The Monty Hall Problem

[rehmwa]

So, basically now I have a 50% chance to get the car.

no - you started with a 33% chance, that didn't change - the only thing that changed was the 66% chance moved from belonging to two doors, to one door.

(your pick of door was random

his pick of door is NOT random)

there's no psychology behind Monte's strategy - He ALWAYS shows an unpicked goat and allows the option to switch

so there's 100 doors - you pick one - he reveals 98 goats - do you switch to that last door?

[boogielicious]

Quote:

Originally Posted by Abu3baid

So, basically now I have a 50% chance to get the car.

no - you started with a 33% chance, that didn't change - the only thing that changed was the 66% chance moved from belonging to two doors, to one door.

(your pick of door was random

his pick of door is NOT random)

there's no psychology behind Monte's strategy - He ALWAYS shows an unpicked goat and allows the option to switch

so there's 100 doors - you pick one - he reveals 98 goats - do you switch to that last door?

I only want one goat, not 98 goats.

[Golfingdad]

I was one of these people for the longest time too. Eventually got the right explanation and crow was eaten.

[rehmwa]

I only want one goat, not 98 goats.

I understand

[CarlSpackler]

I understand

If this is the goat, I would rather have it than just a car.

[iacas]

I thought Tiger was the GOAT.

[CarlSpackler]

I thought Tiger was the GOAT.

Great. You had to bring Eldrick into this. Now it's going to turn into a is he better than ______ thread. ;-)

[Abu3baid]

no - you started with a 33% chance, that didn't change - the only thing that changed was the 66% chance moved from belonging to two doors, to one door.

(your pick of door was random

his pick of door is NOT random)

there's no psychology behind Monte's strategy - He ALWAYS shows an unpicked goat and allows the option to switch

so there's 100 doors - you pick one - he reveals 98 goats - do you switch to that last door?

The reason I mentioned it is a 50% chance is because I have 2 doors and I know that one of them has a car and the other has a goat..

As to your question, this is a very difficult thing now IMO.. without knowing the hosts intention in allowing me to switch how can I make another pick that is not random??

I'm sure there is something I am missing.  let me think about it more.

[Golfingdad]

The reason I mentioned it is a 50% chance is because I have 2 doors and I know that one of them has a car and the other has a goat..  As to your question, this is a very difficult thing now IMO.. without knowing the hosts intention in allowing me to switch how can I make another pick that is not random?? I'm sure there is something I am missing.  let me think about it more.

Read tees2trees explanation ... I believe it to be the most straightforward. You had 33% originally, you have 67% of you switch cuz you're effectively getting two doors for the price of one.

[CarlSpackler]

The gambler's fallacy says that the outcome of previous events does not have bearing on the outcome of future events. Each flip of a quarter has a 50% chance of being heads even if it has come up heads the last 50 times. It depends on whether you look at this as one event or two separate events. From one point of view, you have a new event when given the opportunity to change that is now 50/50. I'm assuming that only an idiot would pick the door that already revealed the goat.

[saevel25]

Read tees2trees explanation ... I believe it to be the most straightforward. You had 33% originally, you have 67% of you switch cuz you're effectively getting two doors for the price of one.

Exactly.

I get the scenario were people think,  it is just two doors so it is 50/50 now.

The premise of the problem is you have 3 doors to choose from. You pick one, you are shown one of the remaining two, should you switch? The math shows when considering a 3 door scenario switching gives you a better chance.

As to your question, this is a very difficult thing now IMO.. without knowing the hosts intention in allowing me to switch how can I make another pick that is not random??

I'm sure there is something I am missing.  let me think about it more.

The Host's intention is to always show a goat and ask you to change. Not really anything deceiving about it.

Here is a diagram explaining the situation. You have 3 doors. Each scenario as your 2nd option asks you to switch or change. As shown, if you stick with your door then you only win the car 1 out of 3 times. If you switch you win a car 2 out of 3 times.

I think the situation that leads it to be skewed away from 50/50 chance is that you original pick a car under the 1/3 premise. So right from the start you have a 1/3 chance of having that car. If you sit on that pick you have a 1/3 chance no matter if the host shows you a goat behind one of the doors.

