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Libertine 12-17-2004 02:48 PM

Just remembered something I encountered the other day. For all those who think 50% is the right answer, here's a master's thesis from a guy from my faculty proving you wrong:
http://www.philos.rug.nl/~barteld/monty.ps

It's in postscript though, so get a viewer for that (e.g. ghostscript).

ytcracker 12-17-2004 02:49 PM

Quote:

Originally posted by galleryseek
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.

dude

arghhhhhhhh
no

its nothing to do with 50%

there are three boxes
ahghghgihsiughdfiugiudfhiguhdfiughihwrihehhfdsoghf dghodfh shit

SilverTab 12-17-2004 02:51 PM

Fuck that I can't believe people don't understand it...


Here we go again:
Case 1:
THE PRIZE IS BEHING DOOR B

Let's say you pick door A.....The host opens door C... you STAY with A...YOU LOSE

Case 2:
THE PRIZE IS BEHIND DOOR C

Let's say you pick door A.....The host opens door B... you STAY with A...YOU LOSE

The only chance you would win is if the prize is behind door A (1/3)

BUT IN BOTH CASE IF YOU STARTED WITH DOOR A AND SWITCHED, YOU WOULD WIN (2/3 chances of winning)

come on it's not that hard to understand

ytcracker 12-17-2004 02:51 PM

Originally posted by ytcracker
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


^- brilliance at work believe that

chodadog 12-17-2004 02:56 PM

RECAP!

In all this back and forth shit, everything has been kind of spread out. I won the ebay auction i was waiting for so i'm going to make one last post before i fuck off to bed for you guys to read and think about.

3 boxes. 1 has a prize in it.

Initially, you have a 1 in 3 chance of picking the correct box.

An incorrect box is then removed, leaving you with 2 boxes. One is the prize. The other is not. Two boxes. 50 percent chance. That makes it 1 in 2. You're then offered the ability to switch. If you choose not to switch, you're left with a 1 in 2 chance. If you elect to switch, you now have a 2 in 4 chance, which is the same as a 1 in 2 chance.

Let me explain.

At first, there are 3 boxes and a total of 3 outcomes each with a 33.3333% chance of occuring.
A box is then removed. The 3 becomes 100% irrelevant. You now have two boxes and a total of 2 possible outcomes. These outcomes each now have a 50% chance of occurring.
You are then offered a switch. This gives each of the 2 existing possible outcomes another possible outcome each, leaving you with a total of 4 possible outcomes. 2 of them will win if you elect to stay. 2 of them will win if you elect to switch giving you a 2 out of 4 chance, i.e., 1 in 2 chance of choosing the correct box. Illustrated below:

Outcome Number 1: If A is right
then the host reveals B
and switching does NOT pay off
OR
Outcome Number 2: the host reveals C
and switching does NOT pay off

Outcome Number 3: If B is right
then the host reveals C
and switching does pay off

Outcome Number 4: If C is right
then the host reveals B
and switching does pay off

2 out of 4 = 1 out of 2.

That should explain it enough.

But to further prove just how irrelevant the third box is, ponder this. You're given 2 boxes. You get to choose one. Then after making your choice, you get told you can change your mind. There are 4 possible outcomes.

The third box is absolutely irrelevant. It serves no purpose whatsoever. This is further illustrated by the fact that NO MATTER WHAT YOU DO, an incorrect box will always be taken away and you'll be left with one right, and one wrong. No matter fucking what! Can't you see how that completely negates the third box?

Anyways. I'm off. Bedtime for me.

SilverTab 12-17-2004 02:57 PM

Let's take a look at the 3 cases again LOL
1 = prize, 0 = no prize

CASE 1:
A B C
1 0 0

You pick A, the host open eighter B or C...you stay with your choice, you win....you switch, you lose... ONLY CASE WHERE YOU WILL WIN

CASE 2:
A B C
0 1 0

You pick A, the host open C....you switch,
YOU WIN...you stay with your choice YOU LOSE

CASE 3:
A B C
0 0 1

You pick A, the host open B....you switch,
YOU WIN...you stay with your choice YOU LOSE


Fuck....it's easy to understand! there's no 50/50 here damnit!

Libertine 12-17-2004 03:00 PM

Quote:

Originally posted by chodadog
RECAP!

