Monty Hall Problem

[Phred]

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 other 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 take the switch?


Does it matter?

This is a problem that has sparked a great many arguments, and controversy. Let's see who can get it.

[Phred]

Here is the answer for those who want it...



...........DON'T Read if you haven't voted yet........



The key is that the host knows what is behind each door.

What if I asked you this: Pick a door - you pick A. Now would you like to stay with door A, or pick door B AND C - if it is behind either of them you win. What would you do? Most likely you would see it to your advantage to switch. If you stay with your original pick you have a 1/3 chance. If you switch to both doors, you have a 2/3 chance right?

It doesn't change if I show you a door that definately doesn't have a car behind it. When you are shown the door you will never be shown your door or the winning door. Only the other bogus door. Which means the odds of the Car being behind the door you didn't choose is 2/3 and 1/3 of being behind your door.

Put another way, imagine that the car is behind C. You have three choices, you can choose A, B, or C. Lets look at each.

1. If you choose A - you will be shown B, if you switch to C you win :)
2. If you choose B - you will be shown A, if you switch to C you win :)
3. If you choose C - you will be shown either A or B, if you switch to the other door you will loose :(

in two of the three choices you win, only in the third will you loose.

For those that still don't get it, what it there where 1000 doors. I asked you to choose a door and you choose say door 17. I said, okay, I am going to show you every door except door 17 and door 796. The car is now behind either door 17 or 796. What are the chances that you got it right on your first guess? 1/1000. What are the chances that it is behind door 796? 999/1000.

[Ael]

I could be wrong about this, but let's see..

If you have three doors, it's safe to say that you only have a 1/3 chance of getting the car and 2/3 chance of choosing a goat.

If someone shows you one door, well that solution is no longer available to you. If Door C has the goat, you take it out of the picture - you are now left with two doors, Door A and Door B...one has a goat, one has car, therefore, your odds actually increase in this case, cause before you had a 33% chance of getting the car and 67% chance of a goat...with the third door gone, you only have two options, "win or lose", and increases your odds of getting the car to 50% and brings the chance ot getting a goat at the same level...

It's the same thing in your last problem there...if you have a 1000 doors, and ONE door has the prize, then the chance you would get that prize, is 1/1000.

If you show me EVERY door but two of them...well now that I have two doors to choose from, it's again 50/50...

[Enigma™]

I see what he's saying though:

Do you have to assume that the host is trying to make you lose or win?
I mean, again I see what you mean though: between 3 doors, you have a 33% chance of winning. If you're shown another door that is wrong, then your odds were still that you only had a 33% chance of getting it right in the first place... BUT:
As Ael said though- If you change any options, remove all the doors but two, you're still left with a 50% chance of winning/losing, no matter WHAT came before it.

The 1000 doors makes more sense however in that WHY would he only choose door 700s door? But that's more about "intuition" then it is the mathematical portion of the problem. I'd probably choose the 700s door too to be honest, however mathematically, no matter how you break it down, if you have 2 choices and only one of them are correct... Then it's a 50% chance either way.

[Phred]

Yeah, this problem is hard to get your brain around. If we can assume that we believe that the 1000 door reasoning is the same as the 3 door reasoning then let me use that for now on to try and explain it.

Pick a door between 1 and 1000. say you pick 539. Now I say. Do you it is more likely that you got it right or wrong?

Ok, obviously it is more likely that you got it wrong. In fact you had a 1/1000 change to get it right.

Now someone says, okay I am taking away the guess work and showing you that it is either behind the door you guessed (which we just agreed that you probably guessed wrong) and another door 317.

Why would it be under the door that you guessed? It was a one and 1000 shot to get it right, unless you were just super lucky, you got it wrong.

Even though you only have two doors left, the odds are not 50/50 at all. They are 999/1000.

Look at it this way: Let these be 10 doors.

A B C D E F G H I J

Guess one: Say F.

So now we have.
F | A B C D E G H I J (I just moved the guessed door to one side).

