*the* smart question

[Bozos of Bones]

OK, let me see...

CASE 1
I have three boxes, one of them has a coin in it. Let's say box B has a coin in it. I take box A, and say it's mine. I open up box C, which comes out empty. So it's down to boxes A and B now. I open up my box, comes out empty.

CASE 2
The coin is now in box A. I choose box A for my box, and I open box B which comes out empty. I remain with box A, open it, and find the coin in it. Box C is then empty.

CASE 3
The coin is in box C. I choose box A for my own, and open box C. I find the coin inside, which makes boxes A and B empty.

CASE 4
Let's say I choose box A, which has the coin, and that box B is the opened one. I switch, and find no coin in my box.

CASE 5
Now I choose to switch again, only now in my possesion there is an empty box A and an opened and empty box C on the side. I take box B, open it and find the coin.

When there are three chances, and one of them is proven taken, the other two chances are recalculated without regard to the third chance, unless the third option was the right one, in which case no recalculation takes place. Nothing fixes the 1:3 odds to the box you choose, it is recalculated with the rest, and you made the mistake of adding the chance of the third box to the second box, rather than distributing it evenly.

Oh and mad=dog, you couldn't be more wrong. Well, you could have, but to be more wrong you'd have to disregard math, facts and logic and say "hot dog" or "rabbit porridge" or some other food stuff.

[ 06-23-2005, 02:50 PM: Message edited by: Bozos of Bones ]

[Morgeruat]

[Aragorn1]

Originally posted by Bozos of Bones: When there are three chances, and one of them is proven taken, the other two chances are recalculated without regard to the third chance, unless the third option was the right one, in which case no recalculation takes place. Nothing fixes the 1:3 odds to the box you choose, it is recalculated with the rest, and you made the mistake of adding the chance of the third box to the second box, rather than distributing it evenly.

You are still treating it as two separate events, which t is n fact not, as wella s ignoring conditional probabiliy, but that does not apply here in its normal sense as there are not two events.

I know where you are coming from (just look at my previous posts!!!), but i have gone through the maths and it is correct. I'm not too bad at maths either. Here it is if anyone is interested: http://astro.uchicago.edu/rranch/vkashyap/Misc/mh.html

Think of it this way: imagine there are 1000 doors, and you can choose one, the chances of sucess are 1/1000. The probability of being behind one of others is 999/1000. THe host then opens all these doors except for one. The chances of the prize ebing behind this are 999/100, as the other doors have been eliminated.


P.S.

You didn't actually bother with the whole drawing thing, did you? You just dismissed what i said as wrong. It is understandable, as the solution goes against normal logic.

[ 06-23-2005, 03:15 PM: Message edited by: Aragorn1 ]

[Aragorn1]

Here's a diagram i found of it, i hope this helps to clear it up.

[Bozos of Bones]

Nah, I drew it in my head. Now try doing it more than once. For each of the boxes the chance is 1/3, and a 2/3 for the other combined two.

1/3 2/3
A B C

1/3 2/3
B A C

1/3 2/3
C A B

See? Perception from all three boxes yields a normal result.

Now...
In the time of my first post and the second one, I failed to read through the second page. I read it now, and I see that we are talking about a gameshow in which the host always opens the empty box. Taking that into consideration means there is a fixed 1:1 chance the empty one gets thrown out, and that makes the logic plausible using the correlation theorem, but only in the event that the empty one is the one to always be discarded.

[Kyrvias]

Everyone is thinking too far into the question!!!


To my mind, it should be simple, then everybody starts saying, "The contestent doesn't see the locket" and everybody is in a heated argument withwhich thier is no end, and if the argument does end, my answer stays the same; answer 3


That is a *really* long sentence.

[Aragorn1]

The answer is 2, see above.

There is an end, it is maths, there must be one and only one correct answer, and the maths proves that the correct option is to switch, statistically speaking.

Here this gives a good explaination:
http://en.wikipedia.org/wiki/Monty_Hall_problem

[Seraph]

On a slightly different topic here is another probability problem:
A bag contains three coins.
One coin is red on both sides
One coin is blue on both sides
One coin is red on one side and blue on the other.
Without looking you pick a coin and put it on a table. You look at the coin and you can see one side of the coin, which is blue, what is the odds that other side is also blue.

[ 06-23-2005, 07:05 PM: Message edited by: Seraph ]

[Bozos of Bones]

Well. since the very fact that the first side is blue, the full red coin is out of the game. Meaning, 50:50 chances, as there are only the full blue and red-blue coins remaining. Unless there is something you're not telling me...

[mad=dog]

It's one third. The other side will also be blue if and only if you picked the all blue coin (which has one third probability of being picked).