In all this back and forth shit, everything has been kind of spread out. I won the ebay auction i was waiting for so i'm going to make one last post before i fuck off to bed for you guys to read and think about.

3 boxes. 1 has a prize in it.

Initially, you have a 1 in 3 chance of picking the correct box.

An incorrect box is then removed, leaving you with 2 boxes. One is the prize. The other is not. Two boxes. 50 percent chance. That makes it 1 in 2. You're then offered the ability to switch. If you choose not to switch, you're left with a 1 in 2 chance. If you elect to switch, you now have a 2 in 4 chance, which is the same as a 1 in 2 chance.

Let me explain.

At first, there are 3 boxes and a total of 3 outcomes each with a 33.3333% chance of occuring.
A box is then removed. The 3 becomes 100% irrelevant. You now have two boxes and a total of 2 possible outcomes. These outcomes each now have a 50% chance of occurring.
You are then offered a switch. This gives each of the 2 existing possible outcomes another possible outcome each, leaving you with a total of 4 possible outcomes. 2 of them will win if you elect to stay. 2 of them will win if you elect to switch giving you a 2 out of 4 chance, i.e., 1 in 2 chance of choosing the correct box. Illustrated below:

Outcome Number 1: If A is right
then the host reveals B
and switching does NOT pay off
OR
Outcome Number 2: the host reveals C
and switching does NOT pay off

Outcome Number 3: If B is right
then the host reveals C
and switching does pay off

Outcome Number 4: If C is right
then the host reveals B
and switching does pay off

2 out of 4 = 1 out of 2.

That should explain it enough.

But to further prove just how irrelevant the third box is, ponder this. You're given 2 boxes. You get to choose one. Then after making your choice, you get told you can change your mind. There are 4 possible outcomes.

The third box is absolutely irrelevant. It serves no purpose whatsoever. This is further illustrated by the fact that NO MATTER WHAT YOU DO, an incorrect box will always be taken away and you'll be left with one right, and one wrong. No matter fucking what! Can't you see how that completely negates the third box?

Anyways. I'm off. Bedtime for me.

You're 100% wrong. I know exactly what your mistake is, I've tried to explain it, but I've failed.
I'll blame myself, though, and assume that I haven't explained it clearly enough yet.

So, to quote myself:

Quote:

Originally posted by punkworld
Just remembered something I encountered the other day. For all those who think 50% is the right answer, here's a master's thesis from a guy from my faculty proving you wrong:
http://www.philos.rug.nl/~barteld/monty.ps

It's in postscript though, so get a viewer for that (e.g. ghostscript).


SilverTab 12-17-2004 03:00 PM

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!


Let's say you pick B.....

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


Let's say you pick C.....

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


Daaaaaaaaaaaaaaaaaamn

ytcracker 12-17-2004 03:02 PM

chodadog

your analysis is incorrect

the third box is incredibly significant it is the key to this whole thing

if the host chose any box at random, then it wouldnt matter

he doesnt though

he chooses an empty box 100% of the time

when you made your initial guess you chose the box based on a 1/3 assumption

not knowing anything about what your box or any other box contains

now when an empty is taken out of the picture, you are now switching based on a 66% chance because before you didnt know the box the host removed was empty

by switching to the other box, you are now basing your opinion because of the presense of the known empty, not 50/50 chance because you picked that box based on 1/3

im having a hard time explaining this...fuckfukcukcukf

ytcracker 12-17-2004 03:03 PM

silvertab your explanation is fucking easy thanks

StuartD 12-17-2004 03:04 PM

Quote:

Originally posted by ytcracker
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

Ok, let me put this another way...

Let's have an invisible man be at box #2... ok? just standing there.

Now, the person takes a box... invisible man get box #2 and stands there and the announcer opens box #3 to reveal that its' empty.

Ok, no the person is standing there thinking "ok, that was opened, so the other box must have a higher chance of being the one with the prize."

BUT WAIT.

What would the invisible man be thinking?? Huh?? Would he be standing there thinking "ok, I have box #2, and the box #3 was empty, that must mean that the box infront of the other guy has a higher chance of having the prize!"

Wait a sec.... does that mean that BOTH boxes have a 66% chance of having the prize?

Can it possibly be... that just because you're standing with one box in your hands, that it DOESN'T AUTOMATICALLY MEAN THAT THE OTHER BOX HAS TO GET THE EXTRA 33%??? Who is person #1 to decide WHICH box has the higher probabillity?? huh?? What about the poor invisible man behind box #2?? His is automatically the lower probability because person #1 says so??

