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alexg

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#1

What do you think the answer to that math riddle is?

12-17-2004, 01:12 PM

here: http://www.gofuckyourself.com/showth...0&pagenumber=1

in short, there are 3 curtains. there's a car behind one of them (you don't know which one), and the 2 others are empty.
You chose one..

the host of the show reveals an empty curtain (there must be at least one left, or even 2 if you chose the one with the car behind it), and offers you to switch to the one left.

if you switch, what chance would you have to win?

27 Voters

50% - It doesn't matter if I switch or not...

0%

14 votes

66.666% - I should definitely switch...

0%

11 votes

I have another answer...

0%

1 vote

I have no idea what you're talking about...

0%

1 vote

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#2

12-17-2004, 01:22 PM

You have two things to choose from. Half of a 100% is 50%. Am I missing something here?

Your post count means nothing.
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#3

12-17-2004, 01:24 PM

Originally posted by psili
You have two things to choose from. Half of a 100% is 50%. Am I missing something here?

indeed

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#4

12-17-2004, 01:24 PM

Originally posted by psili
You have two things to choose from. Half of a 100% is 50%. Am I missing something here?

yea

its not two things to choose from

im getting tired of explaining

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#5

12-17-2004, 01:26 PM

And it can't hurt to post this again: http://www.philos.rug.nl/~barteld/monty.ps

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#6

12-17-2004, 01:28 PM

Originally posted by psili
You have two things to choose from. Half of a 100% is 50%. Am I missing something here?
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#7

12-17-2004, 02:05 PM

mmmmmm.. I have read the other thread

and

If u have 2 boxes (No matter what has happened before) and 1 is a winner and 1 is not, u have a 50/50 chance of picking the right one.

That can not be changed. Its a fact..... Like black is black and white is white, the Earth is round and 2 comes after 1.

If I hold up 2 fists and 1 has a dollar in it and 1 dosnt u have a 50/50 chance of getting getting the dollar

Nothing that has happened before can change that.

If I toss a coin and it lands on heads a million times, it is still 50/50 that it will land on TIALS when I throw it for the 1,000,001st time.

The past has no influance over the outcome.
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#8

12-17-2004, 02:16 PM

Originally posted by damian2001
mmmmmm.. I have read the other thread

and

If u have 2 boxes (No matter what has happened before) and 1 is a winner and 1 is not, u have a 50/50 chance of picking the right one.

That can not be changed. Its a fact..... Like black is black and white is white, the Earth is round and 2 comes after 1.

If I hold up 2 fists and 1 has a dollar in it and 1 dosnt u have a 50/50 chance of getting getting the dollar

Nothing that has happened before can change that.

If I toss a coin and it lands on heads a million times, it is still 50/50 that it will land on TIALS when I throw it for the 1,000,001st time.

The past has no influance over the outcome.

With all respect...You sir are an idiot!

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#9

12-17-2004, 02:18 PM

Originally posted by damian2001
mmmmmm.. I have read the other thread

and

If u have 2 boxes (No matter what has happened before) and 1 is a winner and 1 is not, u have a 50/50 chance of picking the right one.

That can not be changed. Its a fact..... Like black is black and white is white, the Earth is round and 2 comes after 1.

If I hold up 2 fists and 1 has a dollar in it and 1 dosnt u have a 50/50 chance of getting getting the dollar

Nothing that has happened before can change that.

If I toss a coin and it lands on heads a million times, it is still 50/50 that it will land on TIALS when I throw it for the 1,000,001st time.

The past has no influance over the outcome.

you simply dont understand the nature of the problem

its not 50/50

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#10

12-17-2004, 02:19 PM

Originally posted by damian2001
mmmmmm.. I have read the other thread

and

If u have 2 boxes (No matter what has happened before) and 1 is a winner and 1 is not, u have a 50/50 chance of picking the right one.

That can not be changed. Its a fact..... Like black is black and white is white, the Earth is round and 2 comes after 1.

If I hold up 2 fists and 1 has a dollar in it and 1 dosnt u have a 50/50 chance of getting getting the dollar

Nothing that has happened before can change that.

If I toss a coin and it lands on heads a million times, it is still 50/50 that it will land on TIALS when I throw it for the 1,000,001st time.

The past has no influance over the outcome.

The people who think otherwise are probably the same ones that search for trends on the roulette history wheel.

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#11

12-17-2004, 02:29 PM

The poll results show what I always assumed was true ... that 50% of GFYers are a little slow (or at least 50% of those who voted in this poll)

If there are 2 curtains and you have a choice at 1 of them, you have a 50% chance.

