Interesting Math Riddle...

[SpaceAce]

This is, like, the oldest math riddle known to mankind. They dug this shit out of the pyramids in Egypt.

If you think you have a 50/50 chance of being right after they open one of the doors, press your ear to the monitor: THE CHANCE THAT YOU PICKED THE CORRECT DOOR THE FIRST TIME DOES NOT GO UP WHEN THEY TAKE THE THIRD DOOR AWAY. IT DOESN'T MATTER IF THEY REMOVE THE DOOR WITH A TACTICAL NUCLEAR STRIKE, YOU ARE STILL BETTER OFF CHANGING YOUR CHOICE.

If there are 1,000,000,000 doors and you select one then they take away 999,999,998 do you really think there is now a 50/50 chance that you picked the right door the first time? Of course not. There is still a one-in-a-billion chance that you picked the right door. There is also a 999,999,999-in-a-billion chance that the other door is the correct door.

SpaceAce

[Fake Nick]

300 ignorant bitches

[SilverTab]

that would be you

[Fake Nick]

Originally posted by SpaceAce

If there are 1,000,000,000 doors and you select one then they take away 999,999,998 do you really think there is now a 50/50 chance that you picked the right door the first time?
SpaceAce

THAT IS WHERE YOU ARE WRONG

[Fake Nick]

the point everyone wants to make only makes sense with more than 3 doors it never does with 3


the maths behind this base the anwser on the fact that it also has to be right with 100 doors and 100 tries


with one try and three doors changing or not changing does not matter

[SilverTab]

Originally posted by Fake Nick
no its not statistisc win by doing it multiple times


66% = 1

33% = 0


it will be either 1 or 0 if you try once >> 50 % says 1 50 % says 0

i'm quoting it again just for fun LOL


how to magically change 66/33 into 50/50 and call other people ignorant bitch LOL

I just hope you are kidding...

[Fake Nick]

Originally posted by SilverTab
i'm quoting it again just for fun LOL


how to magically change 66/33 into 50/50 and call other people ignorant bitch LOL

I just hope you are kidding...

I wil write a thesis on it and try and explain the world that 1 out of 3 is not equal to 1 out of 1 000 000

[ytcracker]

Originally posted by Fake Nick
THAT IS WHERE YOU ARE WRONG

i dont see how anyone could be this bad

i seriously invision you laughing at us trying to explain this to you when you already know the truth

if that is the case i feel punk'd and u rocked me

[SilverTab]

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!



tell me what part y ou don't understand

[Fake Nick]

A /
B /
C /


I say A / it isn't B / therefore there is 66% chance it is C/

is what you folks claim ?

[WarChild]

I liked it better when the equation involved invisible men. It all seemed so much simpler then.

[SilverTab]

Originally posted by Fake Nick
A /
B /
C /


I say A / it isn't B / therefore there is 66% chance it is C/

is what you folks claim ?

you don't understand the problem...

if you say A...for your first guess, and you switch, you will have it, whater the prize is behind B or C...only case you will MISS by switching is if you got it right the first time

[Fake Nick]

Originally posted by SilverTab
you don't understand the problem...

if you say A...for your first guess, and you switch, you will have it, whater the prize is behind B or C...only case you will MISS by switching is if you got it right the first time

yeah and there is 33% chance you get it right


and 66% chance you get it wrong


but changing will change everything and that is not taking in the equation

[SilverTab]

To put it simply:

If your first pick is:

A:
By switching:
You will win if it's behind door B or C
By staying:
You will win if it's behind door A


If your first pick is:

B:
By switching:
You will win if it's behind door A or C
By staying:
You will win if it's behind door B

If your first pick is:

C:
By switching:
You will win if it's behind door A or B
By staying:
You will win if it's behind door C

[SilverTab]

try it on a piece of paper LOL nothing more I can tell you....

it's been explained every possible way....


even tested on a million iterations....


if the probabiliy is 66/33 after a million attempt, it is also 66/33 after 1000 attempt

and believe it or not...ont your FIRST attempt...the probability is still 66/33

there's no way a 50/50 possibility is being transformed into a 66/33 probability on a million attempt...sorry....it's just impossible....

