Interesting Math Riddle...

**ytcracker** · 12-17-2004, 04:17 PM

Originally posted by psili
I read it, don't understand it, but running the numbers of "always switching" to "never switching" don't lie -- always switching gets you more wins.

hit me up on AIM

"brycec"

i can explain it to you

**SilverTab** · 12-17-2004, 04:18 PM

Originally posted by Fake Nick
thats where you are wrong , this case is about 3 not about 100 or any other number in this case it doesnt matter if you switch , I havent looked at the graph yet , but I do know it wont be correct if doesn't support my claim

listen, if you are THAT convinced it's a 50/50....just try it here

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

you will see the numbers for yourself! ;)

**alexg** · 12-17-2004, 04:19 PM

I was sure that this thread would be on page 10 the day after I started it...

**ytcracker** · 12-17-2004, 04:20 PM

Originally posted by Fake Nick
I grasp it now how they spin it mathematically but in a real life situation it wouldn't matter if you switched

you know that they always pick the one that is empty first , so you know you start out with a 50% chance

not really

ok assume the host is completely honest
hard to believe
but he will offer you the choice to switch regardless

there are 2/3 empty boxes
and 1/3 prize boxes

if you select the prize on your first try
the host has the discretion to reveal EITHER empty box
therefore when you switch you will lose

if you select an EMPTY box, there is only ONE other box that the host can possibly reveal - there is ALWAYS going to be two empty boxes

when you switch on EITHER empty box you will get the prize

2/3 = 66% chance of winning

**Fake Nick** · 12-17-2004, 04:20 PM

Originally posted by SilverTab
listen, if you are THAT convinced it's a 50/50....just try it here

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

you will see the numbers for yourself! ;)

I forgot to mention I dont believe in statistics , I know the statistics would tell you should change

**ytcracker** · 12-17-2004, 04:21 PM

Originally posted by SilverTab
listen, if you are THAT convinced it's a 50/50....just try it here

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

you will see the numbers for yourself! ;)

^^^^

just click the radio buttons and play the game

it will show you

**ytcracker** · 12-17-2004, 04:23 PM

Originally posted by Fake Nick
I forgot to mention I dont believe in statistics , I know the statistics would tell you should change

well frankly this whole problem is based on understanding of mathematics

if it isnt something you are considering, then your answer will be incorrect

this is why when i play blackjack and poker i whip the shit out of your average layman

**Fake Nick** · 12-17-2004, 04:24 PM

Originally posted by ytcracker
not really

ok assume the host is completely honest
hard to believe
but he will offer you the choice to switch regardless

there are 2/3 empty boxes
and 1/3 prize boxes

if you select the prize on your first try
the host has the discretion to reveal EITHER empty box
therefore when you switch you will lose

if you select an EMPTY box, there is only ONE other box that the host can possibly reveal - there is ALWAYS going to be two empty boxes

when you switch on EITHER empty box you will get the prize

2/3 = 66% chance of winning

nope not true ,

it is 50%

I compare this to asci code back in the day 10101010101

1 = you win

0 = you lose

you claim there is 66% chance it will become 1 if you change

but there are only two options 1 or 0 and thus it could be either way since it is ONE case and not hundreds of them

if you would play this prize game 100 times you should indeed switch

if you play the prize game once you have no reason too

**Fake Nick** · 12-17-2004, 04:24 PM

your idea of having 66% chance is based on playing more than once

**SilverTab** · 12-17-2004, 04:25 PM

LOL whatever..it's a proven fact....it has been subject to a lot of studies before

it is easy to prove and you still won't believe it..not much I can do....

**ytcracker** · 12-17-2004, 04:26 PM

Originally posted by Fake Nick
your idea of having 66% chance is based on playing more than once

no

even in an independant testing
66% is accurate
you are considering "luck"

you have to understand representative odds

sure i can hit on 20 and pull a miracle ace out of the deck to hit 21

it did work that one time

but that doesnt mean odds are 100% of pulling that ace

**Fake Nick** · 12-17-2004, 04:26 PM

Originally posted by SilverTab
LOL whatever..it's a proven fact....it has been subject to a lot of studies before

it is easy to prove and you still won't believe it..not much I can do....

they dont base it on playing once ,

think I never been in this discussion before ? ? think again

95% of the people always dissagree with me and say I should learn math 5% doenst have a clue

**SilverTab** · 12-17-2004, 04:28 PM

Originally posted by Fake Nick
they dont base it on playing once ,

think I never been in this discussion before ? ? think again

95% of the people always dissagree with me and say I should learn math 5% doenst have a clue

no...it WORKS only on ONE shot....

here's a quote from the other thread:

when picking randomly out of 3 boxes, you ALWAYS have a 1:3 chance of winning = which means you are going to be wrong 2:3. Eliminate 1 of the wrong answers AFTER you have taken, with 66% probability, the other wrong answer - and you have your explanation.