[Golfingdad]

The gambler's fallacy says that the outcome of previous events does not have bearing on the outcome of future events. Each flip of a quarter has a 50% chance of being heads even if it has come up heads the last 50 times. It depends on whether you look at this as one event or two separate events. From one point of view, you have a new event when given the opportunity to change that is now 50/50. I'm assuming that only an idiot would pick the door that already revealed the goat.

That would be true if the door the host opened was random - if sometimes he revealed a car before asking you to switch, but he doesn't. Its not at all random, it's going to be a goat 100% of the time. And because of that, mathematically, it's completely irrelevant as to whether or not he opens the door before or after you make the switch. Even before you switch, you KNOW the door he opens is s goat, which is why you're trading one for two.

[CarlSpackler]

I decided to put this to the test. You can play a simulator at http://www.stayorswitch.com/ . I randomly picked the first door and never won a car. I always switched and...

Won a car 9 times out of 10.

Edit:

I played 10 more times staying every time and...

It's mind blowing.

[Abu3baid]

Some very cool stuff.. Thanks

[Golfingdad]

I decided to put this to the test. You can play a simulator at http://www.stayorswitch.com/. I randomly picked the first door and never won a car. I always switched and...

Won a car 9 times out of 10.

Edit:

I played 10 more times staying every time and...

It's mind blowing.

Or ... (assuming initial door chosen was door #1 - although it works exactly the same regardless)

Behind door 1	Behind door 2	Behind door 3	Result if staying at door #1	Result if switching to the door offered
Car	Goat	Goat	Wins Car	Wins Goat
Goat	Car	Goat	Wins Goat	Wins Car
Goat	Goat	Car	Wins Goat	Wins Car

Stay = 1 out of 3, switch = 2 out of 3.

[CarlSpackler]

Or ... (assuming initial door chosen was door #1 - although it works exactly the same regardless)

Behind door 1

Behind door 2

Behind door 3

Result if staying at door #1

Result if switching to the door offered

Car

Goat

Goat

Wins Car

Wins Goat

Goat

Car

Goat

Wins Goat

Wins Car

Goat

Goat

Car

Wins Goat

Wins Car

Stay = 1 out of 3, switch = 2 out of 3.

It made a lot more sense how this is not a second decision after playing it a bunch of times.

[boogielicious]

Quote:

Originally Posted by CarlSpackler

The gambler's fallacy says that the outcome of previous events does not have bearing on the outcome of future events. Each flip of a quarter has a 50% chance of being heads even if it has come up heads the last 50 times. It depends on whether you look at this as one event or two separate events. From one point of view, you have a new event when given the opportunity to change that is now 50/50. I'm assuming that only an idiot would pick the door that already revealed the goat.

That would be true if the door the host opened was random - if sometimes he revealed a car before asking you to switch, but he doesn't. Its not at all random, it's going to be a goat 100% of the time. And because of that, mathematically, it's completely irrelevant as to whether or not he opens the door before or after you make the switch. Even before you switch, you KNOW the door he opens is s goat, which is why you're trading one for two.

I don't understand what you have against goats.

[turtleback]

The reason I mentioned it is a 50% chance is because I have 2 doors and I know that one of them has a car and the other has a goat..

As to your question, this is a very difficult thing now IMO.. without knowing the hosts intention in allowing me to switch how can I make another pick that is not random??

I'm sure there is something I am missing.  let me think about it more.

There is no intent on the part of the host.  There are always 2 unchosen doors, at least one of them has to have a goat, and he always reveals a door that has a goat.  The host's role is, essentially, mechanical.

It is fairy easy to verify the correctness of the answer using monte carlo simulation.  I even did it using Excel a few years ago, to convince someone who said it just couldn't be right but who would only believe empirical evidence..

To me the easiest way to see the answer is to consider that I have a 1/3 chance of hitting the correct door the first time and a 2/3 chance of hitting the wrong door the first time.  Monty showing one of the unchosen 2 doors does not chance the probability that I got it right the first time.  And it means that the probability of the car being behind one of the 2 unchosen doors is 2/3, and revealing the one the car is NOT behind does not change that.

Now, if Monty had simply said "you can keep your door or I will give you both of the other doors" then it would be obvious that you switch.  And that is EXACTLY what he is really saying, but he confuses things, psychologically, but revealing the unchosen goat door.  That gives the contestant the illusion that he is choosing between 2 single doors whereas the real underlying reality is that he is choosing between the initial door and BOTH unchosen doors.