[Fake Nick]

yeah I understand how you put it silvertab no need to do that again over and over , not good for your heart

[SilverTab]

Originally posted by Fake Nick
yeah I understand how you put it silvertab no need to do that again over and over , not good for your heart

indeed...this is obviously going nowhere!

[ModelPerfect]

Originally posted by SilverTab
Ok here's the deal....

NOW EVERYONE WHO DOUBT IT JUST READ THIS AND TRY TO UNDERSTAND

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

Now that does make sense. What I didn't catch the first time I read the problem is that this is NOT random. What I was trying to simulate is a set selection practice BY THE HOST to show the empty box, i.e. the host always opens B, regardless of your selection. The end result would give you 50/50 odds if you stayed or switched.

But after rereading the original problem, it's NOT random, because the host purposely avoids showing C and and always opens another box (in the fixed set practice, nothing happens if you select B). That throws off the randomness of the problem enough to change the odds. The host is effectively giving more information in round 2, because he ONLY eliminates empty boxes.

It's also interesting to note that your odds are also 66.7/33.3 if the host RANDOMLY selects which box to open out of the 2 left. Again, this is because he gives more information to you and thus throws off the randomness. i.e. The difference comes in when he randomly opens C to reveal the cash...you obviously will choose this one.

[alexg]

so how many people here are still not convinced?

[Twe Russ]

[ytcracker]

Originally posted by ModelPerfect
It's also interesting to note that your odds are also 66.7/33.3 if the host RANDOMLY selects which box to open out of the 2 left. Again, this is because he gives more information to you and thus throws off the randomness. i.e. The difference comes in when he randomly opens C to reveal the cash...you obviously will choose this one.

as long as he doesnt select the one that you initially picked and eliminate/reveal it, youre right.

if he showed you the one you picked initially, you don't gain any new information.

[Libertine]

Originally posted by ModelPerfect
It's also interesting to note that your odds are also 66.7/33.3 if the host RANDOMLY selects which box to open out of the 2 left. Again, this is because he gives more information to you and thus throws off the randomness. i.e. The difference comes in when he randomly opens C to reveal the cash...you obviously will choose this one.

Wrong.

If he *randomly* (!!!) selects which box to open of the boxes you didn't pick, and the one he selects happens to be empty, the ones that are left both have a 50% chance of containing the prize.

You see, in the original problem, you don't get any new information, because he will never reveal the box containing the prize, and there is always a box to open which doesn't contain a prize.

However, if a box is randomly selected, there is a 1/3 chance that it's the real one. If the box happens to be empty, that does not magically transfer 1/3 chance to the other box you didn't pick. It's just eliminated as a genuine possibility.


If a genuine possibility gets eliminated, chances get redistributed entirely.


The actual problem is:
X = A (1/3 chance) or Y (2/3 chance)
Y= (B or C)
Y will always contain at least 1 empty box, and an empty box will always be revealed. No genuine possibility will get eliminated by revealing a box.

The problem when used with randomness is:
X = A (1/3) or B (1/3) or C (1/3)
if X = B or C it might get revealed. A genuine possibility is confirmed or eliminated.


Non-random revealing: 1/3 vs 2/3
Random revealing: 1/2 vs 1/2

[Libertine]

Originally posted by ytcracker
as long as he doesnt select the one that you initially picked and eliminate/reveal it, youre right.

if he showed you the one you picked initially, you don't gain any new information.

et tu, ytcracker?

[m4rk4s]

wow
if you believe that you obviously failed math

[Fake Nick]

Originally posted by m4rk4s
wow
if you believe that you obviously failed math

not all math problems can be solved with math that is the lesson we should remember out of this


and now excuse me I am off to see wether my lotto numbers won or not !

[Twe Russ]

Wow you guys still on this crap?

