## Monty Hall Problem

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posiduck
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### Monty Hall Problem

I didn't see this one up here yet, figured it was worth posting.

You are on the game show "Let's Make a Deal", and you are faced with three doors.

Monty Hall tells you (and you know that he is not lying or anything), that behind two of the doors there is no prize, and behind one door, there is a brand new car.

Monty then explains that you get to choose a door. Monty will then reveal what was behind a different door from the one you chose. After he reveals that door, you will be given the chance to either stick with the door you initially picked, or switch to the remaining door.

Does switching alter your chances of winning, and if so, which is better, and why?

(remember not to post the answer here but in a solution thread as per spoiler policy, this thread should be for questions about the setup, etc.)
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wisnij
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### Re: Monty Hall Problem

Hah, that's a classic. The answer seems so wonderfully counterintuitive. I still haven't been able to convince my roommate it's correct though...
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posiduck
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I was about to reply to that here, but I'll reply in the solution thread, so it doesn't give anything away.
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GreedyAlgorithm
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### Re: Monty Hall Problem

posiduck wrote:Monty will then reveal what was behind a different door from the one you chose.

Just to be clear, he opens a door without a car behind it.
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Torn Apart By Dingos
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Never mention this problem to someone who hasn't already heard it and knows the answer. You'll regret it.

Marrow
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I have actually never heard it before, and I answered it, and figured out why on my own. Yay me.
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Verysillyman
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I read it last night in a book. It makes sense when it's worked out, but doesn't make real sense.

Marrow
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Its just too bad for me I was wrong, haha.
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Factitious
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Verysillyman wrote:I read it last night in a book. It makes sense when it's worked out, but doesn't make real sense.

Why doesn't it make real sense?
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posiduck
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Because everyone knows you can get better winnings from "Press Your Luck", so the situation shouldn't come up. The car is probably a dodge dart or something anyways.
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Verysillyman
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It doesn't make real sense because it's not intuitive. It has to actually be logicked out for it to work. At least that's how I see it.

Aeroplane
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This was on CarTalk, and they had so many people writing in insisting that this was the wrong answer that they actually set up a little example of it on their website and compiled statistics. Oddly enough, the results came out exactly as predicted.

DaveFP
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This one is indeed a classic. When I first had it explained to me I spent ages arguing that the answer was wrong. Then I realised that it was I who was wrong, and I felt sad.

EndofEternity
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according to my quick tree diagram... you have equal probability if you switch or not.

posiduck
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EndofEternity wrote:according to my quick tree diagram... you have equal probability if you switch or not.

Check out the solution discussion thread to see other people's takes on it.
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Penguin
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This problem also featured in an episode of Numb3rs, if anyone besides me watches it.
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Spider
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Penguin wrote:This problem also featured in an episode of Numb3rs, if anyone besides me watches it.

Indeed it did, I love that show!

I think it's because I desperately want to be Charlie...

RealGrouchy
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In the solution discussion, Tacroy wrote:As it opens, you are invited to play a game in which you have a 1 in 3 chance of winning. You pick one of the three doors, and there is a bit more than a 33% chance that it is the one with the car.

Then, Monty Hall opens a door and reveals a goat.

Of course, in Kazakhstan, this would indicate that you are unsuccessful, and did not choose the door with the goat behind it! [/borat]

- RG>
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Penguin
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I just saw this now, sorry for not responding earlier.

Spider wrote:Indeed it did, I love that show!

I think it's because I desperately want to be Charlie...

I think we should be friends. I'm in school learning how to be Charlie now. My jock brother wants to be a Fed and everything!
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Larissa
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### Interesting book

I missed the Numb3rs episode , but I know the book that was mentioned. "The Curious Incident of the Dog in the Night-time" by Mark Haddon. It's told from the viewpoint of an autistic boy. Good book, but you might not like it if you hated "The Old Man and the Sea" by Hemingway.

