Monty Hall Dilemma - Winning a GTI on a Game Show

[Rocket36]

No BTJ, YOU are wrong. Think about it... You actually agreed with the 1 in 2 chance in your post by saying "although obviously there are only two doors left to open" which is a 50/50 chance of winning regardless of whether you stay with your initial choice or switch to the other door. You can't factor in the first choice as the second choice is all that counts, since the circumstances have changed, which is 50/50.

It really isn't something people should be arguing about as it's quite a simply philosophy. You have two doors to choose from and there is a prize behind one. That's 50/50. The fact that there were 3 choices to start with is completely irrelevant when it comes to the chances of winning as you can't win or lose with that first choice.

[Timbo]

At the outset, the chance you would pick the door with the car is 1 in 3, and a door with a goat, 2 in 3. OK...lock that it.

A door is opened, revealing a goat. Now, here's the thing -- nothing has changed in terms of the odds of your original choice; it still stands at 1 in 3 for the car, and 2 in 3 for the goat, even though one door has been opened.

To reduce the odds that your door has a goat behind it, you must make the switch decision. It is a two decision process, and not making the switch does not alter the original odds, whereas making the switch does.

There is a lot of confusion about this because it is so counterintuitive, and people try to rationalise, as above. But, if you don't believe the explanation, try it with 3 cards, and see how you it plays out

[Rocket36]

LOL Timbo... That's the funniest post in this thread so far. Do you really think that when there are only two options left and you have to make another decision to stay with your original door or switch, you still have a 1 in 3 chance? Or are you just being funny? Hahahaha

Because it really is this simple. There are two doors left, pick one. That's a 1 in 2 chance. REGARDLESS of whether or not you switch doors.

[pixl]

Rocket, the thing you're forgetting is the that the game show host knows what door has the car, which means he ALWAYS has to reveal a goat door. Because we know he ALWAYS has to reveal a goat door, this is what tips the odds in our favour. ie. 66.6% of winning a car by switching.

You can't exclude the first door simply because there's two left. The first door is and has to be included simply because we know 100% that the door the host must reveal is a goat door.

The only way the it can work as a 50/50 choice in the end is if the host forgot which door has the car or if he decides to reveal a door randomly. The key word is random, if the host is randomly picking a door to reveal, then the last two doors are 50/50.

The problem explained in the TV show 'Numb3rs' and the movie '21' -- http://www.youtube.com/watch?v=5e_NKJD7msg

[Rocket36]

But what you're failing to realise is once there's only two doors remaining it's ALWAYS a 1 in 2 chance. FORGET ABOUT THE DOOR THAT'S BEEN OPENED!!! It not longer has any baring on your chances. You have two to choose from and that makes it a 1 in 2 (or 50/50) chance you will choose the right one whether or not you switch.

That youtube explanation takes into account the first decsion. Which is why so many people are fooled into thinking that there's a better than 50/50 chance the second time round. There isn't. When faced with two choices, you will only EVER have a 50/50 chance of making the winning choice. I'm not taking into account "variable change" like people want you to do so you agree with them. I'm simply trying to point out that it's 1 in 2 when faced with the second choice. 2 options, 1 choice = 1 in 2. SIMPLE!

[schoona]

At the outset, the chance you would pick the door with the car is 1 in 3, and a door with a goat, 2 in 3. OK...lock that it.

A door is opened, revealing a goat. Now, here's the thing -- nothing has changed in terms of the odds of your original choice; it still stands at 1 in 3 for the car, and 2 in 3 for the goat, even though one door has been opened.

To reduce the odds that your door has a goat behind it, you must make the switch decision. It is a two decision process, and not making the switch does not alter the original odds, whereas making the switch does.

There is a lot of confusion about this because it is so counterintuitive, and people try to rationalise, as above. But, if you don't believe the explanation, try it with 3 cards, and see how you it plays out

x2

the counterintuitive-ness f it had me second guessing myself too, but after a while it lcicked

[schoona]

But what you're failing to realise is once there's only two doors remaining it's ALWAYS a 1 in 2 chance. FORGET ABOUT THE DOOR THAT'S BEEN OPENED!!! It not longer has any baring on your chances. You have two to choose from and that makes it a 1 in 2 (or 50/50) chance you will choose the right one whether or not you switch.

Thinking micro yes, but from the outset thinking macro , it would be better to swap.

[pixl]

Ok, Rocket, imagine there are now 30 doors. 1 car, 29 goats.

You pick any door and your chance of winning the car is 1 in 30, right?

Now, the host opens 28 doors revealing goats, leaving only the door you picked and one other door closed.

