A little riddle

[bibim]

Here are the rules of the game:

There are 3 closed doors. There is money behind one of these doors, but there is nothing behind the others. If you open the door with the money, you win it.
First, you choose one door without opening it.
Then, someone who knows where is the money comes and opens one of the 2 other doors to show you there is nothing behind it.
Lastly, you are offered to open the door you had chosen previously, or change and open the last door.

What would you do ?

1) keep the door initially selected
2) change
3) It doesn't matter

[koshi]

change

[rattle]

How is that a riddle

It doesn't matter because it's a 50/50 chance anyway.

[Zpock]

It's not, 1/3 if you keep, 1/2 if you change. Think of 1000 boxes, you pick one, he opens all but one, do you change?

[Machiosabre]

ìt actually isn't cause that means that theres a chance left that there's nothing if you open both doors, there you go.

[bibim]

It's not, 1/3 if you keep, 1/2 if you change.

Where is the remaining 1/6 ?

[koshi]

if you swap your chances of winning are 2/3
if you don't swap it's 1/3

if you're not familiar with bayes' theorem, do a decision tree

[Dragon45]

http://en.wikipedia.org/wiki/Monty_Hall_problem

Make the problem simple:

Suppose you had a million doors, and you pick one. Of the doors you haven't selected, all but one is knocked down (999,998 doors gone). Do you switch or not?

[Zpock]

I'm not sure at all but I think it's 1/3 to win 2/3 to loose if you keep, 1/2 to win 1/2 to loose if you change, it's two different situations so they don't necessarily add to one as you think.

Edit: I checked the tree it's 2/3 if you change... Basically if you picked the wrong from the start (2/3) and then change you are guaranteed to win, if you picked the right one (1/3) and change your sure to loose.

[Erom]

Did you read the link Zpock? It's a pretty classic problem. Think of it this way- there are three jugs, each with 1/3 Win in them and 2/3 Fail. You pick a jug. The other two jugs are mixed together and then half of the liquid is removed. All of the removed liquid is Fail, so the jug is left with 2/3 Win and 1/3 Fail.

So now you have a choice between your jug (1/3 Win, 2/3 Fail) and the other jug (2/3 Win, 1/3 Fail). You should switch.

Edit: ninja'ed by your edit, which is also the simplest explanation, I just wanted to try to explain it using my wacky Win Liquid and Fail Liquid analogy.

[Zpock]

The 1/3 to win if you keep and 2/3 to win if you change adding to one is still a "coincidence" I think, consider 4 boxes, and one taken away... if my calculations are correct you would have 1/4 to win by keeping and 3/8 to win by changing.

[imbaczek]

another classic:

does 0,999999... = 1?

[koshi]

did achilles overtake the turtle?

[Erom]

The 1/3 to win if you keep and 2/3 to win if you change adding to one is still a "coincidence" I think

Yeah, it is. Your math (1/4 keep 3/8 swap) is correct for the 4 door case.

[KDR_11k]

Zpock: When the problem gets generalized to larger numbers of choices it's "remove all except one of the unchosen set", thus the chances MUST add up to 1.

another classic:

does 0,999999... = 1?

Yes.
1. Completeness property of real numbers (the open interval between any two real numbers contains an infinite set of real numbers exactly when the two real numbers are not equal). To have a number that's between 0.999.. and 1 you'd need to have a number that is either one digit longer or has a digit larger than the lower one. As 0.999... has an infinite number of 9s and there is no higher digit than that there is no way to create a number between 0.999... and 1, if we assume 1 != 0.999... we'd have a contradiction with the completeness property and thus it follows that 1 = 0.999...

2. 1 = 3*0.333... = 0.999...
Transitive property means that the above can be written as 1 = 0.999...

QED

[manored]

Now I know what to do if I ever get to a game show :)

I have a riddle too, altough it is easy (and old):

What is it broken, then its name is spoken?

And a hard one:
3 men will pass through this test: one disk will be attached to the back of each, and then they must discover what color is the disk is in their back, and provide logical explanaition (guess not allowed). They know that there are 3 white disks and 2 black disks. The first man could see the disks in the other 2 before answering, but failed. the second could only see the one from the last man and know that the first had missed, but also failed. the third could not see anybodys disk but knew both had missed, and concluded his disk was white through logical deduction. Explain his line of tough.

[Zpock]

When the problem gets generalized to larger numbers of choices it's "remove all except one of the unchosen set", thus the chances MUST add up to 1.

And when further generalized to a problem with X out of Y choices removed it doesn't, was my point.

[KDR_11k]

And a hard one:
3 men will pass through this test: one disk will be attached to the back of each, and then they must discover what color is the disk is in their back, and provide logical explanaition (guess not allowed). They know that there are 3 white disks and 2 black disks. The first man could see the disks in the other 2 before answering, but failed. the second could only see the one from the last man and know that the first had missed, but also failed. the third could not see anybodys disk but knew both had missed, and concluded his disk was white through logical deduction. Explain his line of tough.

If the first had seen two black ones he would have said white and been correct, as he hasn't answered that at least one of the two remaining must have a white disc. The second could have said white and be correct had the last a black disc (as that would leave only the second for having a white disc). As the second didn't say that the third must have a white disc.

[rattle]

The 1/3 to win if you keep and 2/3 to win if you change adding to one is still a "coincidence" I think, consider 4 boxes, and one taken away... if my calculations are correct you would have 1/4 to win by keeping and 3/8 to win by changing.

How does the chance improve at all when one of the empty doors is taken care of? 1 in 3 chance initially, 1 in 2 then... in my book.

[KDR_11k]

The chance to win if you don't change is 1/3rd because that's the chance of picking the right door right away. By eliminating one wrong door you are then pretty much asked if your door is the winner or not (because you can keep it or change). As you know there's a 1/3 chance it's a winner there's a 2/3 chance it's not so you'd pick the "not" option. Or to phrase it differently, you pick between saying that your door is the one that contains the prize or that either of the two others contains the prize (since if either contains it the change would automatically lead to that door). If it helps, imagine the wrong door is not opened until after the change but you still get the unopened door, that way it's easier to see that you're switching between one door and the better of two doors (because if you say you take the two doors you get the prize if either of them contains it).

Your intuitive approach fails to take into account that the removed door is guaranteed to be empty, if one door was removed at random (as it happens in the real gameshow) then yes, it would not matter if you change.