I think not.

SilverTab 12-17-2004 03:04 PM

Quote:

Originally posted by ytcracker
chodadog

your analysis is incorrect

the third box is incredibly significant it is the key to this whole thing

if the host chose any box at random, then it wouldnt matter

he doesnt though

he chooses an empty box 100% of the time

when you made your initial guess you chose the box based on a 1/3 assumption

not knowing anything about what your box or any other box contains

now when an empty is taken out of the picture, you are now switching based on a 66% chance because before you didnt know the box the host removed was empty

by switching to the other box, you are now basing your opinion because of the presense of the known empty, not 50/50 chance because you picked that box based on 1/3

im having a hard time explaining this...fuckfukcukcukf


I think I will give up! :(

although the proof has been made...(I even think it's quite easy to figure out) some will just never admit it LOL

alexg 12-17-2004 03:06 PM

Quote:

Originally posted by kowntafit
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 WAS a flaw in my first post but I then corrected it..

there is a 66% chance to win if he switches, not 50%...

if you think it's 50-50, you're wrong... it's pretty simple

SilverTab 12-17-2004 03:08 PM

Quote:

Originally posted by alexg
there WAS a flaw in my first post but I then corrected it..

there is a 66% chance to win if he switches, not 50%...

if you think it's 50-50, you're wrong... it's pretty simple

LOL I tried like 5 different ways of explaining it...seems like people are just NOT reading it LOL...fuck that....

ytcracker 12-17-2004 03:09 PM

Quote:

Originally posted by MaskedMan
I think not.
silvertabs explanation is probably as simple as it gets

your choice is not 50/50

your initial choice was based on 3 boxes

the host will always show you that one box is empty

Quote:

Let's take a look at the 3 cases again LOL
1 = prize, 0 = no prize

CASE 1:
A B C
1 0 0

You pick A, the host open eighter B or C...you stay with your choice, you win....you switch, you lose... ONLY CASE WHERE YOU WILL WIN

CASE 2:
A B C
0 1 0

You pick A, the host open C....you switch,
YOU WIN...you stay with your choice YOU LOSE

CASE 3:
A B C
0 0 1

You pick A, the host open B....you switch,
YOU WIN...you stay with your choice YOU LOSE
this makes perfect sense

do not think of this in terms of only two boxes

Libertine 12-17-2004 03:11 PM

SilverTab, ytcracker, I applaud your hard work but I fear that it will yield no results :(

StuartD 12-17-2004 03:12 PM

Quote:

Originally posted by ytcracker
silvertabs explanation is probably as simple as it gets

your choice is not 50/50

your initial choice was based on 3 boxes

the host will always show you that one box is empty



this makes perfect sense

do not think of this in terms of only two boxes

I've seen that explanation before, and believe me... it does make sense, but is still incorrect.

Try not to think of it in terms of one person... think of it being two boxes.

Read my post... if there's someone at box #1 and someone at box #2... which one has the higher probability?

You assume that the chances are higher so long as one person is making a choice, but when two people are, you can easily see... the odds are even.

alexg 12-17-2004 03:17 PM

let's see how much gfyers really suck at math


ytcracker 12-17-2004 03:18 PM

Quote:

Originally posted by MaskedMan
I've seen that explanation before, and believe me... it does make sense, but is still incorrect.

Try not to think of it in terms of one person... think of it being two boxes.

Read my post... if there's someone at box #1 and someone at box #2... which one has the higher probability?

You assume that the chances are higher so long as one person is making a choice, but when two people are, you can easily see... the odds are even.

masked man

you have to think of it in terms of one person one guess

the reason why is because if the one person has INITIALLY chosen an empty box the host cannot open HIS box to show him his guess was wrong

he will open the empty box that the guy hasnt selected

in which case if he switches he will win

however

lets say he picks the prize box first

the host has discretion over either empty box to open

if he switches away from the prize he loses

this is the only condition where he loses

otherwise the host will always be taking away an empty box in favor of the contestant

im trying to think of how i can illustrate this point in an easy fashion

SilverTab 12-17-2004 03:20 PM

This is my last attempt at explaining it

http://www.americanthumbs.com/silvertab/fordumb.jpg

Fred Quimby 12-17-2004 03:20 PM

Quote:

but if he stays with his old choice, he still has a 33.333% chance of course...
your logic is warped

WarChild 12-17-2004 03:25 PM

Quote:

Originally posted by MaskedMan
I've seen that explanation before, and believe me... it does make sense, but is still incorrect.