Which side of the '|' is the car most likely on? way more likely that it is on the right side of the '|' right? The odds that it is one the right side are 9/10 right?

Now I say, I will show you that the car is either behind your choice: F or it is behind C.

F | - - C - - - - - -

Just because there are two doors left, the odds be would not be 50/50. They stay the same. 1/10 for F and 9/10 for C.

It works the same way for 3 doors. 1/3 and 2/3.

Better?

[Ael]

No, it would be 50/50...if you're left with two doors, your odds increase now because you don't have to go through the other 8 doors...if you know one door has your prize, and the other one doesn't, then you have an even chance.

Your point would only work if you had doors:

A B C D E F G H I J

Choose F, let's say.

F | A B C D E G H I J

Okay, your door is either behind F or C, BUT, I'm gonna leave the other doors as options to choose from...wlel then you have your 9/10 shot of getting it wrong...

If you disclose other options nad give someone two to choose from, and one door is the prize and the other door makes you lose, that's 50% no matter which way you look at it.

You can give me a million doors, narrow it down to two, and STILL have that 50% chance of getting it right, because your sample space is 2, you have 2 things to choose from, and just an equal chance of getting it wrong as you do getting it right.

[Enigma™]

The only error in reasoning I can see that you're making is that you're assuming the one you chose is incorrect - where it VERY well could be entirely correct.

You say that you have a 1/1000 chance of being wrong, but you know what? You also have 1/1000 chance of choosing ANY other door as well. It's not restricted, and as soon as you change the options down to 2 doors, then here's where the flaw in your logic comes in:

You're making the assumption that the math is incorperating the previous possibilities. HOWEVER when you have 2 doors, no matter how you got there, they're still 2 doors.

If you were to apply the same logic to the same idea, but let's change something:

1000 doors, and 500 doors are cars and 500 doors are sheep.

Now apply the exact same logic and you'll see the flaw. I'd keep explaining but I have to get to work now lol :)

[Phred]

The only error in reasoning I can see that you're making is that you're assuming the one you chose is incorrect - where it VERY well could be entirely correct.


Enigma, without even knowing it, you may have helped proove my point. Worded another way you are saying that the only way that I can not win the car by always switching is if I guess right the first time right? What are the odds of me guessing right the first time? 1/the number of doors right? The means my odds for guessing wrong the first time are (number of doors - 1) / number of doors.

Which means that my odds to win the car are that.

so in the case of 1000 doors, you have a one in 1000 chance of getting it right on the first guess. Which means you have a 999/chance of getting it wrong, which means if you get it wrong in the first guess, you are switching to the car.

The 50 50 argument doesn't work. Just because I narrowed in down to two doors, it doesn't mean it is 50 / 50. The host (me in this case) is only going to show you doors without cars. So the host knows that all the doors he is about to show you don't have the car. Right?

So assuming that and assuming that once you make a guess, the car doesn't teleport to another door, then the probability is based on the number of doors.

Still don't believe me, Try this:

Ael or Enigma (or anyone else for that matter), pick a number between one and 1000. Do it now, and if you don't cheat you may start to believe me. Did you do it? Okay, really no cheating. Now using this method we have been talking about, I am going to try and guess your number.






I am going to guess that it is eight hundred sixty three.

Now you give me two choices, eight hundred sixty three and one other choice. If your number is really eight hundred sixty three, you just make another one up. If it is a different number, then you tell me that number. If you didn't cheat, that I have a 999/1000 chance of guessing your number right?

The only way that I won't get it right is if I got your number off the bat.

[Enigma™]

right, but when you get me to TELL you what my number was, the thing is that when I choose a door, they don't tell me if I'm right or not... it's slightly different man. I see what you're getting at and I think it's a good psychological thing, but not a good mathematical one ;)

[Ael]

lol...i'm gonna stop, i have a bad headache...need to find me som biaxident ;)

[Liquid Shadow]

I don't see how you guys are getting all this mathematical stuff out of it...