Are you going to swap or stay with the door you chose? How confident are you that you picked the right door out of 30? Still think you've got a 50/50 chance of winning now? We're not talking about the fact that there are only two doors left, we're talking about the chances of you winning the car.

You have to remember that your first choice, you had a 1 in 30 chance of picking right. Which means, you also have a 29/30 chance of picking wrong!! The fact that there are only two doors left does NOT make it 50/50 since you chose a door with the initial odds of 1 in 30.

Same with the original problem, it's not about two doors left closed -- The fact remains that your first choice, you've got a 66% chance of being WRONG!! Not good odds if you want to win a car.

If you still not getting it, think of 1,000,000 doors. You pick a door and the host opens 999,998 all showing goats -- do you REALLY think your chances of winning are still 50/50?

[Rocket36]

If you still not getting it, think of 1,000,000 doors. You pick a door and the host opens 999,998 all showing goats -- do you REALLY think your chances of winning are still 50/50?

Yes of course they are. Two doors left to open, one choice. THAT is a 1 in 2 chance (or 50/50 chance) of winning. You have to stop taking the host into account. You have to forget about the open doors. Two are closed, there is a prize behind one, you have to pick one. THAT is 1 in 2.

[pixl]

You have to stop taking the host into account.

And you NEED to start taking the host into account. You can't just pick and chose what you want to include. The host makes all the difference, he's the one who stops it becoming a random choice -- This isn't a flip of the coin.

Your argument for a 50/50 outcome is only valid if the host randomly selects a door to open or if there isn't a host at all.

[Timbo]

Your argument for a 50/50 outcome is only valid if the host randomly selects a door to open or if there isn't a host at all.

Correct!

The host is always going to pick a door with a goat. That's 1 in 1. But that happens after your 3 in 1 decision, and if you do nothing, your odds remain 3 in 1. The odds of picking the door with the car, IF you make the switch decision, now become 2 in 3!

Rocket, you are trying to rationalise the natural intuition we have for this game, but it is wrong. Hard to accept, but wrong. As I said, get a partner to lay out sets of 3 cards in front of you, and track the outcome. You will discover that switching gives you the highest outcome of "the car"

[The_Hawk]

Lies, damn lies and statistics, thats what's killing this, and lets not forget the original question. It was, if given the choice, should your switch your choice after they eliminate a goat?

So lets look at this backwards, in order to WIN the car you should pick a goat the first time (which is more likely at 2/3) then switch to the car after they eliminate one of the goats!

So you had a 2/3 chance of being wrong the first time which in turn means that in 2/3 cases you would be better off switching the second time around. So YES, you are statistically better off switching.

Again, you were MORE LIKELY to be wrong the first time which means your *statistically* MORE LIKELY to win if you switch the second time (even though the choice in isolation is still 50/50).

[Rocket36]

even though the choice in isolation is still 50/50

Finally! Someone gets it... Thank you!

[schoona]

Finally! Someone gets it... Thank you!

No-one was disagreeing that point were they? lol

Certainly good to have a bit of a think once in a while.

Do we start on schroedinger's cat?

[Timbo]

The choice in isolation is 1 in 2, but the choice is not in isolation.

The initial choice is made when there are 3 doors and no knowledge as to what lies behind them. At that point, the odds are 2 in 3 for a goat, but 1 in 3 for a car.

After you've chosen your door, the host opens a door, and reveals a goat (this is critical to the game: the host always opens a door that reveals a goat). Now, it seems "so obvious" that the odds have just shifted in your favour, that your choice of door has a 1 in 2 chance of being the car (or the other goat). But that's just your brain playing tricks on you, rationalising something that seems oh sooo intuitive. But it's wrong. The probability that your original chosen door hides a goat is still 2 in 3, because the fact that the host opens a door revealing a goat (which he'll always do) does not alter the original probability associated with your choice when there were three closed doors.

So you need to counter the rationalisation of your natural intuition with the following thought pattern: "when I picked that door, there was a 2 in 3 chance it would be a goat; one door has been opened, revealing a goat, but there's still a 2 in 3 chance that the door I originally chose hides a goat, therefore I should switch to the other door, which has now has a 2 in 3 probability of being the car.

This thinking process lies at the heart of what is known as Bayesian probability theory, which are central to some fairly significant decision support systems. For example, in the context of this forum, I'm pretty sure VAG logs all warranty claims worldwide into a system that tests the likelihood that the claim is genuine or not, based on a Bayesian neywork. So next time your dealer service manager says that he's never heard of your model car having that problem, chances are he's blowing smoke to push you off (!), but if he has access to, and has actually consulted the VAG warranty system, he may be correct...statistically