Are you smarter than a pigeon?

[CoccoBill]

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?

[flomo]

i want the goat

[philly and the phanatics]

yes ldo its to your advantage

[d0zer]

can't milk a car

[Stacks]

I've already watched 21!

[BennyLaRue]

The answer is Wayne Brady.

[boost]

If you chose to stick with the first pick you are making a 50/50 gamble. If you change your door you are making a 50/50 gamble. It doesn't either way.

[bikes]

i've also seen 21. take his offer and switch doors

[a500lbgorilla]

If you chose to stick with the first pick you are making a 50/50 gamble. If you change your door you are making a 50/50 gamble. It doesn't either way.

This was my initial thought when someone posed this question to me. The real answer is not this and therefor really interesting.

I forget how it was explained to me, but imagine you buy a lottery ticket. Someone else in the world then reveals that every other lotto ticket but one has lost. He then offers you to swap your ticket for this other ticket.

Do you switch? I don't think this is the best example because it still *seems* like it's 50/50, but it's closer to 100% that switching nets you the win.

[jyms]

Monty hall problem

Monty Hall problem - Wikipedia, the free encyclopedia

[Stacks]

It's known as the Monty Hall Problem/Paradox.

Edit: Damn you beat me to it Jyms. I was just about to edit and relink the wiki article as well =(.

[CoccoBill]

Way to spoil!

Well yes, indeed, this is the Monty Hall problem. I'll be the first to admit this really sprained my brain when I first heard of it, makes you reconsider how much you should trust your intuition. I've asked this question from a number of people and I've yet to meet one who intuitively figured this one out correctly, it also doesn't bolster my self esteem to know that pigeons outsmart us here: Pigeons outperform humans at the Monty Hall Dilemma | Not Exactly Rocket Science | Discover Magazine

[Stacks]

Easiest explanation I saw on wiki:

"Monty is saying in effect: you can keep your one door or you can have the other two doors." The player therefore has the choice of either sticking with the original choice of door, or choosing the sum of the contents of the two other doors, as the 2/3 chance of hiding the car has not been changed by the opening of one of these doors.

[a500lbgorilla]

Proof of free will, imo

[Lukie]

The OP is missing a lot of assumptions that makes the Monty Hall problem a true Monty Hall problem.

Here is a more complete explanation for the nits-at-heart like myself:

Suppose you're on a game show and you're given the choice of three doors [and will win what is behind the chosen door]. Behind one door is a car; behind the others, goats [unwanted booby prizes]. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door, the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one [uniformly] at random. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you "Do you want to switch to Door Number 2?" Is it to your advantage to change your choice?

Staying has a 1/3 chance of spiking a car, switching is 2/3

simple explanation in layman's terms: Your original pick has a 1/3 chance of getting a car. The host opening another door showing a goat does not change this because there will always be 1-2 remaining door with a goat and you will always be shown a goat; there is no new information.

Of the remaining subset of 2 doors before being shown a goat, there is a 2/3 chance that there is the car among those 2 doors and when the host is kind enough to open one of them, the remaining door must must have a 2/3 chance of being the car.

[givememyleg]

it's a lot easier to see when you change 3 doors to 1,000 doors. at first you had 1/1000 shot, then if they opened all but 2, you have a 1/2 and should obv switch.

[rong]

Easist way to understand it:

I have 3 cards, A, K & Q.

I deal you one.

I then show you the lowest card I have.

You still only have 1 in 3 chance of having the Ace.

[Lukie]

Easist way to understand it:

I have 3 cards, A, K & Q.

I deal you one.

I then show you the lowest card I have.

You still only have 1 in 3 chance of having the Ace.

I don't see how this is an easy way to understand it because there are 3 different cards/prizes. In fact, it would be an entirely different problem.

Example assuming that I am shooting for the ace: Say I'm dealt a king, you are forced to show me the queen, then I can switch with 100% certainty of getting the ace.

edit: your example is correct for the first part, i.e. showing that you are still only at 1/3 chance of having the ace after being shown a card but before deciding to switch. Still confusing though..

[rong]

You don't get to see your card, and you're after the ace

[kingnat]

it's a lot easier to see when you change 3 doors to 1,000 doors. at first you had 1/1000 shot, then if they opened all but 2, you have a 1/2 and should obv switch.

This. Though it's not a 1 out of 2 chance when you switch, it's a 999/1000 chance for reason's that Lukie mentioned, but taking it to 1000 doors is the easiest intuitive way to think of the problem imo.