**Fake Nick** · 12-17-2004, 04:28 PM

Originally posted by ytcracker
no

even in an independant testing
66% is accurate
you are considering "luck"

you have to understand representative odds

sure i can hit on 20 and pull a miracle ace out of the deck to hit 21

it did work that one time

but that doesnt mean odds are 100% of pulling that ace

how can you have numbers for ONE single experiment ?

you are giving numbers on this experiment done more than once

**Fake Nick** · 12-17-2004, 04:30 PM

Originally posted by SilverTab
no...it WORKS only on ONE shot....

here's a quote from the other thread:

when picking randomly out of 3 boxes, you ALWAYS have a 1:3 chance of winning = which means you are going to be wrong 2:3. Eliminate 1 of the wrong answers AFTER you have taken, with 66% probability, the other wrong answer - and you have your explanation.

I wouldnt change and thats the end of it

**Fake Nick** · 12-17-2004, 04:31 PM

either way me and math dont get along

the exception confirmes the rule

they tried to teach me this in school and even then I strongly dissagreed with that , but appearantly all people consider this to be the truth and I think its crap

dunno if I translated it correctly

**alexg** · 12-17-2004, 04:31 PM

Originally posted by Fake Nick
they dont base it on playing once ,

yes they do..

it's easy...

if you chose not to switch.. you have 1/3 chance of winning..
if you switch, you only LOSE if you chose the curtain with the prize behind it initially, thus you have a 1/3 chance of losing, and a 2/3 chance of winning...

**X37375787** · 12-17-2004, 04:33 PM

I ran the following c-program off this page .

It iterates 100,000 times through the random possibilites and here's the result.

Code:

$ ./a.out 
Switch won 33217 (66%), lost 16723 (33%)
Stay won 16904 (33%), lost 33156 (66%)

**ytcracker** · 12-17-2004, 04:34 PM

Originally posted by Fake Nick
how can you have numbers for ONE single experiment ?

you are giving numbers on this experiment done more than once

such is the nature of probability and mathematics
this is the kind of logic you are using

lets say i picked 6 lottery numbers
your logic is saying that there are two outcomes = win or lose
therefore i can either do one or the other
win or lose 50/50

i think get what you are saying - in one shot you could get lucky and beat the odds - but its the *illusion* of 50/50

the actual mathematical probabilityis 66%
it is irrefutable

**SilverTab** · 12-17-2004, 04:35 PM

Originally posted by Fake Nick
I wouldnt change and thats the end of it

ok so you wouldn't change huh?

Let's say the prize is behind door B

On your first guess...you pick door A....
Host opens the door C...he ask you if you want to stay or switch

you STAY with A...you lose (you should've switch)

Let's say it's behind door C

On your first guess, you pick A again...Hosts open door B (knowing it's behind door C) he ask you if you want to stay or switch....

You stay with A, you LOSE....

in BOTH CASE you would've win by switching!
Only time you would win is if it's behind door A (your first pick) sooooo 1/3 of winning by staying... 2/3 of winning by switching!

Repeat the process by assuming you take B as your first pick...same thing will happen! How can you fail to see this!

**Libertine** · 12-17-2004, 04:35 PM

Originally posted by Fake Nick
either way me and math dont get along

they tried to teach me this in school and even then I strongly dissagreed with that , but appearantly all people consider this to be the truth and I think its crap

dunno if I translated it correctly

Laat me raden... ontsnapte vmbo'er?

**SilverTab** · 12-17-2004, 04:36 PM

Originally posted by Equinox
I ran the following c-program off this page .

It iterates 100,000 times through the random possibilites and here's the result.

Code:

$ ./a.out Switch won 33217 (66%), lost 16723 (33%) Stay won 16904 (33%), lost 33156 (66%)

here ya go!

it's no fucking philosophical debate LOL

it's a proven FACT....you have MORE chance of winning by switching! thanks equinox...

