haha you are an idiot

the real idiot is YOU!!!
i can bet you $1M you stupid ignorant

read it again and try to UNDERSTAND

after the first box is taken away, you are left with two boxes. the first box is now out of the problem. forget about it.

one is empty, one is not.

you must pick one. you have no control over if you picked the right one or not out of the first three. the odds are changed to 50% that you pick the right one out of the two.

switching or not does nothing to increase chances.

If you're correct, then your theory would be reflected in real life test results. See quoted post below and maybe you'll figure it out.

The chance of each box being the correct box improves as others are eliminated. So just as the box not selected is now 50 percent likely to be the winning box, so is the box already chosen.

Are you really this retarded? If i had a million dollars, i would take that bet in an instant. You're an idiot.

LOL you are FUCKING WRONG MAN!!!!!!!!

read patiently and try, just try, to understand

you got 3 boxes, you have to choose the right one
the chances are 33%

let's say you chose the wrong one

when there are 2 boxes left, you can switch
in the beginnig you had 33% change of being right; if you switch now, you have 66% of being right;
you also had 66% of being wrong; if you change now, you have only 33% of being wrong;

sorry if my explanation is not clear enough, i can't exphasize it more!

Fuck you you ignorant; don't call me retarded before you finish your 11th grade homework;

for all the extremely low IQ ppl around the board, i've google the answer for you
http://www.comedia.com/hot/monty-answer.html

or just google "monty hall problem"

chodadog, smit: sorry but you're stupid!!!

haha no way that makes sense, you don't know if you picked the correct box or not, you don't gain probability switching the boxes.... haha this isn't even worth it, keep living in your idiot world.

i agree and that most of the time your first choice is the best one

read the post above this one... the one with the explanation why YOU ARE WRONG

take it easy

you're wrong too

gahhh ive never heard of this "famous" problem, the fact that monty deliberately picks an empty box plays the key factor. anyways, here's a computer simulation:

http://math.ucsd.edu/~crypto/Monty/monty.html

hey alexg my friend good luck w/ the test tomorrow!!!

Assuming that the removed box(es) always was/were empty, you would indeed have a higher chance of winning by switching, because the box you switch to would gain the probabilities of the now removed box(es).

Assuming that the removed box is random, you have an equal chance of winning no matter what you choose.

It's simple. End of debate.

ah.. no it's not 50% chance either way

You are a fucking moron my friend, it is not interesting and it is not a riddle, you simply do not understand probability.

You also have a flaw in your explanation, at first it was the host that opened the box, then at the end you said it was the participant that opened it. That is not you major misunderstanding but another error.

I don't beleive how stupid some people here are, I guess you don't have to be smart to be a webmaster huh? :glugglug

There is a very simple rule when it comes to probability. It all has to add up. If you have two choices, it's simply not possible to have 2 different odds.

1 in 2 = 0.5
1 in 3 = 0.3 recurring.

Add them up and you have 0.83 recurring. It needs to add up to 1. Or else 17 percent of the time, you'd be having no result, and that's simply not possible. How do you account for the remaining 17 percent? Or am i missing something here?

I mean, 1 in 2 is the same as 3 in 6. And 1 in 3 is the same as 2 in 6. So what about the remaining 1 in 6? What is the probability that it'll be the prize or won't be the prize?

3 boxes (ABC) and 3 different ways to arrange the winning box (xyz)

ABC
x 100
y 010
z 001

You pick A box each time and don't switch:
x: host reveals C is not a winner, you don't switch, you WIN
y: host reveals C is not a winner, you don't switch, you LOSE
z: host reveals B is not a winner, you don't switch, you LOSE

You win 1/3 of the time.


You pick A box each time and you switch:
x: host reveals C is not a winner, you switch, you LOSE
y: host reveals C is not a winner, you switch, you WIN
z: host reveals B is not a winner, you switch, you WIN

You win 2/3 of the time.

Total probability: 1/3+2/3=1


Initially picking B or C doesn't effect this result so by switching
you increase your chances to 2/3.

But from what everyone has been saying, you're supposed to end up with a 50% chance of winning instead of a 33% chance. You've got a 66% chance of winning on the switch.

And what you just wrote up there made no sense to me whatsoever.

Let me be the first person to fully solve this dilemma.

The 2/3 chance on switching holds, but only if the game show host does the same thing every time, regardless of the choice of the contestant.