If you don't understand just re-state the question, but start with 10 boxes instead of 3. If the host removes 8 boxes, according to the logic of the post, you only have a 10% when there's 2 left? Hmmmm.
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#12

12-17-2004, 02:29 PM

Originally posted by uno
The people who think otherwise are probably the same ones that search for trends on the roulette history wheel.

U just said in 1 sentance what it took me to say in about 14

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#13

12-17-2004, 02:34 PM

Originally posted by uno
The people who think otherwise are probably the same ones that search for trends on the roulette history wheel.

you fail to understand something really EASY here....

the host KNOW where the price is...

LOL you can even try a simulation
http://www.cut-the-knot.org/hall.shtml

Damn.....fucking try it yourself and WATCH THE NUMBERS....THIS IS NOT SOME KIND OF GUESS....IT IS A PROVEN FACT....

and an easy one...

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#14

12-17-2004, 02:37 PM

Originally posted by uno
The people who think otherwise are probably the same ones that search for trends on the roulette history wheel.

There is a roulette theory about betting black (or red) that sounds great on paper:

You start by betting $5 on black.
If you win, fine, you pocket $5 profit.
If you lose, you double the bet ($10).

Since it pays out 100% of the original bet, a win on the 2nd try will net you $5
(You bet $5, lost, then $10, won. $15 spend but you won $20 = $5 net)

If you lose the 2nd (and so on) you keep doubling your bet. No matter how long it goes red it will eventually go black and you will $5 no matter what the bet ($5,$10,$20,$40 ... $512, $1024)

A friend had this idea our first trip to Vegas. The thinking was, what are the odds it will go black 9 times in a row? (Which would have meant you gambled $2555 total so far)

Well, mathematically, 1/512 chance. Fairly small right? Wrong. As the above post points out, past history has no effect so really its 50/50 each time (1/2).

As my friend found out $2555 later - and imagine having that much on a roulette wheel? If he last the next bet would have been $2560 (and over $5k total) which he didn't have.

We won, luckily, but sh1t his pants along the way. Lesson learned.
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#15

12-17-2004, 02:39 PM

You phrased that question like shit.

If you have N-1 chances to select from N boxes you have (N-1)/N odds of selecting the correct box. Your odds on that itteration are 50/50. Over all still (N-1)/N
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#16

12-17-2004, 02:39 PM

Originally posted by Kevsh

If you don't understand just re-state the question, but start with 10 boxes instead of 3. If the host removes 8 boxes, according to the logic of the post, you only have a 10% when there's 2 left? Hmmmm.

ok let's do it with 10 fucking boxes LOL

The prize is behind door #5
so you pick let's say.... box #2

the host open 8 boxes...leaving the #5 (with the prize) , the number 7 (a random one) and your guess, box #2.....

you think you have more chances by staying with number 2 than switching to number 5 ?? LOL

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#17

12-17-2004, 02:41 PM

you 50/50 guys are just ignorant

seriously

i can chat with you guys one on one and break you off

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#18

12-17-2004, 02:43 PM

i cant remember the name of the theory
some english guy who died before it was ever even looked at kindly
this shit was on the discovery channel
you should always switch you answer in that situation for the best odds
if you do
every time
in the long run you will always be in a situation to make more than lost
since it bumps the percentage up above 50% in the long run of course

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#19

12-17-2004, 02:43 PM

Here's another way of looking at it.

There are a million doors. You choose one of them. Monty opens 999,998 empty doors leaving the one you chose and another one. Do you really think you picked the car on your first try and shouldn't switch.

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#20

12-17-2004, 02:45 PM

Originally posted by Kevsh
There is a roulette theory about betting black (or red) that sounds great on paper:

You start by betting $5 on black.
If you win, fine, you pocket $5 profit.
If you lose, you double the bet ($10).

Since it pays out 100% of the original bet, a win on the 2nd try will net you $5
(You bet $5, lost, then $10, won. $15 spend but you won $20 = $5 net)

If you lose the 2nd (and so on) you keep doubling your bet. No matter how long it goes red it will eventually go black and you will $5 no matter what the bet ($5,$10,$20,$40 ... $512, $1024)

A friend had this idea our first trip to Vegas. The thinking was, what are the odds it will go black 9 times in a row? (Which would have meant you gambled $2555 total so far)

Well, mathematically, 1/512 chance. Fairly small right? Wrong. As the above post points out, past history has no effect so really its 50/50 each time (1/2).

