A mate told me about this and I just didn't understand it.

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There's a room with three doors. Behind one of them is a prize.
Q. If you open one door, what is the probability of finding the prize?
A. One in three.

Now say you open a door and there is no prize.
Q. What, now, is the probability of finding the prize?
A. One in three!
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Okay, I completely accept that in the first instance, there are three doors and one prize, so the probability of finding the prize is 1 in 3.

In the second instance, you know that, say, Door A has no prize. That means the prize is either behind Door B or Door C. That, IMO, means the probability is now 1 in 2, since you've already eliminated one of the doors.

So why is the probability still 1 in 3?

No idea, unless they are allowed to move the prize after you have chosen incorrectly...

Google the Monty Hall problem.

Whilst outlined differently, it is essentially the same thing.

what about the door you entered the room by? or is this a "theoretical" problem?

It's the Monty Hall Problem, or something similar. Well, the MH problem is to decide whether having made your pick (Door A) and been shown a door that contains a goat/lump o' coal/cabbage/nothing (Door B) you should change your pick to the remaining door (Door C) in order to maximise your chances of winning the gold/car/parsnips/prizes.

Essentially, in the two cases where you've picked the non-prize first, you win if you switch (having been shown only other non-prize door), and only lose in the case where you switch from the prize door (odds of 2:1). If you don't switch then obviously you only win if you pick the door correctly first time (1:2).

As to your question, I don't understand. The second answer seems incorrect. Or it is incorrect based on my understanding of the problem. I can't work out whether me or your friend is wrong.

Yes. Although I'm not convinced I understand the problem. A (very) brief outline of the MH problem is in my above post.

Nah, as you enter the room you're knocked unconscious. Then a prize is placed behind one door. The other two are concreted up and you have no means of escape through them. All the doors are locked, and you only have two exploding lock-picks. Clearly.

Derren Brown covers this in his book Tricks Of The Mind. If I can be bothered, I'll copy out what he says on the matter.

The problem is insufficiently defined, so there is no correct answer. Important details have been omitted. Either you or your friend forgot half the story!

"What now is the probability of finding the prize?"

You don't even say if you're allowed a further choice! If not, then the probability is Zero since you already lost the game.

I was told this story less than an hour after posting the thread, so it's highly unlikely (though not impossible) that I forgot. More likely that matey forgot half the story. We had spent the last hour of work arsing around and discussing riddles. Like the one about the closet with a light bulb and three switches.

The problem is this:

You have three doors. behind one is a prize. Behind the other two is nothing.The quiz master asks you to choose a door.

Before you see if you are right, the quiz master opens one of the other doors. It will be empty. He then gives you the chance to change your mind.

He is basically giving you a 1 in 2 chance of picking the correct answer. Apparently, if you change your mind, you are more likely to win than if you don’t. This is because there is more chance of the un-chosen (and unopened door) containing the prize.

Now see, this version makes perfect sense

Although, it makes no "common sense" and is therefor a proper riddle type thing

It's accurate though. Try it yourself

I think it's possible that cloaked_wolf's friend has misinterpreted the Monty Hall problem or not provided enough information.
In the MH problem the sequence of events and the actions of the host are very, very important.
With a different sequence of events and without the host then result is different.

In the Monty Hall problem, you pick one of the three doors (which remains closed for the time being).
You have, as stated in question 1, a 1 in 3 chance of having the correct answer. However, looked at another way, you have a 2 in 3 chance of being wrong.
Once you pick your door the host of the game (Monty Hall, who knows what is behind each door) removes one of the incorrect doors from the equation.
The odds that the door you have picked (but haven't seen behind) is the one with the prize are the same as they were at the start (1 in 3 you win or 2 in 3 that you loose).
Now, the host asks whether you want to change the door you originally picked. Your original choice still has a 1 in 3 chance of being correct but also still has a 2 in 3 chance of being wrong so it is always better to switch doors as you have a better chance of winning.

The actions of the host are key here. Without the host, or with a host who has no idea what is actually behind each door the game is significantly different.
Assuming you haven't pre-selected the sequence you will open the doors in then opening the first door gives you a 1 in 3 chance of being right. If you are wrong then it is an even choice as to which remaining door the prize is behind (i.e. 50/50).
If however you decided the sequence you would open the doors in and your first choice was not the prize then it is best to switch as in the original MH problem.

I think. My brain hurts.
Monty Hall Problem

That's the version I described above. Have we got any clarification on the actual problem yet???

I suspect the actual problem was the MH as described by several of you knowledgeable chaps, but CW's mate only half remembered it.

I confess, it's not one I remember. It's quite a pretty demonstration of counter-intuitive statistics, and given that I've studied maths a lot I'm quite surprised I've either not seen it or I've forgotten. It's clearly quite well known, and I suspect I'll remember it better than "matey"