He opened one door and there was a goat. That means that of the two remaining doors, there is a goat and a car.

1 goat.
1 car.
2 doors.

You have a 1/2 chance to get the car.

[Phred]

Okay. I did a quick search on Monty hall problem (which is the name of this riddle). Here is what I found

http://www.cut-the-knot.org/hall.shtml

Maybe this will convince you ;)

There is even a simulation

(edit - I just tried the simulation and it sucks. First you have to press reset. Then, just choose a door and it shows you at the bottom if you would have won if you switched or if you would have if you didn't switch. You don't get to actually switch or anything. You can do it over and over and check out your percentage. I did it 200 times(I tried to be random) (it really doesn't take long to do that many - and got the following:

If you switch 132 wins 68 losses
If you do not switch 68 wins 132 losses

That means that this simulation gave me a 132/200 odds which is slightly less than 2/3s. (66/100 instead of 66.666... /100).

But more than 50/50.

I am not sure I can do any better at explaining this, so if you still don't believe me after this simulation, and are interested, do a google on Monty Hall Problem, or Monty Hall Dillema. It is a famous Problem in the stats circles.

[Enigma™]

Although I see the mathematics in the logic now, even after a "trial run" of 20 games here:
http://math.ucsd.edu/~crypto/Monty/monty.html
I won 5 of 10 times by switching, and 5 of 10 times by NOT switching...

I'd be tempted to still say 50% even though I understand HOW to get the other percentage.

[Ael]

This is why I'm going into music and not math...

[John]

If the host picks the door at random then you have a 50/50 chance and it doesn't mater whether or not you switch.

If the host knows what door has the car and automatically chooses it for you to make the game interesting, you better switch.

- John O.

The key is that host knows which door has the car & picks it for you. I'm going to guess that in a game show the host's pick would be random. Otherwise it wouldn't be very interesting and everone would always switch.

[Alessandro]

Guys I did not read all posts anyway:
I can tell you for sure that changing the door increases the chance to win

I can tell you this because we demostrated it 2 months ago during a statistics course I attended at the faculty of Economics in Padua.

Cheers

[hyeok]

Yes, it is indeed true that your odds are better by switching. Probability shows unexpected things. The trickiest concept seems to be the idea of entanglement, where the probability of consecutive events actually depend on one another. Here is a very simple example that will illustrate the point:

If you flip a coin once and get heads, what are your odds of getting heads again the next time? Initial thought may lead to the conclusion that the odds of getting heads or tails is the same again the next time, since the first flip is independent of the second flip. And hence, the odds are 50/50 each time.

But this is where the laws of probability prove your instincts wrong! Since the OVERALL average between heads and tails you get in multiple coin flips have to be 50/50, if you flip the coin twice, it is more likely that you will get two different outcomes than two same outcomes. So, if you get heads the first time, it is more likely that you'll get tails the second time.

Strange, isn't it. Two separate events' outcomes depend on one another. In probability, you can't simply judge each even independently. That is why your odds are improved by switching doors in that game. I think this page explains it elegantly:
http://en.wikipedia.org/wiki/Monty_Hall_problem

[Liquid Shadow]

I still say it's 50/50


One of the remaining doors has a car, the other has a goat.

One car.
One goat.
Two doors.

1/2 chance.

[hyeok]

I still say it's 50/50


One of the remaining doors has a car, the other has a goat.

One car.
One goat.
Two doors.

1/2 chance.
The important thing you have to consider is which door you picked initially.
You can't just assume that you have chosen the door without the car. What if you chose the door with the car? Than you still have

One car.
One goat.

The difference is, you're either on the goat and about to switch, or your on the car and about to switch.

In the beginning you have 1/3 chance of getting the car, and 2/3 chance of getting the goat.

If you chose the car, now you have 1/2 chance of keeping the car. You may choose to switch or you may choose to stay.

