h2g2 The Hitchhiker's Guide to the Galaxy: Earth Edition

Here's why:

Prize is behind A:

You choose A, gamehost discards B -> switch to C - lose
(1/3) x (1/2) x (1/2) = 1/12

You choose A, gamehost discards C -> switch to B - lose
(1/3) x (1/2) x (1/2) = 1/12

You choose B, gamehost discards C -> switch to A - win
(1/3) x (1) x (1/2) = 1/6

You choose C, gamehost discards B -> switch to A - win
(1/3) x (1) x (1/2) = 1/6

Win: 1/3
Lose: 1/6

You have twice the chance to win.

Notice the middle number I'm multiplying by? In the choice A version, there is a 1/2 chance the host will pick either B or C. In the other two versions, the host MUST pick B or C, only one of which is available, therefore the probability is only 1.

(The probabilities will be the same if the prize is behind B or C)

To put it simply, you have twice the chance of winning by switching, because you have twice the chance of picking the wrong door to begin with. (1/3 chance of picking the right door, 2/3 chance of picking the wrong door)

-DirewolfX

Right, DirewolfX, though I couldn't follow your logic there, but then agian I couldn't follow the original one either!

I think Mummy's faulty reasoning here was that there are only 6 possible equal outcomes, not 8. He had the person picking door A TWICE! That screws up the equality of the choices. When there are three doors to pick from you have an equal chance of picking each only one time. You can pick from the original 3:

1. prize
2. goat
3. goat

If you don't switch that's exactly what you get. If you switch, after the host opens one of the goat doors, you get:

1.goat
2.prize
3.prize

I think that people (including me at first) forget that you always have a 1/3 chance of picking the prize if you stay with the one you picked first. You CAN'T have a 1/2 chance when you have only 1 out of 3 winners. If you don't switch you can't take advantage of the host removing one of the losers from your choices. You are stuck with the original 1 in 3 odds.

But in the end, the host will always remove one of the wrong answers, right? And you get to choose between the two remaining doors. Whether you switch or not, shouldn't it still be one in two?

The basic problem with The Mummy's reasoning is that he's managed to produce eight different - and perfectly valid - outcomes... but he's forgotten that some of them are less likely than others because they involve an additional 1/2 choice. DirewolfX has shown the *correct* probabilities... which, surprise surprise, add up to 1/3 and 2/3...

26199

.... and, as I say in http://www.h2g2.com/forumframe.cgi?forum=38570&thread=49430 , common sense (which tells you that you do have a 50/50 chance), is not totally wrong, if you look at the bigger picture. It's all in how you phrase the question.

Look at it this way. What if there are TWO players, player A and player B. Player A's job is to pick a door in the first round. Player B is "off-stage" and has NO KNOWLEDGE of which door player A picked. The host then removes one of the remaining (wrong) doors. Player B then comes onstage, and, with no idea of what already happened, is told to pick a door. What are his chances of winning? 50%, of course. Two doors, one with a prize, one without.

Now, why would Player B have a 50% chance, and player A (if Player B did not exist) have either a 33% or 66% chance? That makes no sense by any logic. The answer is, of course, that BOTH players have, overall, a 50% chance of getting a prize (if they were to continue on to Round 2, that is). But, those chances are weighted differently for Player A, who's reasoning is changed by the phrase "switch or don't switch", instead of the phrase "One of these two doors has a prize, which do you pick?" IF player A goes into round 2, and instead of looking at it as "Switch or don't", does not base his choice in round 2 in any way on his choice in round 1, his chances are, of course, 50%. WITHIN that subset lie the odds that 33% of the time, the correct door will be the original door. 66% of the time, the correct door will be the OTHER door.

The truth is that his chances of winning ARE 50%, just as The Mummy said. His chances of picking the correct door the FIRST time ARE 33%, just as 26199 and C Hawke says. BOTH sets of statistics are correct, and any professor of logics who uses this example to teach his students that common sense is "wrong" is only looking at PART of the picture, and is, in effect, using statistics to confuse the students. Shame on him.

One thing I learned in Psychology classes in university is that statistics can be used to prove or disprove nearly anything. This is a classic case of that.

I think I'll post this in the other forum too, since that seems to be where the main debate is going on.... Just to avoid me having to go back and forth between the two.

*grin* I'm going to post a slightly different reply here just to make you switch back and forth between the two.

Surely the original question was 'switch or don't switch', the answer to the question is 'switch' - it's as simple as that? If you switch, you're twice as likely to win. What more d'you want?

26199

Yet would not the problem be fundamentally unchanged if you were to eliminate the part where the first choice is made? Regardless of whether you pick the correct or the wrong door the first time, a door will be eliminated, and you will still have absolutely no knowledge of which of the two remaining doors is correct. Meaning that they still both hold equal possibilities of success and failure. Since no knowledge is gained from the initial choice, it might as well not be made at all.

oops, I see this one's been solved. shoulda checked the other forum first.

You *do* gain some information in the first round... when the quiz-show host opens the door, you know which door of the pair you haven't picked the prize definitely isn't behind. This means that if the prize was behind either door, you now know exactly where it is. Which means that if you made the wrong choice in the first place, you know the location of the prize...

26199

Yes, I see that now. By making the first choice and assuming it was worse than the one you will make when there are less choices, you increase your chances.