[Libertine]

Originally posted by m4rk4s
wow
if you believe that you obviously failed math

You talkin' to me?

[ModelPerfect]

Originally posted by punkworld
Wrong.

If he *randomly* (!!!) selects which box to open of the boxes you didn't pick, and the one he selects happens to be empty, the ones that are left both have a 50% chance of containing the prize.

You see, in the original problem, you don't get any new information, because he will never reveal the box containing the prize, and there is always a box to open which doesn't contain a prize.

However, if a box is randomly selected, there is a 1/3 chance that it's the real one. If the box happens to be empty, that does not magically transfer 1/3 chance to the other box you didn't pick. It's just eliminated as a genuine possibility.


If a genuine possibility gets eliminated, chances get redistributed entirely.


The actual problem is:
X = A (1/3 chance) or Y (2/3 chance)
Y= (B or C)
Y will always contain at least 1 empty box, and an empty box will always be revealed. No genuine possibility will get eliminated by revealing a box.

The problem when used with randomness is:
X = A (1/3) or B (1/3) or C (1/3)
if X = B or C it might get revealed. A genuine possibility is confirmed or eliminated.


Non-random revealing: 1/3 vs 2/3
Random revealing: 1/2 vs 1/2

If the host randomly chooses which of the 2 boxes to open, then you get tons more information. Here's the possible outcomes, presuming for this example the cash is in C and you always select a different box (if the host selects C that doesn't eliminate it):

(1) You select A, host shows B, you choose C = WIN
(2) You select A, host shows C, you obviously choose C = WIN
(3) You select B, host shows A, you choose C = WIN
(4) You select B, host shows C, you obviously choose C = WIN
(5) You select C, host shows A, you choose B = LOSE
(6) You select C, host shows B, you choose A = LOSE

Your odds of winning by always switching, even though the host RANDOMLY chooses between the 2 remaining boxes = 66.7%
It's because you get the additional information when the host randomly chooses C that alters your odds.

Maybe you just didn't understand my original statement: You aren't randomly selecting the boxes...only the host is randomly selecting which of the 2 remaining boxes to show you in round 2. Sometimes, he will show you exactly which box to choose. Now if the host revealing a box eliminates it, then your odds of winning drop to 33.3% no matter what you do (because instances #2 and 4 will then be loses). So the end result is your odds are never 50% if the host randomly reveals one of the 2 remaining boxes.

[ytcracker]

Originally posted by punkworld
et tu, ytcracker?

what im saying

is lets say the host picks the boxes at random and picks the one you picked and opens it

and its empty

you havent gained any information because here was no guarantee he was picking a known empty

youre going to have to switch anyway

then it becomes a matter of 50/50

maybe i misunderstood

[Libertine]

Originally posted by ModelPerfect
If the host randomly chooses which of the 2 boxes to open, then you get tons more information. Here's the possible outcomes, presuming for this example the cash is in C and you always select a different box (if the host selects C that doesn't eliminate it):

(1) You select A, host shows B, you choose C = WIN
(2) You select A, host shows C, you obviously choose C = WIN
(3) You select B, host shows A, you choose C = WIN
(4) You select B, host shows C, you obviously choose C = WIN
(5) You select C, host shows A, you choose B = LOSE
(6) You select C, host shows B, you choose A = LOSE

Your odds of winning by always switching, even though the host RANDOMLY chooses between the 2 remaining boxes = 66.7%
It's because you get the additional information when the host randomly chooses C that alters your odds.

Maybe you just didn't understand my original statement: You aren't randomly selecting the boxes...only the host is randomly selecting which of the 2 remaining boxes to show you in round 2. Sometimes, he will show you exactly which box to choose. Now if the host revealing a box eliminates it, then your odds of winning drop to 33.3% no matter what you do (because instances #2 and 4 will then be loses). So the end result is your odds are never 50% if the host randomly reveals one of the 2 remaining boxes.

I see what the problem is.