As my friend found out $2555 later - and imagine having that much on a roulette wheel? If he last the next bet would have been $2560 (and over $5k total) which he didn't have.

We won, luckily, but sh1t his pants along the way. Lesson learned.

thats why casinos have a max bet on every table
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#21

12-17-2004, 02:45 PM

Originally posted by Kevsh
There is a roulette theory about betting black (or red) that sounds great on paper:

You start by betting $5 on black.
If you win, fine, you pocket $5 profit.
If you lose, you double the bet ($10).

Since it pays out 100% of the original bet, a win on the 2nd try will net you $5
(You bet $5, lost, then $10, won. $15 spend but you won $20 = $5 net)

If you lose the 2nd (and so on) you keep doubling your bet. No matter how long it goes red it will eventually go black and you will $5 no matter what the bet ($5,$10,$20,$40 ... $512, $1024)

A friend had this idea our first trip to Vegas. The thinking was, what are the odds it will go black 9 times in a row? (Which would have meant you gambled $2555 total so far)

Well, mathematically, 1/512 chance. Fairly small right? Wrong. As the above post points out, past history has no effect so really its 50/50 each time (1/2).

As my friend found out $2555 later - and imagine having that much on a roulette wheel? If he last the next bet would have been $2560 (and over $5k total) which he didn't have.

We won, luckily, but sh1t his pants along the way. Lesson learned.

Your friend's theory was correct (if there were no 0 and 00). That is the reason they have 0 and 00 on the wheel. The house always wins. Black and Red are not 50% bets. They may pay 2:1 but its not a 1:2 bet. Its a 9:19 bet.

Over enough spins (a million+) the # of red hit equals # of black hits, but # red hits + # of black hits != total # of hits.

Last edited by bret; 12-17-2004, 02:46 PM.
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#22

12-17-2004, 02:45 PM

Originally posted by Kevsh
The poll results show what I always assumed was true ... that 50% of GFYers are a little slow (or at least 50% of those who voted in this poll)

If there are 2 curtains and you have a choice at 1 of them, you have a 50% chance.

If you don't understand just re-state the question, but start with 10 boxes instead of 3. If the host removes 8 boxes, according to the logic of the post, you only have a 10% when there's 2 left? Hmmmm.

Yes, let's start with 10 boxes instead of 3. You choose 1, so you have a 10% chance.

Now, what you are saying is that if 8 empty boxes are removed non-randomly - which is completely possible whether you have picked the right box or not - by someone who knows what's in all boxes, your chance magically grows to 50%?

Think about what you're saying dude... imagine there being 3 boxes. You get one, I get two. In one of the boxes is a can of beer, in the others nothing.
If I look inside both my boxes, and toss one of them away, does that raise your chance of having the beer to 50%?

Ofcourse not!

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#23

12-17-2004, 02:46 PM

basicly running off the fact that if you bet one one to begine with your accepting 1:3 odds if you dont change after the first one is exposed you still sticking with the 1:3 odds

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#24

12-17-2004, 02:47 PM

Originally posted by punkworld
Yes, let's start with 10 boxes instead of 3. You choose 1, so you have a 10% chance.

Now, what you are saying is that if 8 empty boxes are removed non-randomly - which is completely possible whether you have picked the right box or not - by someone who knows what's in all boxes, your chance magically grows to 50%?

Think about what you're saying dude... imagine there being 3 boxes. You get one, I get two. In one of the boxes is a can of beer, in the others nothing.
If I look inside both my boxes, and toss one of them away, does that raise your chance of having the beer to 50%?

Ofcourse not!

I think we're trying in vain bro

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#25

12-17-2004, 02:47 PM

Originally posted by Alky
thats why casinos have a max bet on every table

no. that is why roulette wheels have 0 and 00

They have max bets to control losses from cheating.
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#26

12-17-2004, 02:48 PM

Originally posted by SilverTab
ok let's do it with 10 fucking boxes LOL

The prize is behind door #5
so you pick let's say.... box #2

the host open 8 boxes...leaving the #5 (with the prize) , the number 7 (a random one) and your guess, box #2.....

you think you have more chances by staying with number 2 than switching to number 5 ?? LOL

Of course!! Whats to understand??

U have 2 boxes... u know that 1 has a prize and one dosnt.... what r your chances of choosing the correct box?

What else do you know, tell me what u know that I dont? U have NO other info

U just know that the other X amount of boxes dont have the prize!!

