Ok, some of you may know this but it's a pretty cool counter-intuitive problem.



So, you are faced with 3 boxes. You know 2 of them have a goat in them, and 1 has a Car. The host, monty, asks you to choose a box. You choose one, after you do so Monty opens one of the 2 remaining boxes and reveals a goat. However, Monty knows which box the car is in, so no matter what after the player makes there first choice he always opens the box which has a goat in it. At this stage, you are asked whether or not you would like to switch the box you chose with the other, non opened box.

The question is, what should you do? Switch, or not switch? Why?

Not switch.

Your original choice has a 50% chance of being the car whilst if you switch it's 0%, since if you do switch, Monty will go for the box with the goat in either case.

Hey man don't think you understand fully. After you choose 1 of the 3, then monty opens one of the remaining 2 boxes but you know he always opens the goat, never the car, so you don't know if the one you picked is the car, or the other box is the car.

So what should you do?

Pick the box that isnt moving, making noise and smelling like ass.

Actually I screwed it up, in the original they are doors but it doesn't really matter.

You should always switch.

In the 1st instance, you have a 1 in 3 chance of choosing the car. In the 2nd instance, after Monty has gotten rid of one of the goats, you now have a 1 in 2 chance of getting the car. Since Monty will always avoid picking the car, you should go for the one he didn't pick coz there's a 50% chance you'll be right. It may well turn out that the box/door you originally picked had the car, but you only have a 33% chance of being right. However, in scenario #2, you've increased the chances to 50% and should switch.

http://www.youtube.com/watch?v=Wky3WqpAzXw...feature=related

gis very close but not quite.

I'm pretty sure as soon as he opens the first box you have a 50/50 chance whether you switch or not.

ARRRRRRRRRRRRRRRRRRRRRGGGGGGGGGGGGGGGGGHHHHHHHHHHHHHHHHHHH!

So you're saying that you pick a box, and before Monty opens the one you picked, he opens a box with goat in it?

and then he asks you if you would like to switch your choice with the remaining box. I dont think it matters if you switch or not. Out of the two unopened one is definitly a car and one is a goat, no matter which one you picked first there will always be atleast one goat for monty to reveal so its 50/50 after he opens the box whether you switch or not.

The only reason I doubt that is because its the obvious answer and would make the question kinda pointless.

Haha yeah, like I said it's really counter-intuitive. The first time it was written about there was heaps of complaints/objections/misunderstanding, etc.

alright whats the answer? I'm not gonna get any work done today unless you tell me, PM it to me.

Ok, you have a 1/3 chance at the start, so there is a 2/3 chance it's not the box you picked. Then when he eliminates a box you switch, coz your original box is still only 1/3 chance. However, the other box is still a 2/3 chance. Is that it?


If it isn't, I'm done with this shit. My brain can't cope with problems any more.

Essentially, what I'm trying to say, is that you are effectively picking the other 2 boxes when you switch.

Gis is right.

Scenario 1: You pick a goat, Monty shows you the other goat and offers you the car
Scenario 2: You pick the car, Monty shows you a goat and offers you the other goat

Since there are two goats and one car, Scenario 1 is twice as likely. Therefore, you should switch.

Sounds like the right answer...

It is. What do I win?

I dont buy it, him revealing the goat makes it like that box never existed and giving you another choice makes it a whole new game like your first choice never existed, you are now picking between 2 boxes and 2 boxes only the third has been eliminated. So your chances are 50/50.

You know you will have 2 chances to guess so how is the equation ever 1/3? The act of him revealing the goat you DIDNT pick makes your next guess 50/50. If he were to reveal the box you did pick and let you choose again your overall chances would be 2/3 but since he revealed the one you didnt pick you have no way of knowing which of the remaining TWO boxes contain the car making your first pick meaningless it can have no effect on the odds of your next choice you might aswell have never made the first choice or even included a third box because you now have ONE choice for TWO boxes or 1/2.

I tried to explain that a few ways so you might get one.

Monty might aswell just reveal one of the goats at the beginning and say now choose which of the remaining two has the car. What would be the odds then? 50/50.

My friend asked me this at the end of last year. You switch boxes. Your original odds are 1/3 so if you stay with the first box, you have a 1/3. Once there's only 2 boxes you switch because odds are it wasn't the first one you chose and now you have a 1/2 chance. He explained it a lot better than this. I think once the first box is taken away that box has a 0% chance, your original has a 1/3 (or 33.3%) chance, and the final box has a 2/3 (or 66.6%) chance - if that makes any sense. But yes, you do switch.

watch the movie 21.