Try not to think of it in terms of one person... think of it being two boxes.

Read my post... if there's someone at box #1 and someone at box #2... which one has the higher probability?

You assume that the chances are higher so long as one person is making a choice, but when two people are, you can easily see... the odds are even.

That makes no sense at all. The variables are not the same with two people.

One person selects two boxes (by actually selecting 1 box) while the other person is stuck with only 1 box.

Look at it like this: You are trying to select 1 winning item out of 3 possible choices. If I give you one unknown box and keep two for myself, who has the better odds of winning? I have two chances of winning and you only one.

Adding a fictional third person completely changes the equation.

ytcracker 12-17-2004 03:26 PM

Quote:

Originally posted by WarChild
That makes no sense at all. The variables are not the same with two people.

One person selects two boxes (by actually selecting 1 box) while the other person is stuck with only 1 box.

Look at it like this: You are trying to select 1 winning item out of 3 possible choices. If I give you one unknown box and keep two for myself, who has the better odds of winning? I have two chances of winning and you only one.

Adding a fictional third person completely changes the equation.

exactly

because say the invisible man was sitting at the other empty box

when the host takes that box from the invisible man he cant just switch boxes
thats not the fucking point

StuartD 12-17-2004 03:32 PM

sheesh.. you guys go on picking that other box then. Good luck on that one.

I say one of you guys puts it to a test. Have someone know where the prize is so that he can remove a box every time....

and you go on picking the other box every single time for about 500 times.

See whether or not you get it right 250 times, or more.

SilverTab 12-17-2004 03:33 PM

http://www.americanthumbs.com/silvertab/fordumb2.jpg

ytcracker 12-17-2004 03:34 PM

Quote:

Originally posted by MaskedMan
sheesh.. you guys go on picking that other box then. Good luck on that one.

I say one of you guys puts it to a test. Have someone know where the prize is so that he can remove a box every time....

and you go on picking the other box every single time for about 500 times.

See whether or not you get it right 250 times, or more.

http://www.cut-the-knot.org/hall.shtml

scroll down about a quarter of the way
and play the simulation

just hit the radio buttons to guess where the prize is

try this yourself
it will become clear to you

SilverTab 12-17-2004 03:37 PM

Quote:

Originally posted by ytcracker
http://www.cut-the-knot.org/hall.shtml

scroll down about a quarter of the way
and play the simulation

just hit the radio buttons to guess where the prize is

try this yourself
it will become clear to you

you beat me to it...

http://www.cut-the-knot.org/hall.shtml
http://www.cut-the-knot.org/hall.shtml
http://www.cut-the-knot.org/hall.shtml
http://www.cut-the-knot.org/hall.shtml
http://www.cut-the-knot.org/hall.shtml


Just try it damn it!...the numbers are there....can't see why people won't understand it....

if you stay, you have 1/3 chances....(if you picked it correctly at your first pick)

IF YOU DIDN'T PICKED CORRECTLY ON YOUR FIRST PICK, YOU WILL ALWAYS HAVE IT BY SWITCHING!

ytcracker 12-17-2004 03:39 PM

Quote:

Originally posted by SilverTab
you beat me to it...

http://www.cut-the-knot.org/hall.shtml
http://www.cut-the-knot.org/hall.shtml
http://www.cut-the-knot.org/hall.shtml
http://www.cut-the-knot.org/hall.shtml
http://www.cut-the-knot.org/hall.shtml

IF YOU DIDN'T PICKED CORRECTLY ON YOUR FIRST PICK, YOU WILL ALWAYS HAVE IT BY SWITCHING!

the reasoning behind this is because the host will ALWAYS remove an empty box

if you have selected EITHER one of the TWO empty boxes, he will remove the other empty box and you will switch to the prize

you have a 2/3 chance of selecting an empty box the first time

you have a 1/3 chance to select the prize

this is not fucking hard

Mr. Mike 12-17-2004 03:40 PM

Wow, there's gotta be a better way to write this "Riddle". I need some Tylonol

SilverTab 12-17-2004 03:56 PM

Quote:

Originally posted by ytcracker

this is not fucking hard

apparently it is!.... :(

Repetitive Monkey 12-17-2004 03:57 PM

Stop being such retards and read my post on page two again. If the removed box is random, you don't gain any chance of winning by switching, because the removed box might be the "winner" and is as such still "in effect" (just not pickable).