Please tell me what u know other than that?
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#27

12-17-2004, 02:50 PM

Originally posted by damian2001
Of course!! Whats to understand??

U have 2 boxes... u know that 1 has a prize and one dosnt.... what r your chances of choosing the correct box?

What else do you know, tell me what u know that I dont? U have NO other info

U just know that the other X amount of boxes dont have the prize!!

Please tell me what u know other than that?

so...logically....

if you have 3 box..... ONE with the prize...

You have more chances of winning by SWITCHING than by staying with your current pick!....

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#28

12-17-2004, 02:51 PM

There's no 50/50 LOL....

if you switch it's a 66% chance of winning

if you stay it's 33% chances of winning!

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#29

12-17-2004, 02:52 PM

ok found more info on it
it was Bayes' theorem

found at http://argyll.epsb.ca/jreed/math9/strand4/4201.htm

The Monty Hall Problem

Suppose you're on a game show, and you're given the choice of three doors: Behind one door is the Grand Prize; behind the others, Booby Prizes. You pick a door, say Door A, and the host, who knows what is behind each door, opens another door, say Door B, revealing a Booby Prize. The host then offers you the opportunity to change your selection to Door C. Should you stick with your original choice or switch? Does it make any difference? (This is similar to the routine on the TV game show Let's Make a Deal, hosted by Monty Hall, hence the name of the problem.)

Assuming that the host always chooses to open a door with a Booby Prize, and would never reveal the Grand Prize, the possibly surprising answer is that you should switch to the third door, which is now twice as likely as your original choice to be hiding the Grand Prize. This problem can be analyzed using Bayes' theorem or trees (see "You're the Expert" at the end of Chapter 6 of Finite Mathematics Applied to the Real World), but here is an intuitive argument. When you chose Door A, the probability that you chose the Grand Prize was 1/3 and the probability that it was behind one of the other doors was 2/3. By showing you which of Doors B and C does not hide the Grand Prize (Door B, say), the host is giving you quite a bit of information about those two doors. The probability is still 2/3 that one of them hides the Grand Prize, but now you know which of the two it would be: Door C. So, the probability is still only 1/3 that the Grand Prize is behind Door A, but 2/3 that it is behind Door C.

If you find this result counterintuitive (and even most mathematicians do), try running the simulation below. Choose a door by clicking on it. The host (your computer) will then open one of the other doors, revealing a pig. You may then, by clicking on the appropriate door, choose to stick with your choice or switch to the remaining door. After a moment the doors will close to allow you to try again. Below the doors are shown two running calculations: the experimental probability that you will win if you stay with your original choice and the experimental probability that you will win if you switch. After many tries, will these numbers be close to 1/2, or will they be close to 1/3 and 2/3 respectively?

here is the link
http://argyll.epsb.ca/jreed/javaMath...tyHallSim.html

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#30

12-17-2004, 02:53 PM

Originally posted by SilverTab
I think we're trying in vain bro

I'm afraid you're right. It's just so frustrating to have to accept that people simply refuse to see the obvious

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#31

12-17-2004, 02:53 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!

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#32

12-17-2004, 02:56 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

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#33

12-17-2004, 02:58 PM

Originally posted by SilverTab
so...logically....

if you have 3 box..... ONE with the prize...

You have more chances of winning by SWITCHING than by staying with your current pick!....

No, If u have 3 boxes then u have a 33.33333.....% chance of getting the correct one

Switch as many times as u like, it will always be a 1/3

If someone takes away an empty box u have 2 boxes.... one has the prize, one dosnt

u r back to 50/50 again
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#34

12-17-2004, 03:00 PM

Originally posted by damian2001
No, If u have 3 boxes then u have a 33.33333.....% chance of getting the correct one

Switch as many times as u like, it will always be a 1/3

If someone takes away an empty box u have 2 boxes.... one has the prize, one dosnt

u r back to 50/50 again

LOL whatever....

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#35

12-17-2004, 03:01 PM

seem like people here can't understand simple english...FUCKING TRY IT ON A PIECE OF PAPER

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

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#36

12-17-2004, 03:01 PM

Originally posted by punkworld
Yes, let's start with 10 boxes instead of 3. You choose 1, so you have a 10% chance.

Now, what you are saying is that if 8 empty boxes are removed non-randomly - which is completely possible whether you have picked the right box or not - by someone who knows what's in all boxes, your chance magically grows to 50%?

