The problem with your thinking is the or. That or means there are two possible outcomes if option A is right and the contestant chooses option A. For some reason, you've lumped two outcomes into one to prove your point, when two outcomes are just that, two seperate outcomes.

Explain to me why this isn't the case:

Contestant picks A.


If A is right (1/3 chance of that):
then the host reveals B
and switching does NOT pay off

If A is right (1/3 chance of that):
then the host reveals C
and switching does NOT pay off

If B is right (1/3 chance of that):
then the host reveals C
and switching does pay off

If C is right (1/3 chance of that)
then the host reveals B
and switching does pay off

It makes no sense to me that the first two outcomes i've shown above shouldn't be considered as likely as the rest.

You just used a total of 4/3. Think about that for a sec.

But, if it makes you happy, I'll also show how you SHOULD post your version:

If A is right (1/3 chance of that):
then the host reveals B (1/2 of 1/3)
and switching does NOT pay off
or
the host reveals C (1/2 of 1/3)
and switching does NOT pay off

If B is right (1/3 chance of that):
then the host reveals C
and switching does pay off

If C is right (1/3 chance of that)
then the host reveals B
and switching does pay off

The two variations are a subset sort of thing. I just copied and pasted your answer and forgot to remove that. This will make it more clear.


If A is right (1/3 chance of that):
then the host reveals B
and switching does NOT pay off
OR
the host reveals C
and switching does NOT pay off

If B is right (1/3 chance of that):
then the host reveals C
and switching does pay off

If C is right (1/3 chance of that)
then the host reveals B
and switching does pay off

Now you're just being pedantic. :winkwink:

Well, there you go. You also posted it. Now explain to me what's wrong with it.

Wrong? There's nothing wrong in the version I posted. But look at the probabilities of the several options:

A is right and B is revealed => 1/6
A is right and C is revealed => 1/6
B is right and C is revealed => 1/3
C is right and B is revealed => 1/3

still going? :Graucho

i totally understand this

but i still hate it
math is hilarious

I know where you're getting that from. You think that the 1 in 3 "contains" both possibilities and therefore each one is only a 1 in 6 chance. I totally disagree, and also, this is way out of line with the explanation given on all these websites these other idiots keep linking to. They seem to think it's a 50% chance for one, and a 33% chane for the other.

I think if i'm wrong, then you're right, but those other 'tards are way out of it.

But anyways. On to my explanation. Sure, there is a 1 in 3 chance that the person is going to choose the right one initially, but there are actually four possible outcomes, and thus, you're dealing with a 1 in 4 chance of each outcome.

A is right and B is revealed => 1/4
A is right and C is revealed => 1/4
B is right and C is revealed => 1/4
C is right and B is revealed => 1/4

It doesn't matter how many boxes there are. It matters how many outcomes there are. I'm starting to confuse myself. =\

Think about what you just said:
A is right and B is revealed => 1/4
A is right and C is revealed => 1/4
B is right and C is revealed => 1/4
C is right and B is revealed => 1/4

That would mean that there is 1/2 chance of A being right in the first place, and with 3 equal options, that would be really weird.

You said it yourself, the "b or c" is a subset. A subset, by definition, only takes part in the set it is a subset of.

Yikes ... What about B is right and A is revealed or C is right and A is revealed?

A can't be revealed because A is the correct box, and only an incorrect box is revealed before the switch option is given.

A won't be revealed because it's the one the contestant has chosen. A getting revealed would make the choice of switching not really an issue, since if A is revealed and is empty, not switching yields 0%.

try out this simulation...see the numbers for yourself!
http://www.cut-the-knot.org/hall.shtml

Ahhh I didn't see that you guys were using A to represent the contestant's choice.

It doesn't mean that. It only means there's a 1 in 2 chance after the third box has been revealed and a switch has been offered. Prior to that, it's a 1 in 3 chance, just the same as the others.

There is a 1/3 chance that you'll hit the prize door, and a 2/3 chance that you'll miss the prize. If you do not switch, 1/3 is your probability to get the prize. However, if you missed (and this with the probability of 2/3) then the prize is behind one of the remaining two doors. Furthermore, of these two, the host will open the empty one, leaving the prize door closed. Therefore, if you miss and then switch, you are certain to get the prize. Summing up, if you do not switch your chance of winning is 1/3 whereas if you do switch your chance of winning is 2/3.

Think about what you're saying now for a moment. You are claiming that probability on your original choice increases, even though regardless of the correctness of your choice the host can always open one door.

That doesn't make sense, now does it?

See, when you first choose, there are 3 possible outcomes. A, B, or C. You have a 1 in 3 chance of picking the correct box.

However, once a box is removed on the condition that said box is not a prize winner, and then given the chance to swap your choice, there are now 4 possibilities.

Everything changes when the third box is removed. 3 has nothing to do with it anymore. You know have 2 boxes and 2 possible variations to each. A total of 4 possible outcomes.