People have trouble with conditional probability

shoot the cunt, sell the car, and buy more goats.

Spot on man! Very good work!

No but Kane remember I stressed that he always shows you a goat. If you pick the car, he shows you a goat, if you picked a goat he still shows you a goat. The showing of the goat changes things but for a different reason, just that it eliminates one of the boxes.



At the start you had a 1/3 chance to pick the car right? So when he asks you to swap, you know that originally you either picked, the car, goat 1 or goat 2. Now monty opens you a door and asks if you should switch or not. Now, the only time you don't get the car is if you originally chose the car in your first guess and thus when you switch you lose it and get a goat. However, there is a 2/3 chance that you chose a goat in your first guess, now every time you originally chose a goat in your first guess and then you switch when asked to, you will win right? Because if at first you chose a goat [ 2/3 chances ] and then he shows you a goat, the other box must be the car.

So basically you had a 1/3 chance of being correct at the start, showing a goat doesn't mean anything because you chose the original door BEFORE you knew about the goat. So it's a 2/3 chance you didn't get the car first up, which means if you switch you will get the car.

Ask if you need further clarification.

who cares i'll just buy my own car if i dont win and if i lose i still got a goat, i can milk that shit and make a profit off goat milk

LMAO

Good thinking

Yay me!

It's all about the probability of you picking the right door initially. That's what's important.

Look at it this way. You pick a door, let's say Door 1. Monty then says you can either open that door or BOTH of the other doors. You would switch, wouldn't you? It makes sense, because you're opening two doors instead of one. Your odds increase to 2/3. However, you KNOW that one of those two doors has a goat behind it. That's a fact, regardless of what's behind Door 1. Monty showing you the goat makes no difference at all. He's showing you what you already know to be true, it doesn't affect the probability.

Let's say I take two cards, the Ace of Spades and the Queen of Hearts, and deal us both one card each, face down. Without looking at your card, you have a 50% chance of holding the Ace of Spades. If I offer you the switch, it doesn't matter whether you take it or not, probability-wise. The odds at the time the cards were dealt was 50%.

Let's say I take the Ace of the Spades and three Queens from the deck, and deal us both two cards each, face down. Without looking at your cards, you have a 50% chance of holding the Ace of Spades. If I look at my cards and turn over the Queen of Hearts, does that make any difference? No, because one of my two cards must be a Queen. I'm simply showing you which one it is, which is irrelevant. If I offer to swap my one remaining face down card for both of your face down cards, it doesn't matter whether you take it or not, probability-wise. The odds at the time the cards were dealt was 50%.

Now let's say let's say I take an entire deck of cards and deal you one, at random, face down. I keep the remaining 51. The odds of you having the Ace of Spades is about 2%, and the odds of me having it are about 98%. Given the opportunity, you would switch seats with me. But there's only one Ace of Spades, so (at least) 50 of my cards are worthless. It doesn't matter whether I keep them in my hand or whether I show them all to you. Even though we both only have one card I'm still far more likely to have the Ace of Spades than you are, because I was far more likely to be dealt it in the first place.

Try the online game if you're not convinced.

That says it better than I ever could

Actually it makes all the difference in the world. The fact that you know monty will show you a goat is the reason the probability is 2/3. Otherwise if he just randomly opened a door [and he didn't know it would be a goat or the car] the probability would be 50/50. Similarly, if you didn't originally pick a door and you arrived and he already had one door open with a goat and the 2 other doors, the probability would be 50/50.

Like I explained earlier and JVLN FNGZ explained, it's all in the fact you choose the first door and you know he will show you a goat either way.

At the start there's a 1/3 chance of picking the car and 2/3 of getting a goat. Look at these permutations to understand easier.


Goat > Switch = Car +
Goat > Switch = Car +
Car > Switch = Goat. -

Car=2, Total = 3

Probability of winning with switching= 2/3



Goat > don't switch = Goat. -
Goat > don't switch = Goat. -
Car > don't switch = Car. +

Car=1, Total =3.

Probability of winning without switching = 1/3,


That's it all right there. Hope you understand now. [If not, lost cause haha].