just sayin...

[Kazuya Mishima]

I always thought a better way to conceptualize this was to change the problem. Instead of having 3 doors have 100 and you chose 1 of those 100, door number 17 lets say. Then the presenter opens every door but your door 17 and door number 34 revealing nothing behind them and then asks if you wanna stay with door number 17 or switch to the one he reserved to be opened, number 34.

[Catox]

I always thought a better way to conceptualize this was to change the problem. Instead of having 3 doors have 100 and you chose 1 of those 100, door number 17 lets say. Then the presenter opens every door but your door 17 and door number 34 revealing nothing behind them and then asks if you wanna stay with door number 17 or switch to the one he reserved to be opened, number 34.

It transforms the probability from 1/3 - 2/3 to 1/100 - 99/100 and makes it more certain to win if you change your choice.
But I doubt that by itself this example makes the maths clearer : I think people will keep telling you that you have 1/2 to win either way as long as they don't understand the importance of the first choice.

[Eriottosan]

As a mathematician, this is one of my favourite ways to confuse people :). To me, it makes perfect sense, and it was the answer I gave when the question was first posed to me.

Spoiler

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Think of it like this: as BinoAl said, your original door has a 33% chance of having the big prize, which means that there is a 66% chance that it is behind another door. One door is eliminated, so the chance it is behind that one is 0%, meaning the other one you didn't pick carries the whole 66%. As a result, you switch if given the choice

The real mindfuck comes when trying to apply this logic to "Deal or No Deal" ... ;) (Do you have that in Americaland?)

[MoRmEnGiL]

This is a case of maths failing to comply with logic.

[Eriottosan]

This is a case of maths failing to comply with logic.

It's a case of logical thinking failing to comply with Maths.

Maths is never wrong, unless you use the Einstein proof that 1=2 ... Which in itself has a purposeful mistake :).

[Battosay]

And yet, that's not true.
When you chose the first door, it had a 33% chance of being the right one.
When someone opens an empty door, there's only two left, so each has a 50% chance.
When you're asked to switch or not, if you do, you'll have a 50% chance of chosing the right one. If you keep your choice, it's still 50%.

[Eriottosan]

And yet, that's not true.
When you chose the first door, it had a 33% chance of being the right one.
When someone opens an empty door, there's only two left, so each has a 50% chance.
When you're asked to switch or not, if you do, you'll have a 50% chance of chosing the right one. If you keep your choice, it's still 50%.

At the risk of disagreeing with a moderator, I'm afraid, Battosay, that statistics will disagree with you. In practice, the almost unbelievable mathematic reasoning prevails. It's similar to the way to winning rounds of 3 coin flips :).

[Battosay]

Oh yeah, I know, that's mathematically true.
However, my brain disagree.

[Eriottosan]

Oh yeah, I know, that's mathematically true.
However, my brain disagree.

Don't get me wrong, my mind screams NO! too, but whenever I look at these kinds of things, my mathematic reasoning shouts drowns out the "NO!." Contrasting states of the human brain. "yes, it's true, but it can't be!".

[walker_boh_65]

And yet, that's not true.
When you chose the first door, it had a 33% chance of being the right one.
When someone opens an empty door, there's only two left, so each has a 50% chance.
When you're asked to switch or not, if you do, you'll have a 50% chance of chosing the right one. If you keep your choice, it's still 50%.

Sorry pal, but you are wrong on this one. Let's put out a list of all you possible options if the prize is behind door number 3. (maybe this can help people see it better :) )
If you pick door 1, (then door 2 is opened) you switch to door three, you WIN..
If you pick door 2, (then door 1 is opened) you switch to door 3, you WIN.
If you pick door 3, (then door 2 is opened) you switch to door 1, you LOSE.
so that's a 2/3 chance you win if you switch doors.
If you pick door 1, (then door 2 is opened) you stay door 1, you LOSE.
If you pick door 2, (then door 1 is opened) you stay door 2, you LOSE.
If you pick door 3 (then door 2 is opened) you stay door 3, you WIN.
so that's a 1/3 chance of wining if you stay.

:)

[embirrim]

Doesn't opening the 3rd door narrow down the chances to 50% on each of the remaining doors?
I think you have to consider the problem as a whole, and not by fases. The probability you have to calculate is: which one is right knowing that 3 is wrong? In this case it's 50% I think.

[BinoAl]

Doesn't opening the 3rd door narrow down the chances to 50% on each of the remaining doors?
I think you have to consider the problem as a whole, and not by fases. The probability you have to calculate is: which one is right knowing that 3 is wrong? In this case it's 50% I think.

Yeah, that's the crazy part of it: Testing shows that it is a 33%-66% split. It doesn't make sense at first glance, but its how it works out :)

[embirrim]

Yeah, that's the crazy part of it: Testing shows that it is a 33%-66% split. It doesn't make sense at first glance, but its how it works out :)

There's something wrong with the testing then =P I think you're assuming that you're wrong.

[Catox]

Nothing can be wrong with those tests.
Make as many tests as you want, and you'll find out that first choice was right in 33% of cases and wrong in 66%.

All you need to understand then is that changing your choice means inverting the result.
Whether your first choice is wrong or right, there are only two cases regarding the two other doors :
- two wrongs (first choice right)
- one right and one wrong (first choice wrong)
There is always a wrong door left to open, that mean you are in this situation once it's opened :
- first choice right : remaining door is wrong
- first choice wrong : remaining door is right
If you change your choice and choose the remaining door, you invert the result of your first choice.
If you had 66% chances of being wrong with your first choice, then you have the same 66% choice of being right by choosing the last door.

[Gargantuan_Penguin]

Here is what I have to contribute to the conversation: my mind = blown

That is all, now if you'll excuse me, my nose is bleeding.

[Catox]

there, take a tissue