Kind of a riddle...

...but not really.

Was having a discussion about this at work today, but I'll elaborate on that after some answers.

You're on a game show and presented with three doors. Behind one door is a car. The other two have nothing. You are given the opportunity to select a door. Before your choice is revealed, the host tells you he will open one door that does NOT have the car behind it. He does so, and extends the invite to either keep your current choice, or select the door that he didn't open. What do you choose? Poll included.

If anyone is familiar with this conundrum or its name, please keep it to yourself for now. No explaining of answers yet. I have intentionally not revealed the name of the situation so no one goes tromping off to Google. :D

I'm confused. But I say always trust your gut and stay with your first pick

Anytime I go through a door I press....old habits die hard.

Where's Monty Hall? ;)

Quote:

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Originally Posted by Reca

I'm confused. But I say always trust your gut and stay with your first pick

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I second that. Go with your gut.

I'd switch doors. And seriously Rhino? Geez. :rolleyes:

Quote:

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Originally Posted by jcsdscott

I second that. Go with your gut.

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+1

I know the puzzle so no vote. (and if you know it don't be messing up the fun, durnit!) There is a very critical part that gets left out many tellings, good to see this one does cover it.

Read carefully and good luck :)

Those of you who have already answered, please explain your choice if you haven't done so. Be specific. Those who answer from here forward, also explain why!

After the host eliminated one option, my chances of winning just increased from 33% to 50%.

I know the theorem as well, but even those that know it dispute over the answer, so I don't think it matters.

Switching or not switching doesn't make a difference in the odds. It's 50/50 either way. You know your chosen door wasn't one with nothing behind it, since they open one of the two doors with nothing -- but that means it still might be... or it might be the other one remaining. You really have no new information, other than knowing that you were at least partly right. (Note, I assume that the door he opens is NOT the door you chose. In other words, it was a "you chose door number 1, so now I'll eliminate one of the other doors that has nothing behind it.")

I also know the theorum, but will not answer. I agree with Rhino, I don't think it matters.

As someone has stated, this situation is a statistical dilemma known as the Monty Hall problem, named after the game show host's name. It's a problem we spent a solid week on in a statistics analysis course I once took 10+ years, and it's one I return to often to exercise my brain, as it were. My answer is to switch doors, but I return to it often and reexamine my choice. It came up the other day while we were talking about the statistical chances of finding a piece of specific evidence in a pile, and fighting against the old ingrained Gambler's Fallacy.

For those that believe that the choice is 50/50...is it? You started with 3 doors. That means one door had a 33.3333% chance of having the car behind it. Thus, considered together as one variable, the other two doors had a combined statistical probability that one of them has the car behind it. The host, who has prior knowledge of where the car is, and ideally, is acting completely randomly (that is to say, he isn't acting differently depending on which door hides the car, etc) gives you a freebie by showing you which one doesn't have the car behind it. Although it seems counterintuitive, the door he leaves still has up to a 66.6666% probability of having the car behind it. Two times out of three, switching will be the correct choice.

Don't believe me? Here are all possible outcomes of the game with three doors.

Door 1 Door 2 Door 3
Car Nothing Nothing
Nothing Car Nothing
Nothing Nothing Car

Let's say you choose Door #1 in all three scenarios. In scenario #1, let's say Monty opens Door #2 for you randomly, i.e. he has no discernible pattern as to which door he chooses to open when both are empty. You choose to go to Door #3. Oops, you lose...there's still a 1/3 chance of that happening right? Let's say Scenario #2 comes up, and you again choose Door #1, and Monty shows an empty Door #3. You change your selection and win. Same for scenario #3...you choose #1, he opens #2, and you switch to #3 and win. Out of all possible scenarios, switching yields a win 66.6666% of the time.

I know it is counter-intuitive. I know it is. I railed against the teacher that introduced this to us, and I tried to explain it completely verbally the other day and failed. My last example to drive the point home would be to vastly increase the number of doors. Let's say there are 10,000 doors. Your choice has a 1/10,000 chance of having the car. The other doors collectively have a 9,999/10,000 chance of having a car behind one of them. The host, acting randomly, but having knowledge where the car is, does you a favor by opening 9,998 of those empty doors. Do you agree that switching to the final door gives you a probability of winning greater than 1/10,000, and in fact gives you chances upwards of (depending on some variables, total probability vs. conditional probability, etc) 9,999/10,000?

I'd stick with first gut feeling (not switching), then kick the other door in.

Problem solved, problem staying solved.

Lol

Still sticking with my gut. But this made me think.