Monty Hall

[andrewducker]

[Poll #1770413]

Explanation

I have known what the answer was for ages, but for some reason it only "clicked" in my head today. You can blame sarahs_muse for triggering it.

Comments

[chuma.livejournal.com]

Wrong.

You are not weighting your chances. Your options are MF FM or FF but the chance of having a male are 50-50. You then have a 50% chance of the Male being in place 1 or 2. Thus you have 25% chance of MF and FM and a 50% chance of FF.

[theweaselking.livejournal.com]

Intriguing argument.

Imagine that Monty Hall has three goats and a car, behind four doors. The car is equally likely to be behind each door, such that your odds are:

1: 25%
2: 25%
3: 25%
4: 25%

Monty opens door #1 to reveal no car.

*I* say that your odds are now
1: 0%
2: 33%
3: 33%
4: 33%

*You* say that your odds are now:
1: 0%
2: 25%
3: 25%
4: 50%

I am wondering why you think this is.

[chuma.livejournal.com]

You chose a door before you were shown a goat in one of the other 3. The host KNOWS that that door didn't have the car behind it before offering you a swap. He revealed extra information about the system of 3 doors after chosen. The important distinction is that this information was not a given at the point you chose the door as that the host knows for certain one of those doors was with a goat inside.

[theweaselking.livejournal.com]

You chose a door before you were shown a goat in one of the other 3.

Very true, and the fact of that choice was why the odds acted in a nonintuive way.

In this problem, he's doing the opposite. He's *not* adding information by telling you that the car is not behind door #1, so the odds of the car being behind the other three doors should be equal, right?

[chuma.livejournal.com]

Okay let me try it this way and see if I can distinguish between the Monty Hall problem and not.

You have 3 doors. I tell you that door number 1 has a goat behind it, but a car hides behind door number 2 or 3. The odds in thsi scenario are:-

Door 1 0%
Door 2 50%
Door 3 50%

------------

In the Monty Hall Problem you are told a car is behind one of the 3 doors and you are told to pick. For arguments sake we will say you chose Door number 3. The odds are thus:

Door 1 33%
Door 2 33%
Door 3 33%

Monty then knows that door number 1 has a goat behind it and shows you it. He asks if you want to swap. The odds of it behind behind either door number 1 or 2 remain 66%, but as he has ruled out the first door the odds are now:

Door 1 0%
Door 2 66%
Door 3 33%

If door number 1 contained the car, he wouldn't open it, he would instead open door number 2. If door number 3 contained the car then it wouldnt matter which one he opened. It all revolves around him giving new information at this point. The information is dependant upon your choice and THAT is why it changes things.

[theweaselking.livejournal.com]

That's exactly right. You have the problem defined perfectly and you have the correct answer.

In the baby raptors situation, imagine "FF" to be your "car", and the other three possible results to be "goats".

All four results are equally likely, which corresponds to Monty having four doors, right?

*Without* you picking, Monty eliminates one goat, the "MM" goat, which leaves three doors. Each of those three doors now has a 1/3 chance of being the "FF" car, right?

[chuma.livejournal.com]

Wrong. You are introducing the notion that the position is or importance and then confusing the odds.

Either you do know that the female is in the 1st slot or you don't. If you do the options are:

F* M
F* F

If you don't then the options are:

F* M
F* F
M F*
F F*

Where F* is the one that was identified. You may give these all equal weighting now. You are mixing up the 2 solutions are deciding that the position of the female matters only if there is a Male. Wrong.

[theweaselking.livejournal.com]

So you're once again arguing that

1: 25%
2: 25%
3: 25%
4: 25%

compresses to
1: 0%
2: 25%
3: 25%
4: 50%

once you know that "it's not 1"?