If you chose the goat the first time, you once again have 1/2 chance of keeping the car or getting the goat. Liquid Shadow, you've only considered this stage of the game, but you must also consider the inital probabilities of 1/3 and 2/3!

[John]

Because the host is playing mind games with you, it's basically 50/50.

You could throw 100 heads in a row and the chance of the next one being a head is basically 50/50.

But this is where the laws of probability prove your instincts wrong! Since the OVERALL average between heads and tails you get in multiple coin flips have to be 50/50, if you flip the coin twice, it is more likely that you will get two different outcomes than two same outcomes. So, if you get heads the first time, it is more likely that you'll get tails the second time.


The OVERALL average doesn't have to be 50/50. It's astronomically very unprobable, but there is a probability that every throw you do for the rest of your life will be heads.

The chance of throwing:

2 heads in a row - 1/4
4 heads in a row - 1/16
8 heads in a row - 1 / 256
16 heads in a row - 1 / 65536
32 heads in a row - 1 / 4,294,967,296

You could keep going like this forever and calculate a probabilty of throwing all heads for any number (i.e. 1,000,000,000,000,000). The point is that while the law of averages has a very very good chance of working out, there is a very very very astronomically remote chance that you could throw heads for the rest of your life. It's so remote that it's hardly worth talking about (32 heads in a row = once out of 4.3 billion tries). But that chance does exist. There is no garantee that coin tosses will work out to be 50/50, but chances are they will.

[hyeok]

Because the host is playing mind games with you, it's basically 50/50.

No, the host is not playing mind games with you. The host reveals a door with the goat whether you pick the car or not. The host's action is already predetermined, regardless of what the player does.

But you are correct about the fact that the average will most likely be close to 50/50, but does not necessarily have to be. When I said,

Since the OVERALL average between heads and tails you get in multiple coin flips have to be 50/50

replace "have to be" with "will most likely be" :D

But still, your chances do not stay the same when you switch doors in the monty hall game. Check out the wikipedia page. It explains it better than i could!

http://en.wikipedia.org/wiki/Monty_Hall_problem

This is probably the most eloquent explanation that Wikipedia provides:

By switching, the player is ensuring that he will win if he originally picked a goat. The probability of picking a goat was 2/3, so the player should switch.

[John]

I concure. I'm sure the host isn't trying to play games with you on the game show.

[EdKeys]

ok first of all for ppl that still cant figure this thing out, in terms of PROBABILITIES it does matter and its better to switch.

The flipping coins idea is good, if u really think that u can flip coins and continue getting heads forever then, it dosent matter if u switch doors, but if think the way I do flipping coins will give different results and its more likely to get heads tails heads instead of heads heads heads.

In other words, try not to see this as FACTS, its all about probabilities.

In the game, its obvius that its easier to pick a goat than the car, on the original pick. So then, if the host picks a door with a goat, u should switch beacase odds say u had a 2/3 chanse of being wrong...

[John]

its more likely to get heads tails heads instead of heads heads heads.


Actually the probability of getting heads tails heads is the same as the probabilty of getting heads heads heads. You can try this at home. Take a coin toss 100 series of three tosses/series. Count the number of Heads Tails Heads (not Tails Head Tails) and count the number of Heads Heads Heads. You'll find that most of the time they come out about the same. Sometimes, you'll find where Heads Heads Heads actually has more.

I'm way to lazy to explain it, but you can ask just about any math teacher. If they tell you otherwise, they either didn't understand the question or they're not a very familar with probability.

- John O.

PS - What fun is a game show if just about every contestant interaction is going to go:

Contestant: "I'll take door one"

Host: "Now, I will reveal a door for you"

Contestant: "Ok, I'll switch"

That's really compelling. Everyone is going to want to watch that. It's much more interesting if the host has the ability to play little mind games with you. Then it becomes much more "interesting" for the viewers. Just a wild guess, but I would think the people sponsering the game show would want it to be as "interesting" as possible.

[Phred]

It's nice that I am no longer alone trying to convince the world.

;) - Thanks guys.