Penguin
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It was in the season 1 finale "Manhunt", if you're curious. Good episode... then again, they all are
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william
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To convince somebody, increase the number of doors from 3 to a million, and Monty Hall reveals all but two of the doors, never opening either yours or the door with the car. If Monty revealed every door but yours and door 152387 you'd be a moron not to go to door 152387.

gsail11
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I stumbled on this problem a few years ago when looking into probability, and I have trained my mind for it to be logical. When I brought it up with my highschool math team, laat year, it caused quite a controversy. One of our top students simply wouldn't accept any argument against that it would make no difference to switch. He's a smart guy, but he's just not interested in hearing that he got a problem wrong.

This reminds me of another fantastically counterinuitive problem, but I think i'll make a new thread for it.

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Pratchett solves this puzzle suprisingly well.
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ph33l_da_l0v3
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william wrote:To convince somebody, increase the number of doors from 3 to a million, and Monty Hall reveals all but two of the doors, never opening either yours or the door with the car. If Monty revealed every door but yours and door 152387 you'd be a moron not to go to door 152387.

I used that exact scenario to try to explain the problem to my friend and he STILL didn't understand... I couldn't convince him to ignore his intuition.

GreedyAlgorithm
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ph33l_da_l0v3 wrote:
william wrote:To convince somebody, increase the number of doors from 3 to a million, and Monty Hall reveals all but two of the doors, never opening either yours or the door with the car. If Monty revealed every door but yours and door 152387 you'd be a moron not to go to door 152387.

I used that exact scenario to try to explain the problem to my friend and he STILL didn't understand... I couldn't convince him to ignore his intuition.

Try the same scenario except bet money on it. Then either he'll figure it out or you'll win money. Win/win!
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ph33l_da_l0v3
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Haha good idea = P

steve110
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### Re: Monty Hall Problem

good idea = P? What's the mean?

Sir_Elderberry
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### Re: Monty Hall Problem

I understood like this:

-I pick a door. I've essentially split the doors into two groups "door I have selected" and "doors I haven't". "Doors I have selected" has a 1/3 chance of containing the correct door. "Doors I haven't" has a 2/3 chance. The host, then, gives me a chance to be guaranteed the correct one, if possible, from the 2/3 group.
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Lord Aurora
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### Re: Monty Hall Problem

Sir Elderberry, I like everything about you now, because I noticed the location you've specified in your profile.

Also, that is a very good way to look at the problem.
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squareroot
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### Re:

william wrote:To convince somebody, increase the number of doors from 3 to a million, and Monty Hall reveals all but two of the doors, never opening either yours or the door with the car. If Monty revealed every door but yours and door 152387 you'd be a moron not to go to door 152387.

That's the best ffing intuitive explanation I've heard in my lifetime.
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mike-l
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### Re: Monty Hall Problem

3.5 year necro, impressive!
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Vesuvius
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### Re: Monty Hall Problem

I agree this thread-o-mancy is obscene, but I'm glad for it.

I'm still trying to convince my brother this is true. I've pointed it out to him that the XKCD community agree, and he finally grudgingly accepts it.

He doesn't understand it, but that's not important.

squareroot
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### Re: Monty Hall Problem

Sorry, I hadn't realized how long dead it was.. it was on the front page of the "Logic Puzzles" forum. Heheheh...
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redrogue
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### Re: Monty Hall Problem

I couldn't get this through to a friend even WITH the million doors scenario. He was so convinced it was 50-50, he would not accept any explanation. So I grabbed three playing cards and played "find the ace" for \$1 a round. After only ten rounds (and only two successes with his "not switching" strategy), he was astounded. Only then could I explain the math behind it, and only then did he get it. Best \$6 I'd ever made.
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Puck
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### Re: Monty Hall Problem

The explanation that I've been able to convince people with is:

<Give them 3 doors, say they select door 1>

Now, do you want what's behind door number 1, or what's behind Doors 2 and 3?

They'll protest, but you can then show them that if the prize is behind Door 2, their switch will always be to Door 2, and if the prize is behind Door 3, their switch will always be to Door 3.

---

There is some recent science suggesting that pigeons are better than people at this game, but I sort of doubt they understand what's going on... If a pigeon pushes a button and doesn't get a reward immediately, it would seem natural to try a different button rather than trying the same one again.
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Turtlewing
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### Re: Monty Hall Problem

redrogue wrote:I couldn't get this through to a friend even WITH the million doors scenario. He was so convinced it was 50-50, he would not accept any explanation. So I grabbed three playing cards and played "find the ace" for \$1 a round. After only ten rounds (and only two successes with his "not switching" strategy), he was astounded. Only then could I explain the math behind it, and only then did he get it. Best \$6 I'd ever made.

