21 - I am confused

[Doggy Styles]

What you are doing is looking at the changed door-count, and implicitly imagining this re-allocation; but it is not happening.

What difference would it make if it was happening? You've still got two doors and you don't know what's behind either of them.

[original PM]

Aaaaargh my head hurts

As soon as you eliminate a door from the 3 doors you are looking at a 50/50 chance

you have 2 doors one has a goat one has a car

we have to assume that no matter what door you pick out of the three the host will always pick a door with a goat and so the second choice is always door 1 which contains a goat and door 2 which contains a car.

I am struggling to see how the inital choice can effect the final choice.

2 red balls , 1 blue ball in a bag

first draw - red ball

what are the chances or drawing a red ball on the second draw?

50/50 because your siution is now 1 bag containing a red ball and a blue ball - the colour of the previous ball does not matter

Head hurts

[expat]

What difference would it make if it was happening? You've still got two doors and you don't know what's behind either of them.

If the random re-allocation was happening, then the odds of finding a car behind your first choice of door would be 1/2, same as if there had been only 2 doors to begin with. If no re-allocation happens, then the odds of finding a car behind your first choice of door are not changed by seeing a goat in a doorway, they are the same as when the car was put behind 1 of 3 doors, i.e. 1/3.

So it makes a critical difference: is it a random distribution to 1 in 3, or to 1 in 2?

Now, ISTM that the trap in this one is that people's minds reset for the 2 doors, but the distribution does not. Hence the discord between expectation and reality.

[thelurker]

My head is still spinning, as all i can see is it being 50/50 chance.

2 doors

one door: Car

One door: Goat

[oracleslave]

My head is still spinning, as all i can see is it being 50/50 chance.

2 doors

one door: Car

One door: Goat

Pondlife gave a simple explanation on the first page of the thread.

[Pondlife]

Pondlife gave a simple explanation on the first page of the thread.

Yeah, but he's an idiot.

[original PM]

Film is not as good as the book IMHO. The guy has two others that I'm aware of; one on the creation of the Mercantile Oil exchange in Dubai and the other on the Japanese stock market about the time of Nick Leeson.

As for the logic:

You originally pick a door (a) and you have a 1/3 chance of being correct, therefore a 2/3 chance of the car being behind the other doors (b,c).

Guy shows you door (b) has a goat behind it. So, logically there is a 2/3 chance that door c has a car and a 1/3 chance that door a does.

But this logic is flawed

As soon as you open a door it is eliminated from the probability and thus the probability is re-set.

Once you change the boundary conditions you can use information from previous boundary conditions

Lets say you have 3 cups in front of you each one with a blue ball and 2 with a red ball in - you are asked to pick a cup and you aim is to get a blue ball.

Regardless of which cup you pick you know that a cup is going to be removed which contains a red ball.

You are then asked to pick a cup again - the choice is 50/50 - the inital stage of removing a cup to leave 2 cups is irrelevant and has no bearing on the probability of picking a blue ball or a red ball in the second stage???

Surely?

[snaw]

But this logic is flawed

As soon as you open a door it is eliminated from the probability and thus the probability is re-set.

Once you change the boundary conditions you can use information from previous boundary conditions

Lets say you have 3 cups in front of you each one with a blue ball and 2 with a red ball in - you are asked to pick a cup and you aim is to get a blue ball.

Regardless of which cup you pick you know that a cup is going to be removed which contains a red ball.

You are then asked to pick a cup again - the choice is 50/50 - the inital stage of removing a cup to leave 2 cups is irrelevant and has no bearing on the probability of picking a blue ball or a red ball in the second stage???

Surely?

Isn't it all about the wording - from that wiki on page 1, this is how it should be mathematically stated:

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. The car and the goats were placed randomly behind the doors before the show. The rules of the game show are as follows: After you have chosen a door,the door remains closed for the time being. The game show host, Monty Hall, who knows what is behind the doors, now has to open one of the two remaining doors, and the door he opens must have a goat behind it. If both remaining doors have goats behind them, he chooses one randomly. After Monty Hall opens a door with a goat, he will ask you to decide whether you want to stay with your first choice or to switch to the last remaining door. Imagine that you chose Door 1 and the host opens Door 3, which has a goat. He then asks you “Do you want to switch to Door Number 2?” Is it to your advantage to change your choice?

[TimberWolf]

Isn't it all about the wording

Some PHd's were upset with Marion Vos Servent some 10 years ago over her wording of the problem. I don't think she made it clear that the host must pick a goat, which changed the problem, or at least made it ambiguous enough to gave them an escape route. I was actually on Marion's side at the time.

[Pinto]

But this logic is flawed

As soon as you open a door it is eliminated from the probability and thus the probability is re-set.

Once you change the boundary conditions you can use information from previous boundary conditions

Lets say you have 3 cups in front of you each one with a blue ball and 2 with a red ball in - you are asked to pick a cup and you aim is to get a blue ball.

Regardless of which cup you pick you know that a cup is going to be removed which contains a red ball.

You are then asked to pick a cup again - the choice is 50/50 - the inital stage of removing a cup to leave 2 cups is irrelevant and has no bearing on the probability of picking a blue ball or a red ball in the second stage???

Surely?

You need to take into account the probability of your first choice revealing the Aygo, i.e. only 1/3, but you have a 2/3 chance of ending up with a goat. You would only have a 1/2 chance if you were allowed to switch the car behind the 2 remaining doors thereby resetting the probability.

The car however hasn't moved. There's still only a in 1/3 probability of it having been placed behind your chosen door, which means that there's a 2/3 probability that's in one of the other doors. As one of those other doors was opened to reveal a goat, your 2/3 probability is now on the remaining unchosen door.