You are assuming that when the host reveals a random box, after that you get to switch no matter if the box that was revealed was the right one or not, and the revealed box is not eliminated.

I am assuming that when the host reveals a random box, if it's the right box you just lose the game, and if it isn't you can choose whether to switch or not, with the revealed box being eliminated.
I am obviously counting your chance on switching only if the box revealed turns out to be the wrong one, because if it's the right one, the game is over and you don't get to switch anymore.
Switching, in this case, does not improve your chances.


Guess the problem here was just some differences in opinion on which definition of the rules of the alternative game to use

[broke]

Still????

[canplayer]

Originally posted by alexg
so how many people here are still not convinced?

Actually alex, the first time you made the pick, each box has a 1/3 chance of having the 10k.

In round 2, your initial pick still had a 33% of having the 10k. HOWEVER, switching DOES NOT increase your odds in round 2. Your odds of winning in round 2, is 50/50 , the second empty box had a 1/3 chance of having the 10k in round 1.

It's really simple and logical. Changing your choice will not alter the end result, and assuming because you are left with 2 options that the other empty box is a greater choice is FALSE. It had a 1/3 chance of having the 10k in round 1. In Round 2, with 2 options, each has a 50/50 split. Your chances of winning are 50/50 in round 2.

Your simple riddle is playing on your mind, it is switching the probability in round 2, but telling you that your initial choice had a 33.33% chance of being right. THAT IS ONLY true when no new information was presented. Both probabilities changed when the host showed the empty box, THEREFORE your initial probability changes as well.

THE END.

[Libertine]

Originally posted by canplayer
Actually alex, the first time you made the pick, each box has a 1/3 chance of having the 10k.

In round 2, your initial pick still had a 33% of having the 10k. HOWEVER, switching DOES NOT increase your odds in round 2. Your odds of winning in round 2, is 50/50 , the second empty box had a 1/3 chance of having the 10k in round 1.

It's really simple and logical. Changing your choice will not alter the end result, and assuming because you are left with 2 options that the other empty box is a greater choice is FALSE. It had a 1/3 chance of having the 10k in round 1. In Round 2, with 2 options, each has a 50/50 split. Your chances of winning are 50/50 in round 2.

Your simple riddle is playing on your mind, it is switching the probability in round 2, but telling you that your initial choice had a 33.33% chance of being right. THAT IS ONLY true when no new information was presented. Both probabilities changed when the host showed the empty box, THEREFORE your initial probability changes as well.

THE END.

Not again

[SilverTab]

Originally posted by punkworld
Not again

just don't bother...

[canplayer]

lol i had to chime in it was hard to resist

[SilverTab]

Originally posted by canplayer
lol i had to chime in it was hard to resist

thing is...the odds are really 66/33...it has been proven like 100 times in this thread alone...

[canplayer]

Originally posted by SilverTab
thing is...the odds are really 66/33...it has been proven like 100 times in this thread alone...

i'm a non believer , lol !!

[canplayer]

well math is telling me I am wrong, I'm still going to start a movement with the rest of my believers on this board.

[SilverTab]

Originally posted by canplayer
i'm a non believer , lol !!

oh I don't really care anymore...we even ran a C program and tried it 500k times...

500k iterations:

$ ./a.out
Switch won 167024 (66%), lost 82927 (33%)
Stay won 83561 (33%), lost 166488 (66%)


I did like 2 graphics and countless explainations about it...if you don't bother to read'em, I really won't try to explain it anymore! ;)

[SilverTab]

Originally posted by canplayer
well math is telling me I am wrong, I'm still going to start a movement with the rest of my believers on this board.

like I said before, this is a known problem...not like we're the first one debating it....it HAS a solution...a fairly simple one if you take the time to try it out yourself.....

like I said earlier, it's just like someone poping in and saying gravity doesn't exist, or 2 + 2 is not = 4......

the odds ARE 2/3....it's been proven mathimatically....it's no philosophical debate...it's not one of thoes "Depends on how you look at it" problems...it's a statistical fact...