This is why in Deal or No Deal, switching suitcases at the very end doesn't improve your odds, because you don't gain any "new information" throughout the process like you do in the Monty Hall Problem. In that epically retarded show you are just picking things at random.

[rong]

I really don't see how that's easier than combining my two posts. But w/e

[rong]

Or to (try and) make it clearer:

I have 3 cards: AKQ. I deal you one, you don't see it.

You have 1/3 chance of having the Ace.

I have 2/3.

I then show you the lowest card I have. No matter what card I show you, you still don't know if you have the ace as I always have either a King or a Queen, the lowest of which I show you. So this gives you no new information and doesn't change the chance of you or me having the ace.

[d0zer]

Or to (try and) make it clearer:

I have 3 cards: AKQ. I deal you one, you don't see it.

You have 1/3 chance of having the Ace.

I have 2/3.

I then show you the lowest card I have. No matter what card I show you, you still don't know if you have the ace as I always have either a King or a Queen, the lowest of which I show you. So this gives you no new information and doesn't change the chance of you or me having the ace.

Maybe this will help:

You have to nail one of three midget hookers bareback, and one of them doesn't have aids. You pick one, then the midget pimp who knows them intimately fucks one, very deliberately putting on an 8-layered full body condom after spraying her down with rubbing alcohol and bleach. Do you switch?

[rong]

But you wouldn't switch then.

[d0zer]

But you wouldn't switch then.

enjoy your 67% aids equity,

[rong]

Depends if the midget pimp had a free choice of hooker or if the hookers were hidden behind a door.

[Stacks]

Or to (try and) make it clearer:

I have 3 cards: AKQ. I deal you one, you don't see it.

You have 1/3 chance of having the Ace.

I have 2/3.

I then show you the lowest card I have. No matter what card I show you, you still don't know if you have the ace as I always have either a King or a Queen, the lowest of which I show you. So this gives you no new information and doesn't change the chance of you or me having the ace.

Since you are showing the lowest card you have, if you show a King, that would mean you also have the Ace, and therefore I have the queen?

Whereas, if you show a Queen, then you might have the King, and I have the Ace. Or you might have the Ace, and I have the King.

[rong]

Dam it!

[Stacks]

Now if you show one at random, as in the Monty Hall problem, then I don't see how it would differ really. Essentially the Ace is the car, and the Queen/King are goats.

[Roid_Rage]

Maybe this will help:

You have to nail one of three midget hookers bareback, and one of them doesn't have aids. You pick one, then the midget pimp who knows them intimately fucks one, very deliberately putting on an 8-layered full body condom after spraying her down with rubbing alcohol and bleach. Do you switch?

Protip: pick the ugly one.

[Hawk]

I am still not smarter than a pigeon.

[spoonitnow]

Now if you show one at random, as in the Monty Hall problem, then I don't see how it would differ really. Essentially the Ace is the car, and the Queen/King are goats.

the monty hall situation depends on the person choosing which door to expose to you knowing which door holds the big prize or w/e and never shows that door to you

if the door chooser person picks at random, you're 50/50 the times he doesn't expose the prize by opening a door

[kiwiMark]

lolmaths, I've always found this to be the easiest way to think it through:

1/3 of the time you picked the car. If you change, you get a goat.
1/3 of the time you picked a goat. If you change, you get the car.
1/3 of the time you picked a goat. If you change, you get the car.

So you can see that changing will get you the goat 1/3 of the time and the car 2/3 of the time.


pointless continuation for completion's sake:

1/3 of the time you picked the car. If you hold, you get the car.
1/3 of the time you picked a goat. If you hold, you get a goat.
1/3 of the time you picked a goat. If you hold, you get a goat.

Holding gets you a car 1/3 of the time, and a goat 2/3 of the time.

[Roid_Rage]

Wp kiwi

[Lukie]

the monty hall situation depends on the person choosing which door to expose to you knowing which door holds the big prize or w/e and never shows that door to you

if the door chooser person picks at random, you're 50/50 the times he doesn't expose the prize by opening a door

Right, which is why Boost's response to the question as posed in the OP was at the very least, not wrong. Whether he intended that or not I have no idea.

[Lukie]

You don't get to see your card, and you're after the ace

lol, fail on my part.

But damnit, I want to see my cards when they are dealt to me.

