With your last line it should be clear you made a mistake somewhere. It's impossible for the chances of both doors to be 1 in a million, unless you think there's a 999,998 chance there is no car.

Here, let's try to put it another way - your two choices aren't door #1 and door #2. Your two choices actually are
- Your pick
- Monty's pick

As there are 1,000,000 doors, and we know there's only one car, the chances of you yourself picking the car is 1 in a million.

The remaining chance, Monty's pick, is therefore 999,999 in a million. Think about it:

Case 1: You pick the correct door. The car is in your door. (1 in a million chance)
Case 2: You pick the incorrect door. The car is in the other door. (999999/1m chance)

Does that make sense?

You said earlier that the host deliberately opens a goat door. So, since {chance car was behind two doors you didn't pick before} was 2/3, it stays at 2/3, because the host wouldn't reveal the car door if it was there.

The chance it's your door stays at 1/3. The chance it's one of the other two doors is still 2/3, except that you only have one choice of door there. Think of it as a choice between your door OR every other door.

Out of all the people who posted here you're the one who made the explanation as clear as possible.

Ah, the Marilyn Vos Savant problem... hah.

The trick is... you have a 1 in 3 chance of picking the right door.

and a 2 in 3 chance of picking the wrong door.

The fact that one door is wrong is not really new information because you already knew that one of the two doors was going to be wrong... and the host DOESN'T open the doors randomly.

Farmageddon said: LuisPacheco said: Hi, first let me start by reintroducing myself. I'm the poster formerly known as LuStaysTru. Lost my password.....I came back just for this lol Well anyways, I believe this to be what is called the "Gambler's Fallcy". The odds in fact do not change. From wiki: Now suppose that we have just tossed four heads in a row, so that if the next coin toss were also to come up heads, it would complete a run of five successive heads. Since the probability of a run of five successive heads is only 1⁄32 (one in thirty-two), a believer in the gambler's fallacy might believe that this next flip is less likely to be heads than to be tails. However, this is not correct, and is a manifestation of the gambler's fallacy; the event of 5 heads in a row and the event of "first 4 heads, then a tails" are equally likely, each having probability 1⁄32. Given the first four rolls turn up heads, the probability that the next toss is a head is in fact, . While a run of five heads is only 1⁄32 = 0.03125, it is only that before the coin is first tossed. After the first four tosses the results are no longer unknown, so their probabilities are 1. Reasoning that it is more likely that the next toss will be a tail than a head due to the past tosses, that a run of luck in the past somehow influences the odds in the future, is the fallacy. It's not really the same thing. When Monty opens a door, he knows what he's doing. Say you aways keep the door you first chose. It would be the same thing as just choosing a door and seeing if it has the car. What are the odss of choosing the right one between the trhee doors? 1/3. So if you decide to always keep your door, your chances of winning are 1/3. Now since always changing your door is complementary to that, it gives you 2/3 of getting the car. EDIT: Take a look at this: http://www.andafter.org/images/album/design-grafico/infograficos/Monty_Hall_a.GIF

Hah, that graph reminds me of the year I was on my Junior Highschool math club.

Even though I wrote that the host CLOSES every other door instead of OPENING them ? ^^

Yes, even depsite that. You're good!

Ok, let's rework this a bit.  Let's say instead of doors, there are three boxes that are identical.  You can either keep your first pick, which was 1/3 chance of winning, or give up the box you picked, they get shuffled, and one without the prize in it, is removed.  You then pick one of the two remaining boxes.  Do you do this or not?

In this case, the repicking makes the odds 1/2 of being a winner, as opposed to 1 out 3.  So, I would say you do that.  I understand this to be this way, because the conditions when picking have been reset so it is like a different choice.  The odds are set based on the conditions that were around when the person picked.  These odds do remain consistent.  However, if things change in the environment to the OTHER choices, and those conditions improve, it is possible to get something more favorable by switching.

I think with boxes you can explain it like that:

Pick your box - you can keep this box OR you can instead take all the other boxes.

This is a good analogy.

If you decide beforehand to keep your door, you're essentially negating all the rest of the problem (since it'll make no difference) adn will have the same odds of winning as you had of picking it right at first.

Now, if you decide beforehand to switch, you'll win as long as you don't pick the winning door at first. So your odds are complementary to those of picking it right at first.

It's the chance of the one you choose being right against the chance of any other one being right.