Think about what you're saying dude... imagine there being 3 boxes. You get one, I get two. In one of the boxes is a can of beer, in the others nothing.
If I look inside both my boxes, and toss one of them away, does that raise your chance of having the beer to 50%?

Ofcourse not!

Yes, it does. Because 1 of the 2 boxes left has a beer in it (unless you chugged it ) ....

People get a grip and forget the math, that's what confusing everyone. Think in practical terms, not percentages and raised or lowered chances.

Think about if you were in a beauty pageant (a real stretch for most of us, I know). There are 12 contestants left. One by one they are told by the host they didn't win until there are 2 left.

If you and a friend were betting on who would become Ms.Whatever, and did so when there were 12, logic says that each of you, on his own, would have a 1/12 chance of guessing who the winner would be.

Now, there are 2 contestants left. Each of you were brilliant: You each have 1 of the remaining contestants, but of course only 1 of you will win...

What are the odds that you choose the right one?
Would you think it's logical / mathematically / practical to change your pick?
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Kevsh

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#37

12-17-2004, 03:07 PM

Originally posted by SilverTab

you think you have more chances by staying with number 2 than switching to number 5 ?? LOL

No, you have the same chance. After each empty box is removed the odds are recalculated. As stated before, past history has NO effect on the future in such cases.

8 boxes, 2 boxes, 176.7 boxes removed - if there are 2 left there is no gain nor loss in odds by switching your pick.

I think those that believe you should switch are being confused by the math that seems to suggest you should switch. But math sometimes conflicts with logic (and reality) as it has in this case.

Anyway, those that believe still will and those that don't, still won't. This thread hasn't converted anyone from their original position !!!
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psili

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#38

12-17-2004, 03:07 PM

Here's the deal:

We have one group of people who are saying there are ALWAYS three choices and are explaining their point from that view.

We have another group of people who are saying there are only two choices and are explaining their point from their view.

It's like one group sees a cup half full and the other half empty -- your both right, but wrong in the other group's eyes. The fucked up thing is, the argument just keeps going and going and going until each group can accept the other group's way of looking at it and realize everyone was correct and wrong at the same time.

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SilverTab

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#39

12-17-2004, 03:09 PM

I'm not making some fucking random assumption this is a proven FACT...damn

http://members.shaw.ca/ron.blond/TLE...LET/index.html
http://www.cut-the-knot.org/hall.shtml
http://www.ds.unifi.it/VL/VL_EN/games/games6.html
http://www.math.toronto.edu/mathnet/games/monty.html

just google Monty Game Show or something like that...there are a lot of site about it...and a lot of people already spent time on this problem!...

What's fucking next...some random idiots from GFY telling us the Theory of relativity is false? Wake the fuck up already

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SilverTab

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#40

12-17-2004, 03:10 PM

Originally posted by psili
Here's the deal:

We have one group of people who are saying there are ALWAYS three choices and are explaining their point from that view.

We have another group of people who are saying there are only two choices and are explaining their point from their view.

It's like one group sees a cup half full and the other half empty -- your both right, but wrong in the other group's eyes. The fucked up thing is, the argument just keeps going and going and going until each group can accept the other group's way of looking at it and realize everyone was correct and wrong at the same time.

it's not a deal of seeing a cup half empty or half full and both being right

THIS IS SIMPLE STATISTICS AND IS A PROVEN FACT...fuck....now who's gonna come in and tell us that gravity doesn't exist...

that's it i'm out of this fucking thread! too many clueless people!

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Fletch XXX

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#41

12-17-2004, 03:13 PM

silvertab wins

dialectics!

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Damian_Maxcash

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#42

12-17-2004, 03:14 PM

Originally posted by SilverTab
seem like people here can't understand simple english...FUCKING TRY IT ON A PIECE OF PAPER

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

What r u talking about??

There is no ABC..... one has already gone.... he has opened a window and either A, B or C has gone

So the question only has 2 answers...... A or B

What is 3 - 1 ?

U have 3 possible answers as in the origional question.... take away one.... what is left?

2 answers...... 50/50

U have 2 options
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SilverTab

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#43

12-17-2004, 03:15 PM

Originally posted by damian2001
What r u talking about??

There is no ABC..... one has already gone.... he has opened a window and either A, B or C has gone

So the question only has 2 answers...... A or B

What is 3 - 1 ?

U have 3 possible answers as in the origional question.... take away one.... what is left?

2 answers...... 50/50

U have 2 options

sorry...you don't understand the problem

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SilverTab

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#44

12-17-2004, 03:17 PM

Originally posted by damian2001
What r u talking about??