If the removed box is always empty, you of course gain its probability by switching and so have a higher chance of winning.

ytcracker 12-17-2004 03:58 PM

http://www.ytcracker.com/images/moran.jpg

SilverTab 12-17-2004 03:59 PM

Quote:

Originally posted by Repetitive Monkey
Stop being such retards and read my post on page two again. If the removed box is random, you don't gain any chance of winning by switching, because the removed box might be the "winner" and is as such still "in effect" (just not pickable).

If the removed box is always empty, you of course gain its probability by switching and so have a higher chance of winning.


:thumbsup

power182 12-17-2004 04:02 PM

Quote:

Originally posted by punkworld
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.

Uh no, I already said that. You're the second one, not the first.

Libertine 12-17-2004 04:03 PM

Quote:

Originally posted by Repetitive Monkey
Stop being such retards and read my post on page two again. If the removed box is random, you don't gain any chance of winning by switching, because the removed box might be the "winner" and is as such still "in effect" (just not pickable).

If the removed box is always empty, you of course gain its probability by switching and so have a higher chance of winning.

The box is not random, that would defeat the purpose of the gameshow.

SilverTab 12-17-2004 04:04 PM

So assuming you always pick A as your first choice... (again)

A B C
1 0 0

eighter B or C is removed...you switch, you lose, you stay you win...only case....


now if the prize isn't behind your original pick:

PRIZE IS BEHIND DOOR B:
A B C
0 1 0

you pick A....C is removed, you end up with:

A B
0 1

will you switch to B and WIN or stay with A and LOSE?

now, PRIZE IS BEHIND DOOR C
A B C
0 0 1

Your original pick is A...now B is removed, you end up with
A C
0 1

will you STAY WITH A and LOSE or switch and WIN...

fuck that LOL it's simple

JUST FUCKING SWITCH

Libertine 12-17-2004 04:12 PM

Quote:

Originally posted by power182
Uh no, I already said that. You're the second one, not the first.
No. Read your own post, and the mistake in it:

Quote:

This statistic would hold true if and only if the game show host did not know the results of each box. Knowing game shows this is not the case, they will always reveal an empty box first, this skews your stats and places them at a 50/50 chance. If they could possibilly open the one containing the cash on the first try then it would hold water. Bad example, should have placed it in a senerio where nobody knows the results before hand.
You are saying that because it's NOT random, it's 50/50. However, the opposite is the case... because it's not random, it's 33.3r/66.6r.

garfield81 12-17-2004 04:20 PM

Quote:

Originally posted by alexg
Well, not really a riddle, but a rather interesting and not so easy to understand fact....


Let's say there's a show on TV where the participant has to chose 1 out of 3 closed boxes. In one of them there are $10k, and the other 2 are empty.

He makes a choice, and before opening the box the host of the show opens one of the two other boxes which is empty (there must be at least one, because a total of 2 boxes out of the 3 are empty).

now, should the participant change his choice to the other last box that the host didn't open? or should he stay with his choice?

most would think it doesn't matter at all... but the correct answer is that he should switch...

why?

at the beginning the participant was to chose one of 3 boxes... one of the three contained the cash, therefore he had a 33.333...% of chosing the right one...
now, after the participant has opened one of the empty boxes, if he changes his choice... then he has to chose 1 out of 2 boxes, and has a 50% chance to get it right...
but if he stays with his old choice, he still has a 33.333% chance of course...


it doesn't remain 33.333% it changes to 50% because there are only 2 boxes. no need to get all complicated on it it's just trivial.

Drake 12-17-2004 04:20 PM

Quote:

Originally posted by ytcracker
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

:1orglaugh

Why wasn't the riddle written like this? 99.9% of us would get it right this way.

ytcracker 12-17-2004 04:22 PM

Quote:

Originally posted by garfield81
it doesn't remain 33.333% it changes to 50% because there are only 2 boxes. no need to get all complicated on it it's just trivial.
sorry you are wrong

i have no faith in humanity anymore


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