OK, change it to 1000 doors. You pick one and I get the rest. I open all but one of mine and you have the choice to switch.....

It does matter what happened before. Either you have it or I do, there are only two possible outcomes....but the odds it was in my group of doors is much greater.


Make sense now?

I'll go with Uncle Cecil's answer(s):

http://www.straightdope.com/classics/a3_189.html

In short, it will depend upon whether the person offering you the choice knows which is the "proper" door.

JP

I would pick the door that a car would fit through.

The odds that I have the correct door are 1000 to 1. At first. So sure, you've got better odds. At first.

Once you've eliminated all of your doors except one, and the winner is still hidden, there is no reason to assume yours is better than mine.

In fact, I'd stay with my door, because you've proven that you can't pick a winner after opening 99.8% of the options available to you!

Agreed. And in this case he does.

We can set up a wager:D

We'll set up the doors and after your pick I'll be Monty. I'll open one, then bet you $100 a try that I have the car.

This evades me, why does it matter if he knows or gets lucky? The subset of doors still has the statistical advantage, no?

I've gone over this problem before. The difference between a flipped coin and the monty hall problem is the fact that in the monty hall problem, the first decision affects the second one, where in a flipped coin it does not. Second you're not choosing a specific door per say, you're choosing the best prize of 1 door or 2 doors.

My answer is 1/3 chance of the first door, and 2/3 chance of the second door.

http://www.videosift.com/video/The-Monty-Hall-Problem

Re: Probability Challenge - Can You Solve It?
 

I'm trying to disregard the Monty Hall stuff, and I didn't go to the Wiki site (yet).

I would stick with the same door as my original choice.

My reasoning is that if my first choice was a goat-door, then the host would probably have opened it - with me losing the big prize immediately. But the host didn't open it. Instead, he showed me a goat behind a different door.

This action seems to be a deliberate misdirection to cause me to second guess my choice. So I should stand firm.



Of course, the goat revelation may have been a delaying tactic to allow stage hands time to switch what was behind my original door. This means that I should now choose the other door.

Dammit. :confused:

edit: I just want to say that my reasoning is assuming that game shows are not about pure statistics - that there is always a scam in progress that invalidates any mathematical reasoning.

I suggest everyone here watch a Texas Hold'em tournament on TV and see what the statistics displayed do as more cards are revealed.

Freaky!

Changing doors clearly is the right answer, when the host knows where the car is, but it is so counterintuitive.

Tell me if I've got the algorithm right.

Set up an array of three elements that are randomly populated with a winner, a loser, and another loser. The goal is to pick a winner.

I would run two separate cases of two picks - one that picks the same element two times, and one that changes elements on the second pick.

Keep the original pick:
1- First pick is a winner, one of the loser choices is eliminated. Second pick is still a winner - WINNER! THIS WILL HAPPEN 1 IN 3 TIMES
2- First pick is a loser, the other loser choice is eliminated. Second pick is still a loser- LOSER! THIS WILL HAPPEN 2 IN 3 TIMES

Switch picks on the second pick:
1- First pick is a winner, one of the loser choices is eliminated. Second pick is the remaining loser - LOSER! THIS WILL HAPPEN 1 IN 3 TIMES
2- First pick is a loser, the remaining loser is eliminated. Second pick is the winner - WINNER! THIS WILL HAPPEN 2 IN 3 TIMES.

I may have changed my own mind.

I agree, but what am I missing about the host knowing being a factor? If he picks the car by accident obviously the bet is off, but if he picks a goat by accident you are still better off with his remaining door.

I just don't see how his knowing or not matters as long as he opens a "goat door".

This makes more sense to me now. The odds are you picked the wrong door at first. So, you have (66% of the time) "eliminated" a wrong door by choosing it. The host eliminates the other wrong door. By changing doors you have (66% of the time) now chosen the correct door.

If you keep your choice, the odds are only 33% that you get the right door.

Let's go back to the 1000 door scenario.

Say you pick one door. Odds are it will be wrong (0.1% chance of a winner). Say now that all other 998 losing doors are opened, leaving only one door. I say that by changing you have a 99.9% chance of choosing the right door.

I disagree with everyone that does not say 1/2. There is never really 3 doors to chose from. The third door is an illusion. When you chose a first door you are shown a different door and that takes it out of the equasion before you really start. Your pick of a door doesn't count. If they open the door you choose your odds are 1/3. If the don't and they remove a door your odds are now 1/2.

It's a good question.

When you look at the charts on Wiki, or the algorithm which Amail did below, it is easy to see the correct answer is that you should change.

Like Amail says below, the odds are you chose the wrong door at first. You choose a door. The remaining 2 doors carry a 66% chance of one of them having the car behind it. When the host opens one of those 2 doors, showing the goat, the remaining door ALONE now carries the 66% chance of having the car.

But I can't see where Len is wrong, as long as the door the host reveals has the goat, it doesn't seem to matter whether the host knew or not.

Schrodenger's cat comes to mind. Until one of the doors is opened, the car is and is not behind both of them. The uncertainty principle.

So let's work thru it as though the host doesn't know, but always opens a door.

Keep your pick:
1 - Choose the winner first, host reveals a loser, keep your winner - 33%
2 - Choose a loser, host reveals the other loser, keep your loser - 33%
3 - Choose a loser, host reveals the winner - 33%

It doesn't matter if the host knew or not. You're a loser 66%.

Change your pick:
1 - Choose the winner first, host reveals a loser, switch to a loser - 33%
2 - Choose a loser, host reveals the other loser, switch to a winner - 33%
3 - Choose a loser, host reveals the winner - 33%

Now it matters if the host knows what to show - your still a loser 66%.

I think the explanation as to why it matters if the host knows is anti-climactic.

If he picks the car by accident you lose...that's it..switch all you want and it doesn't matter as you cannot pick the car.

In a common sense approach you would say "hey that's not fair, you implied you would give me a choice with an opportunity to win", but math doesn't know fair or unfair. It's a matter of being so obvious and irrational an outcome that you dismiss it.

I mean if the host opened the car door by accident the crowd and contestant would go nuts about how unfair it was, but barring a change in the rules or a "do-over" the player still loses two out of three times that way. I would prefer the explanation spell it out like that rather than leaving you with the impression that it's 50/50 every-time the host is ignorant.

That's the crux of the biscuit, as they say.

If the host knows, and will always pick the other losing door, that's an important thing for the contestant to know. It essentially doubles their odds of getting the car.

If the host doesn't know, and might pick the car after you've chosen wrong, then you're better off playing The Price is Right.