Interesting Math Riddle... - Page 3 - GoFuckYourself.com

Drake · 12-17-2004, 02:26 PM

Ok, so what's the real answer?

Edit: 100 math riddles

Doctor Dre · 12-17-2004, 02:27 PM

he still have 50 % ..

Doctor Dre · 12-17-2004, 02:28 PM

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 ...

Doctor Dre · 12-17-2004, 02:30 PM

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

you got 3, you remove 1

2 boxes, 50/50.

sixxxthsense · 12-17-2004, 02:33 PM

Quote:

Originally posted by kirupai
What if he chooses the same box, he still has the 50%

exactly!

dum ass riddle!

SilverTab · 12-17-2004, 02:35 PM

Quote:

Originally posted by Doctor Dre
i don't beleive a word of that crap ...

you got 3, you remove 1

2 boxes, 50/50.

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)

ytcracker · 12-17-2004, 02:36 PM

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

Libertine · 12-17-2004, 02:36 PM

Quote:

Originally posted by chodadog
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.

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?

Libertine · 12-17-2004, 02:38 PM

Quote:

Originally posted by ytcracker
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

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

SilverTab · 12-17-2004, 02:43 PM

Quote:

Originally posted by punkworld
Frustrating to argue against people who somehow fail to see the obvious, isn't it?

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....

ytcracker · 12-17-2004, 02:43 PM

Quote:

Originally posted by punkworld
Frustrating to argue against people who somehow fail to see the obvious, isn't it?

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

BlueDesignStudios · 12-17-2004, 02:43 PM

Quote:

Originally posted by punkworld
Frustrating to argue against people who somehow fail to see the obvious, isn't it?

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

StuartD · 12-17-2004, 02:44 PM

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.

alexg · 12-17-2004, 02:45 PM

Quote:

Originally posted by smit
haha you are an idiot

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

alexg · 12-17-2004, 02:45 PM

Quote:

Originally posted by MaskedMan
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.

omg you still don't get it do you?

galleryseek · 12-17-2004, 02:45 PM

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.

chodadog · 12-17-2004, 02:46 PM

Quote:

Originally posted by punkworld
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?

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.

SilverTab · 12-17-2004, 02:46 PM

Quote:

Originally posted by MaskedMan
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.

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!

alexg · 12-17-2004, 02:46 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

ytcracker · 12-17-2004, 02:48 PM

Quote:

Originally posted by MaskedMan
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.

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

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?

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

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

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!....

12-17-2004, 02:26 PM	#101
Drake Hello world! Industry Role: Join Date: Mar 2003 Posts: 12,508	Ok, so what's the real answer? Edit: 100 math riddles Last edited by Drake; 12-17-2004 at 02:28 PM..

12-17-2004, 02:36 PM	#107
ytcracker stc is the greatest Join Date: Dec 2002 Location: rip sean murray Posts: 12,403	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 __________________ www.ytcracker.com \| www.digitalgangster.com i love you

12-17-2004, 02:44 PM	#113
StuartD Sofa King Band Join Date: Jul 2002 Location: Outside the box Posts: 29,903	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. __________________ This is me on facebook This is me on twitter

12-17-2004, 02:48 PM	#121
Libertine sex dwarf Join Date: May 2002 Posts: 17,860	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). __________________ /(bb\|[^b]{2})/

12-17-2004, 02:51 PM	#123
SilverTab Confirmed User Join Date: Nov 2001 Location: MTL Posts: 5,060	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 __________________ mmm my sig was too big... no more cool animation but hey still! need php? ICQ: 94586959 Last edited by SilverTab; 12-17-2004 at 02:52 PM..

12-17-2004, 02:45 PM	#116
galleryseek Confirmed User Industry Role: Join Date: Mar 2002 Posts: 8,234	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.

12-17-2004, 02:51 PM	#124
ytcracker stc is the greatest Join Date: Dec 2002 Location: rip sean murray Posts: 12,403	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 __________________ www.ytcracker.com \| www.digitalgangster.com i love you

12-17-2004, 02:56 PM	#125
chodadog Confirmed User Join Date: Apr 2002 Posts: 9,736	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. __________________ 26 + 6 = 1

12-17-2004, 02:57 PM	#126
SilverTab Confirmed User Join Date: Nov 2001 Location: MTL Posts: 5,060	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! __________________ mmm my sig was too big... no more cool animation but hey still! need php? ICQ: 94586959

12-17-2004, 03:00 PM	#128
SilverTab Confirmed User Join Date: Nov 2001 Location: MTL Posts: 5,060	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 __________________ mmm my sig was too big... no more cool animation but hey still! need php? ICQ: 94586959

12-17-2004, 03:02 PM	#129
ytcracker stc is the greatest Join Date: Dec 2002 Location: rip sean murray Posts: 12,403	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 __________________ www.ytcracker.com \| www.digitalgangster.com i love you

12-17-2004, 03:03 PM	#130
ytcracker stc is the greatest Join Date: Dec 2002 Location: rip sean murray Posts: 12,403	silvertab your explanation is fucking easy thanks __________________ www.ytcracker.com \| www.digitalgangster.com i love you

12-17-2004, 03:11 PM	#136
Libertine sex dwarf Join Date: May 2002 Posts: 17,860	SilverTab, ytcracker, I applaud your hard work but I fear that it will yield no results __________________ /(bb\|[^b]{2})/

12-17-2004, 03:17 PM	#138
alexg IL4L.com Join Date: Aug 2003 Location: Israel - ICQ: 162136565 Posts: 11,287	let's see how much gfyers really suck at math __________________ Find fuck buddies in your area!

12-17-2004, 03:20 PM	#140
SilverTab Confirmed User Join Date: Nov 2001 Location: MTL Posts: 5,060	This is my last attempt at explaining it __________________ mmm my sig was too big... no more cool animation but hey still! need php? ICQ: 94586959

12-17-2004, 03:32 PM	#144
StuartD Sofa King Band Join Date: Jul 2002 Location: Outside the box Posts: 29,903	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. __________________ This is me on facebook This is me on twitter

12-17-2004, 03:33 PM	#145
SilverTab Confirmed User Join Date: Nov 2001 Location: MTL Posts: 5,060	__________________ mmm my sig was too big... no more cool animation but hey still! need php? ICQ: 94586959

12-17-2004, 03:40 PM	#149
Mr. Mike Confirmed User Join Date: Aug 2004 Posts: 913	Wow, there's gotta be a better way to write this "Riddle". I need some Tylonol __________________ Sig for sale. ICQ: 163-545-054