To make my point more obvious, think about it like this. The third box does not matter. Just ignore it. It's a given that you will end up with 2 boxes, one of which is empty, and one of which contains the prize. So why not just start there? Start with two boxes. 1 in 2 chance.

I am 99% sure i've got this right.

And so does the probability of the one you didn't choose. The pool becomes smaller, thus the chance of each box being the prize is increased. Simple. Instead of being 1 of 3 possibilities, each box is now 1 of 2 possibilities. Everything readjusts. 3 has nothing to do with it anymore. 3 doesn't exist anymore in the realm of this problem after the removal of the third box.

Ok, so what's the real answer?

Edit: 100 math riddles

he still have 50 % ..

Removeing one box = 50 % on each others

that's it . he got 50 % of chances to win either box he choose . there are only 2 ...

i don't beleive a word of that crap ...

you got 3, you remove 1

2 boxes, 50/50.

exactly!

dum ass riddle!

Read this SLOWLY

There is a 1/3 chance that you'll hit the prize door, and a 2/3 chance that you'll miss the prize. If you do not switch, 1/3 is your probability to get the prize. However, if you missed (and this with the probability of 2/3) then the prize is behind one of the remaining two doors. Furthermore, of these two, the host will open the empty one, leaving the prize door closed. Therefore, if you miss and then switch, you are certain to get the prize. Summing up, if you do not switch your chance of winning is 1/3 whereas if you do switch your chance of winning is 2/3.

THE ONLY CASE WHERE YOU WILL WIN IF YOU DON'T SWITCH IS IF YOU PICKED THE CORRECT BOX ON THE FIRST ATTEMPT = 1/3

THE ONLY CHANCE YOU WILL NOT WIN BY SWITCHING IS IF YOU PICKED THE CORRECT BOX FROM THE START.... (in other words...1/3 chances OF NOT WINNING, HENCE, 2/3 chances of WINNING)

ugh

watching the replies to this is frustrating

its really simple

unless you are clairvoyant, you only have a 1/3 chance in picking it right the first time

once a KNOWN empty has been eliminated, switching provides you with the advantage because with the additional information of a known empty taken out of the equation, youre now basing your decision off of a 2/3 skew

ok i made it worse

The third box DOES matter, that's the whole point...

But let's look at a bigger-sized example. Perhaps that will make it clear.

Imagine someone has 1000 marbles. One of them contains a diamond, 999 do not.

He says "I will let you pick 1 of my marbles, and after that, I will put aside 998 of my marbles that do not contain diamonds."

Independent of your choice, he will ALWAYS be able to put aside 998 marbles that do not contain diamonds.

You pick a marble, with a 1/1000 chance of getting the diamond.

Then, he says, "I will now put aside 998 of my marbles that do not contain a diamond."

Now, has your marble suddenly gone from a 1/1000 chance of being a diamond to a 1/2 chance of being a diamond?
Even if, have you picked any of the non-diamonds, this very same thing would have happened?

Frustrating to argue against people who somehow fail to see the obvious, isn't it? :glugglug

exactly! It's pretty simple actually!

Let's say you pick A.....

The only case you will win is if it's behind door A....but if it's behind door B or C...you will have it by switching!...as simple as that!

1/3 chances against 2/3 by switching....

yes

i think this will help

**GFY VERSION OF THE RIDDLE**

there are three hot looking women in front of you and two of them are trannies but they are all wearing baggy pants

their pimp says choose the woman and you can have sex with her

so you choose the one with the least defined jawline

the pimp then removes one of the trannies from the lineup

you notice the one you chose has a boner

case closed

:thumbsup

you only need to think about it for a minute to verify... it's not that hard

omg people... this shouldn't be this difficult.

1 - 3 chances.... take away 1.

1 - 2 chances.

That's as simple as it gets. If you have 2 boxes, that's all you have... REGARDLESS of how many are taken away before hand... if you have 2 boxes remaining, that's 2 boxes... the odds are... it's one of the two. That's it.. that's all.

When there are 2 boxes, there is no way that one has 2/3 of chance of being right. Get it? Got it? Good.

dude if i were you i would at least pretend to understand the solution.
dumbass

omg you still don't get it do you?

oh my god i can't believe there is this much chat over this... it's so fucking plain and simple.

god damn people are stupid. he could choose to stay with his current box, or the other one.. he still has the same 50% chance.

That's not a similar problem. You have to remove equally sized portions sort of thing or it throws it totally out of whack and it's a different problem.

In our 3 box problem, you have 3 things that are each equal. In your example, you have 1 marble versus 1 marble versus 998 marbles. It's ridiculous.

LOL so wrong...again, thing about it...
The host KNOW where the prize is...


Let's say you pick A.....

The only case you will win is if it's behind door A....but if it's behind door B or C...you will have it by switching!...as simple as that!

:1orglaugh

MM

fuck man

you dont have 1/2 chances
you have 1/3 and 2/3

the host KNOWS where an empty box is
if he was just arbitrarily choosing a box then it wouldnt matter

but the fact is he is eliminating a KNOWN empty

therefore your new choice gives you the benefit of the revealed empty box