If that is the case, when you choose A [1/3 chance] instead of (B or C) [2/3 chance], the host by removing an empty door from (B or C) and giving you the option to switch, effectively simply gives you the option of switching to (B or C) combined. There is always at least one empty door among (B or C), so the host is always able to remove one, and the act of removing one door is actually insignificant.
Think of it this way: (B or C) has a 2/3 chance, and at least one of (B or C) is empty. By pointing out one of both as empty, the combined chance for (B or C) does not change. All it changes is merge the chances of both doors into one.


Now, that is pretty simple. A problem arises when we give the host the option of not revealing one door.
If the host has the option of not revealing a door and indeed not always does reveal a door, and he has knowledge of which door is the right one, he might intentionally try to take your choice away from the right one by giving you a "more logical" alternative. Or, on the other hand, he might try and guide you towards the right choice.

Simply put, the 2/3 chance on switching is valid if and only if the elimination of a door and the subsequent option of switching are introduced completely independent of the validity of the original choice.

Woj; What you are forgetting is that there are two possible box changes if the correct box is initially chosen whereas there is only one possible change if each of the incorrect boxes is initially chosen. So there are 4 possible outcomes for each prize location, for a total of 12 possibilities.

So look. Here are all the possible occurences if the prize is in box A.

Contestant Chooses: Box A
Removed: Box B
Possible Outcomes: Stays/Wins, Switches/Loses.

Contestant Chooses: Box A
Removed: Box C
Possible Outcomes: Stays/Wins, Switches/Loses.

Below: Box A obviously can't be removed because it's the winning box.

Contestant Chooses: Box B
Removed: Box C
Possible Outcomes:Stays/Loses, Switches/Wins.

Contestant Chooses: Box C
Removed:Box B
Possible Outcomes:Stays/Loses, Switches/Wins.

2 wins for switching.
2 wins for staying.

You're making a huge mistake by counting both the "contestant chooses A" options as options of similar size as the other two.
By that reasoning, if no boxes were removed and no choice to switch was given, option A would somehow have a 50% chance all by itself.

But if no boxes were removed, then it would simply be 1 in 3 chance of each one happening as there would be only be one possible outcome for choosing box A. But because a box is removed, there are 2 possible outcomes for box A. The fourth option would not exist if no boxes were removed and no switch option was given.

That's bullshit. Switching does not increase the likelihood that he gets it right. Logical fallacy, buddy.

But those 4 options don't have equal chances. Just because they're all there, doesn't mean they're all just as likely.

Wrong, it does increase the likelihood. Mathematical fact, buddy.

For one to be more likely than the other, wouldn't it's result have to be more frequent in the pool? I mean, let's take a six sided die. Let's say it has 1, 2, 3, 4, 5 and 5. No six on this dice.

Total possible results= 6.
Chance of a 1 coming up = 1 in 6
....
Chance of a 4 coming up = 1 in 6
Chance of a 5 coming up = 2 in 6 or 1 in 3.

In the 3 box situation we're dealing with, there is only one possible combination for each outcome, and therefore they are just as likely as eachother, aren't they?

I mean, for one outcome to be more likely than any other outcome, it would have to appear more than the others in a list of all the possible outcomes, no?

Simply put, no. This very post of yours in fact makes clear why: there are actually 6 possibilities if you follow your line of thinking. The fact that A being right offers 2 valid ones and B or C being right each only offer 1 valid one, does not change that fact. (this paragraph lacks a bunch of explanations and definitions, and makes some big jumps, so if you don't get it my laziness is probably to blame)


Let me explain the entire issue in the simplest way possible:

A = 1/3 chance
B = 1/3 chance
C = 1/3 chance

Contestant picks A.


If A is right (1/3 chance of that):
then the host reveals B or 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

No.

Heads, tails or side of the coin.

Does "side of the coin" have an equal chance as the other two simply because it occurs just as often as they do in a list of possible outcomes?


However, in the monty hall case, something else is actually the matter: the host revealing the right box is also a logical possibility, just not a practical one. In order to get the right results, you need to include it in your calculations - if you want to work with a list of possible outcomes where revealing a different box constitutes an entirely different outcome.

You DO have more chances by switching...the best explanation, I think, is this one:

If you always stick with your initial guess, when do you win? Only when you pick the car with your initial guess. This is 1/3 of the time.

If you always switch from your initial guess, when do you win? Only when you don't pick the car with your initial guess. This is 2/3 of the time.

Kinda easy to understand....

The host KNOWS where the prize is....

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.