On how to pick goats from cars: The Monty Hall Problem | Hacker News Reader

On how to pick goats from cars: The Monty Hall Problem

garry.posterous.com

17 pointsprakash13 comments

This never made intuitive sense to me until I heard it rephrased this way:

In front of you are 1002 doors. You pick a door, and then Monty Hall opens 1000 doors to reveal goats. You have the option to pick the remaining door or stick with your original choice. Which should you do?

I've always thought the combining doors way of explaining this made the most sense - http://en.wikipedia.org/wiki/Monty_Hall_problem#Combining_do...

After a lot of brain twisting it sinks in.. and I do probability for living!

For doubters: Before the door is opened by the host, you are less likely to have chosen the right one(with prob 1/3). Once he opens the door, he eliminates one "bad door". Now, given that you were less likely to have chosen the right one in the first place and one wrong one has been removed, it is highly likely that the residual door is the right one.

Right, so the first choice is door 1, and the host reveals that door 3 is a goat.

The decision to switch or not is really a decision between door 1 and door 2. At this point isn't your first guess completely irrelevant?

I've heard this problem and the reasoning before, and for some reason I can't get myself to believe that switching is better.

There are two possibilities when you pick door 1: either you've picked a door with the prize behind it (33%) or you've picked a door with a goat behind it (67%).

Assume you stick. In that case, you have a 33% chance you've picked the right door, as before. Opening another door doesn't change whether you've picked the right door or not originally.

Assume you switch. In the case you've picked a door with a goat originally (67%), you will necessarily switch to the door with the prize. In the case you've picked a door with the car originally (33%), you will necessarily switch to a door with a goat. Therefore switching gives you a 67% chance of picking the door with the prize.

Another interesting strategy to consider is "randomize" - what if you picked randomly between the two remaining doors? Then you'd have a 50% of getting the prize, since you're selecting randomly between two doors, one of which you know contains the prize.

Imagine a similar but slightly different scenario: We now have two contestants: Mr A and Mr B. Mr A is the contestant in the original story. He gets to pick a door. The host then opens a door with a goat (and removes it from the stage). Then Mr A gets to choose whether to keep his door or open the third.

Now the second contestant, Mr B, joins the show. He has not seen anything prior and has no knowledge of Mr A or the other door previously removed from stage. All he sees is two doors. He has the option of opening one of the two.

Mr A has 2/3 chance of switching door and coming out with a car. Whereas Mr B has 1/2 chance (from his perspective) of picking either door and coming out with a car.

I think the main problem why so many people have a hard time grasping the solution is that they're looking at the probability as Mr B and ignoring the extra information they do have as Mr A.

Seriously, I had one of my friends ask me this question and I was stymied too !! Neway, the 1002 thing is really intuitive and I definitely agree that we think from the perspective of Mr.B and the point of the matter is that the host removes all the "bad" doors !

classic teaser. If you extend the problem to n doors and you're shown all but one, I think it helps the intuition.

Years ago I recall a lunch conversation with a group of mathematicians, some with PhDs in statistics, that was quite lively.

This is also explained well by Kevin Spacey in the movie 21.

Not really. That was just enough to get me to read the Wikipedia article to figure out what the heck they were talking about in that movie. It made the student sound super smart, but most people I know didn't have a clue what they were saying (including myself).

Yeah. And what was that whole business about the Newton-Rhapson method?

Also in a numb3rs episode. And Allen Paulos in one of his books outlines the script of a movie based in this game which could help to interest people in Mathematics.

you pick door 1. the host does his thing. now the situation is:

if it was behind door 1, you win by not switching.

if it was behind door 2 or 3, switching would win.

seems intuitive to me.

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