[d0zer]

I may just be really baked, but I'd like to go on a rant about how counter-intuitive this whole shit is except I just lost my motivation cuz typing is hard and stuff

[Stacks]

the monty hall situation depends on the person choosing which door to expose to you knowing which door holds the big prize or w/e and never shows that door to you

if the door chooser person picks at random, you're 50/50 the times he doesn't expose the prize by opening a door

I didn't mean to say randomly showing one of the cards. What I meant was he just wouldn't have a system whereby he has to show the lowest card, as first stated by Dan. But that yes he knew which was the Ace, and would show the card that would leave us with 2 cards, one of which was still the Ace.

[drmcboy]

pretty sure it's just a test to see who will post the wiki link first, sorry jyms. But you won? there is some awesomeness to be had in not being the guy who knows the riddle, even if you are.

[Hawk]

ty kiwi

[CoccoBill]

Right, which is why Boost's response to the question as posed in the OP was at the very least, not wrong. Whether he intended that or not I have no idea.

Holding will never improve your odds and switching will never worsen them, it's always correct to switch whether the host opens the door randomly or not.

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?

Non-randomness implied. Boost's response was wrong.

[Lukie]

Holding will never improve your odds and switching will never worsen them, it's always correct to switch whether the host opens the door randomly or not.

If the host picks among the remaining doors randomly but still chooses a door with a goat, then staying and switching both have .5 car EV and based on your wording staying is also correct.

Non-randomness implied. Boost's response was wrong.

No, nothing in your post really hints at non-randomness if you've never been exposed to the Monty Hall problem. The host could easily know where the car is but still choose randomly, and while that wouldn't make a whole lot of sense in the context of an actual game show, that's not the point of the problem or an assumption that someone solving it should have to make. This is especially true because simply adding the words, host must choose a door with a goat makes it completely unambiguous.

[Lukie]

I know it might be a little bit of a stretch, but compare the following 2 sentences:

and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat.

to:

and the host, who must reveal a goat, opens door #3.

[Lukie]

Holding will never improve your odds and switching will never worsen them, it's always correct to switch whether the host opens the door randomly or not.

This actually made me think of an interesting variation that has somewhat to do with weighted ranges in poker. In said variation the host knows what is behind each door but can choose, at his discretion, how often or what % he will reveal each door. Of course, the solution is anticlimactic and it's exactly what you said. always switch.

[couriermike]

I thought this was gonna be about B. F. Skinner. Animal Learning Theory is totally fascinating.

[DropTheBanana]

I thought this was gonna be about B. F. Skinner. Animal Learning Theory is totally fascinating.

I did a research paper in college about him and behavioral psych on the whole. It really is incredible.

[couriermike]

They should create Skinner boxes to teach people poker. 'Oh, did you fold to the turn raise? Good, have a Snickers bar.' (That's a Mars bar in Europe, I think.)

back to monty hall:
kiwiMark and Lukie both explained it really well. The way Lukie explained it is kinda like somebody folding their equity share in a pot. When the host opens the wrong door and we switch, we get the probability of the wrong door for free.

[kevster]

Maybe this will help:

You have to nail one of three midget hookers bareback, and one of them doesn't have aids. You pick one, then the midget pimp who knows them intimately fucks one, very deliberately putting on an 8-layered full body condom after spraying her down with rubbing alcohol and bleach. Do you switch?

I lol'd the loudest I have lol'd in a long time. Very funny.

[kevster]

lolmaths, I've always found this to be the easiest way to think it through:

1/3 of the time you picked the car. If you change, you get a goat.
1/3 of the time you picked a goat. If you change, you get the car.
1/3 of the time you picked a goat. If you change, you get the car.

So you can see that changing will get you the goat 1/3 of the time and the car 2/3 of the time.


pointless continuation for completion's sake:

1/3 of the time you picked the car. If you hold, you get the car.
1/3 of the time you picked a goat. If you hold, you get a goat.
1/3 of the time you picked a goat. If you hold, you get a goat.

Holding gets you a car 1/3 of the time, and a goat 2/3 of the time.

Congrats. You just saved my life and those of my immediate neighbours. A mental breakdown followed by a killing spree was imminent.

[boost]

kiwi's breakdown is the best. My brain still doesn't like it simply because it is so counter-intuitive.

[oskar]

idk how it is counter intuitive. Changing is 1/3(+0) + 2/3(+1) and not changing is 1/3(+1) + 2/3(+0) and 2/3 of a car is often times more car than 1/3 of a car.#
Any of you guys play poker? *trollface*

[a500lbgorilla]

It's intuitive that it's intuitive which is why it's counter intuitive. I can't believe I actually have to write this out explicitly. :neil degrasse tyson face: