now that gives me an idea to go back in school...:1orglaugh :1orglaugh

haha this is bullshit.

I cannot believe that people are still 'debating' this..

though the results appear to go against logic.. when you think about it / look at the solution it's correct..

the links posted are dead, show me another link with the reasoning behind it

Stupid. It is obvious from that point in time he has a 50% chance regardless, so switching does not increase his probability of winning.

dude it's 5 am here and i'm back from a party but i can't sleep; still i got an answer for you so-called riddle

I WOULD SWITCH THE BOXES

why?

if you will change to the other box, you will have 2/3 chances! you'll duoble your first chance!


the chance of you choosing the right box when you have 3 boxes is 33%
the chance of you choosing the wrong box out of 3 boxes is 66%

thus, when you have only 2 boxes (your first choice and what is left) if you change your selection your chances of winning are HIGHER!!!

that makes no sense.

I like to call this Retard-o Math. It's a 50 percent change no matter what. There's no reason to change the box. He's just as likely to fuck up either way.

dude don't be an asshole


when you had you first selection there was 33.3% you were right (1 box out of 3)
and 66.6% you were wrong

now, when there are 2 boxes, if you change you slection, ther is 66.6% you are right (100-33.3) and 33.3% you're wrong (100-66.6%)

That's it!!

Let me explain to you what is wrong with this. Let's say it was 100 boxes. Picks a box and then 98 are opened. By your logic, that means he still has a 1 in 100 chance of having the right box if he keeps his selection, and a 1 in 2 chance of winning if he changes his box.

It's not possible because the probability has to sort of add up. You can't have 1 in 2 and 1 in 100 as all the possibilities. You can have 1 in 100 and 99 in 100. You can have 1 in 2 and 1 in 2.

Go get 100 boxes, try it out with a friend and see how many times you get it right and how many times you get it wrong when you perform the test 100 times without changing your choice. You really think you're only going to win 1 time?

You people suck at math.

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....