- 12 Jun '07 22:56This is an oldie but goodie. Apologies if it's been done here before.

You're on a gameshow. The way the game works is this. There are three doors. The prize is behind one of the three doors. No clues, its random, but the host knows which is the winner.

You pick one. The host doesn't open that door, but he does open one of the other doors, but not the one with the prize behind it, to narrow it down to two. If you got your first guess right then he'll have 2 to choose from (although you would not know this yet) and he'll pick at random, if you were wrong then there's only one he can pick (though again you wouldn't know that).

Then he asks you if you want to stick with your first guess, or switch to the other one which hasn't yet been opened.

What should you do? Or are you chances just as good either way? - 13 Jun '07 04:03

This is correct, switching raises the probability of winning from 1/3 to 2/3, which is very counterintuitive. However, this is assuming that the host does this every time (i.e. will always show one of the doors which does not win, and offer the chance to switch).*Originally posted by coquette***it was a game show for years and loads has been written about it in various statistical sites. switching is the right decision 2/3 over 1/3 chance of winning.** - 13 Jun '07 06:29Switching is NOT the correct answer.

We know that the host knows which one the door's behind, but as he didn't state that he was going to open a door every time, there's little reason to believe that he would. If you had picked an incorrect door, surely he would just have allowed you to enter that one? Instead, he has opened another door, in the attempt to get the contestant to reason that switching is better, and switch. By opening a door, it is rational to reach the decision that switching is better. Thus, by opening a door, he reveals that he wants you to switch. As, he knows which one it's behind, it can be safe to say that when he wants you to switch, you're on the right one.

Thus, you should stay with the door you're on. - 13 Jun '07 06:35I don't get it, there's three doors, you pick one with a 1/3 chance of winning. One more door is shown to be false by the host. Surely, now that the door you have already chosen is one of two doors, one of which has a prize, your chances have already increased to 50% without having to switch.

Once the contestant opens one of the doors, the probability of all doors containing a prize changes, no? - 13 Jun '07 06:49Technically not agryson. An easy way to think about it is this:

You pick a card at random from the deck. Another person then looks through the deck, showing every card apart from the ace of spades, and keeping one card at the end. Who is more likely to have the ace of spades?

In this example, you have a 1/52 chance, he has a 51/52 chance. - 13 Jun '07 08:46

No, he always opens a door, as specified in the question. He has no stated aim of trying to stop you winning.*Originally posted by doodinthemood***Switching is NOT the correct answer.**

We know that the host knows which one the door's behind, but as he didn't state that he was going to open a door every time, there's little reason to believe that he would. If you had picked an incorrect door, surely he would just have allowed you to enter that one? Instead, he has opened another door, in the att ...[text shortened]... you to switch, you're on the right one.

Thus, you should stay with the door you're on.

So you switch. - 13 Jun '07 08:58

No, because the new information that you get is conditional on the door that you chose.*Originally posted by agryson***I don't get it, there's three doors, you pick one with a 1/3 chance of winning. One more door is shown to be false by the host. Surely, now that the door you have already chosen is one of two doors, one of which has a prize, your chances have already increased to 50% without having to switch.**

Once the contestant opens one of the doors, the probability of all doors containing a prize changes, no?

Since the decision to open a door by the host is limited to the two doors left, all of the probability that the prize was in one of those doors (2/3) collapses completely into the door he hasn't opened from those two.

Therefore, the probability of winning by not switching is 1/3 and the probability by switching is 2/3.

This is the Monty Hall problem that I've mentioned in your "Gut" thread. - 13 Jun '07 11:09

It was not specified in the question that he always opens a door. That is the trap that many fall into. We are only on the gameshow for one time, and in that time he has opened a door. There is little reason to believe that he would not have let you open an empty door normally if you had picked an empty door, and so wishes you to use probabilities and switch.*Originally posted by mtthw***No, he always opens a door, as specified in the question. He has no stated aim of trying to stop you winning.**

So you switch.

If he has to open a door every time, then you should switch. However, as this was not specified the answer is:

You should stick on your chosen door. - 13 Jun '07 12:06

Even if he doesn't open a door every time, he did open a door that time. That means that your perceived probability of 2/3 for not having chosen the good door on the first attempt collapse into the remaining door, leaving you with a 1/3 probability in your first choice and a 2/3 probability on switching.*Originally posted by doodinthemood***It was not specified in the question that he always opens a door. That is the trap that many fall into. We are only on the gameshow for one time, and in that time he has opened a door. There is little reason to believe that he would not have let you open an empty door normally if you had picked an empty door, and so wishes you to use probabilities and ...[text shortened]... itch. However, as this was not specified the answer is:**

You should stick on your chosen door.