I'll bite. What is the math behind it? (i tried to find the solution thread but couldn't)

To me it looks like this:
chance of getting the prize on first try 1/3 (irrelevent)
chance of getting the prize after the door is opened 1/2 (choose to stay or change)

The argument that the original choice from 3 doors affects the second phase sounds a lot like "rolling the ones out" of a d20.
Presumably i've missed something but I don't know what.

mike-l
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### Re: Monty Hall Problem

Turtlewing wrote:
redrogue wrote:I couldn't get this through to a friend even WITH the million doors scenario. He was so convinced it was 50-50, he would not accept any explanation. So I grabbed three playing cards and played "find the ace" for \$1 a round. After only ten rounds (and only two successes with his "not switching" strategy), he was astounded. Only then could I explain the math behind it, and only then did he get it. Best \$6 I'd ever made.

I'll bite. What is the math behind it? (i tried to find the solution thread but couldn't)

To me it looks like this:
chance of getting the prize on first try 1/3 (irrelevent)
chance of getting the prize after the door is opened 1/2 (choose to stay or change)

The argument that the original choice from 3 doors affects the second phase sounds a lot like "rolling the ones out" of a d20.
Presumably i've missed something but I don't know what.

Read Puck's reply immediately before yours, that's really what's going on. You are given the choice between {I got it right first} and {I got it wrong first}. In the case where {I got it right first}, Monty is irrelevant and you are wrong if you switch. In the case where {I got it wrong first}, Monty does you the favor of showing you which of the remaining doors has the car.

The reason it's not like 'rolling the ones out' is that the door the car is behind isn't chosen again after Monty reveals a goat, it's still where it originally was.
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douglasm
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### Re: Monty Hall Problem

Turtlewing wrote:The argument that the original choice from 3 doors affects the second phase sounds a lot like "rolling the ones out" of a d20.
Presumably i've missed something but I don't know what.

"Rolling the ones out" doesn't work because every time you reroll the previous result is discarded and made completely irrelevant. "What you rolled last time" simply doesn't appear in any part of the new roll's situation. With the Monty Hall problem, the choice is "Stick with your original choice, or swap." If the problem is restated such that the bolded section is no longer accurate or cannot be determined, then and only then will the first choice be irrelevant to the choice to swap.

The common mistake in reasoning about this problem is to treat it as if the host said "Here's a goat, now close your eyes while I shuffle the remaining two doors. Ok, now pick one." In the actual problem as stated, the shuffling doesn't happen.

Now, the math:
Always stay:
1/3 chance you pick the prize the first time. In this case you stay, and you win the prize.
2/3 chance you pick a goat the first time. In this case you stay, and you get a goat.
1/3 chance of winning the prize.

Always switch:
1/3 chance you pick the prize the first time. In this case the host reveals a goat, you switch to the other goat, and you get a goat.
2/3 chance you pick a goat the first time. In this case the host reveals the other goat, you switch to the only remaining door, and you get the prize.
2/3 chance of winning the prize.

Malle
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### Re: Monty Hall Problem

Let's try to explain it using a dice analogy.

You need to roll a 20 to win a roll. The game master wants to give you a good chance to win, so he does the following.

He takes 20 20-sided dice and orders them so that each shows a unique value, so all values from 1 to 20 are shown, then hides each die under a separate cup.The GM knows which face is showing under which cup, but you do not.

You pick one cup, which might or might not hold the dice showing 20. The game master then says to you "To give you a better shot at winning this roll, I'll tell you that if the die showing 20 is under any of the cups you did not choose, it's under this cup" and points to one of the cups you did not pick.

You know that you had 1/20 chance to pick the right cup from the start.
You also know that the chance that you didn't pick the right cup at the start is 19/20.
You also know that if you didn't pick the right cup from the start, the die you are looking for is under the cup the GM points to.

Thus, since the probability you didn't pick the right cup from the start is 19/20, the probability that the die is under the cup the GM points to is also 19/20.