There is no ABC..... one has already gone.... he has opened a window and either A, B or C has gone

So the question only has 2 answers...... A or B

What is 3 - 1 ?

U have 3 possible answers as in the origional question.... take away one.... what is left?

2 answers...... 50/50

U have 2 options

let's say you pick A.....

if you STAY with your choice, you will only win if the car is behind door A...

if you switch, you will win no matter if the car is behind B or C...TRY IT damn it

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SilverTab

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#45

12-17-2004, 03:17 PM

Ok here's the deal....

NOW EVERYONE WHO DOUBT IT JUST READ THIS

We will assume that there is a winning door and that the two remaining doors, A and B, both have goats behind them. There are three options:

1. The contestant first chooses the door with the car behind it. She is then shown either door A or door B, which reveals a goat. If she changes her choice of doors, she loses. If she stays with her original choice, she wins.

2. The contestant first chooses door A. She is then shown door B, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses.

3. The contestant first chooses door B. She is then is shown door A, which has a goat behind it. If she switches to the remaining door, she wins the car. Otherwise, she loses.

Just fucking try it on your OWN.....

If your first pick is A...
The only time you will by NOT SWITCHING is if it's really behind door A...

Now if the the car is behind eighter B or C...you WILL WIN BY SWITCHING...IN BOTH FUCKING CASE

FUUUUUUUUUUUUUUUUUUUUUUUUCK

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psili

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#46

12-17-2004, 03:18 PM

Originally posted by SilverTab
I'm not making some fucking random assumption this is a proven FACT...damn

http://members.shaw.ca/ron.blond/TLE...LET/index.html
http://www.cut-the-knot.org/hall.shtml
http://www.ds.unifi.it/VL/VL_EN/games/games6.html
http://www.math.toronto.edu/mathnet/games/monty.html

just google Monty Game Show or something like that...there are a lot of site about it...and a lot of people already spent time on this problem!...

What's fucking next...some random idiots from GFY telling us the Theory of relativity is false? Wake the fuck up already

Trippy. I played the first link with the Java app. It still makes no sense to me that when the first option is removed it's not a 50/50 anymore, but by the java application it did prove that always switching ended in more wins than not.

So yea, if I'm ever on a game show and I get this message, I'll plug your name for ya. Even though I still only see two choices and not three.

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SilverTab

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#47

12-17-2004, 03:23 PM

Originally posted by psili
Trippy. I played the first link with the Java app. It still makes no sense to me that when the first option is removed it's not a 50/50 anymore, but by the java application it did prove that always switching ended in more wins than not.

So yea, if I'm ever on a game show and I get this message, I'll plug your name for ya. Even though I still only see two choices and not three.

Damn! Finally someone who decided to try it on his own! ;)

Thanks for being reasonable....

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psili

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#48

12-17-2004, 03:27 PM

Originally posted by SilverTab
Damn! Finally someone who decided to try it on his own! ;)

Thanks for being reasonable....

Still makes no sense. But I also don't have a degree in fluid dynamics, but I can usually drink without spilling too.... (huh ?)

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SilverTab

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#49

12-17-2004, 03:31 PM

here's a better place to test it

http://www-2.cs.cmu.edu/~donna/personal/MontyHall/

be sure to check
Monty always pick losing door

it's pretty easy when you think about it psili:

let's say you pick A....the only way you will win by not switching is if it's really behind A....but if it's behind B, Monty will show door C...you switch to B, you win!...if it's behind door C...monty will open door B, you switch to C, you win

So assuming you pick door A...and you don't switch, you will win 1/3 of the time (if it's behind door A)...if you pick A and you SWITCH...you will win, whater it's behind B or C... (2/3 of the time)

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alexg

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#50

12-17-2004, 03:35 PM

Originally posted by SilverTab
here's a better place to test it

http://www-2.cs.cmu.edu/~donna/personal/MontyHall/

be sure to check
Monty always pick losing door

it's pretty easy when you think about it psili:

let's say you pick A....the only way you will win by not switching is if it's really behind A....but if it's behind B, Monty will show door C...you switch to B, you win!...if it's behind door C...monty will open door B, you switch to C, you win

So assuming you pick door A...and you don't switch, you will win 1/3 of the time (if it's behind door A)...if you pick A and you SWITCH...you will win, whater it's behind B or C... (2/3 of the time)

I can't believe you're still trying to explain it to them...

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