**Fake Nick** · 12-17-2004, 04:37 PM

Originally posted by alexg
yes they do..

it's easy...

if you chose not to switch.. you have 1/3 chance of winning..
if you switch, you only LOSE if you chose the curtain with the prize behind it initially, thus you have a 1/3 chance of losing, and a 2/3 chance of winning...

well , I still dont think that is accurate

let us say

A /

B /

C /

I chose C /

>> B / gets eliminated

so now it is down to A / and C /

you are now saying because it wasn't B / and I choose C / automatically the best option and most winning opportunity will always be A / ?

that just doenst make sense to me

you will have a 50 % chance of it being true and a 50% chance of it not being true

luck is an important factor in such a guessing game and should always be considered when calculating your odds

I cant understand how in ONCE case I would be better of chosing A / because it isn't B / and I choose C / initially

it would be if you dont count in luck , but who is the fool who does'nt count in luck in a guessing game ? ?

**alexg** · 12-17-2004, 04:37 PM

Originally posted by Fake Nick
I wouldnt change and thats the end of it

It is legitimate not to switch if you believe in god and you believe he has guided you to chose the correct curtain, or if you believe you can judge by the expression on the host's face or by his voice which curtain contains the prize (maybe if you have the appropriate education/skills then you can do this with a small error margin).

In my opinion these kinds of judgements are legitimate, but if you believe that the experiment is IDEAL and the results only depend on theoretical statistics, then you should definitely switch. there's no doubt about it...

**X37375787** · 12-17-2004, 04:38 PM

Originally posted by SilverTab
here ya go!

it's no fucking philosophical debate LOL

it's a proven FACT....you have MORE chance of winning by switching! thanks equinox...

No problem. Wasn't sure whether anyone had ran the program before, but I just did for kicks. And although it defies my logic, it truely seems to work out that way.

Here's the code, and it's no cheating involved.

**edit** GFY's gay censorship fucks up the code...

**X37375787** · 12-17-2004, 04:39 PM

With 300k tries:

$ ./a.out
Switch won 100000 (66%), lost 49742 (33%)
Stay won 50083 (33%), lost 100175 (66%)

**ytcracker** · 12-17-2004, 04:40 PM

Originally posted by Fake Nick
it would be if you dont count in luck , but who is the fool who does'nt count in luck in a guessing game ? ?

there is no such thing as "luck"

there is math
thats it

gambling is all math

**SilverTab** · 12-17-2004, 04:40 PM

Originally posted by Equinox
With 300k tries:

$ ./a.out
Switch won 100000 (66%), lost 49742 (33%)
Stay won 50083 (33%), lost 100175 (66%)

it's simple LOL

**X37375787** · 12-17-2004, 04:41 PM

500k iterations:

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

Any more questions?

**SilverTab** · 12-17-2004, 04:41 PM

Originally posted by ytcracker
there is no such thing as "luck"

there is math
thats it

gambling is all math

people fail to see that this is no meta-physical debate LOL

it's pure statistics!...

**X37375787** · 12-17-2004, 04:42 PM

1 Miiiiiiiiiiiiiiiiiiiiiilllliiiiiiiiiiiiiiiiion iterations:

$ ./a.out
Switch won 333294 (66%), lost 165767 (33%)
Stay won 167252 (33%), lost 333687 (66%)

**Fake Nick** · 12-17-2004, 04:43 PM

Originally posted by ytcracker
there is no such thing as "luck"

there is math
thats it

gambling is all math

no its not , there is not math involved in playing once on the lottery

the 66% chance of winning when changing is based on many many tries not on ONE try

on one try it will always be 50% chance of winning

**ytcracker** · 12-17-2004, 04:43 PM

Originally posted by SilverTab
people fail to see that this is no meta-physical debate LOL

it's pure statistics!...

if u do vegas
we can MIT hack the city while the 50/50s lose their money on sucker bets

**X37375787** · 12-17-2004, 04:44 PM

10 Million iterations, and still no 50/50 to be found.

$ ./a.out
Switch won 3334441 (66%), lost 1665671 (33%)
Stay won 1668679 (33%), lost 3331209 (66%)

**Fake Nick** · 12-17-2004, 04:45 PM

you are now saying because it wasn't B / and I choose C / automatically the best option and most winning opportunity will always be A / ?

this is where it all comes down too ,

you have 50% chance of your 66% rule being in effect and 50% chance of it not being in effect on your ONE try thus it makes no difference to change or not on ONE try

thats the math behind my reasoning and I know it is correct

**X37375787** · 12-17-2004, 04:46 PM

This is not goddamn math, it's stats.

If some statistical value is true for 10,000,000 tries, it is also valid for one try of the same experiment, Fake Nick. Get over it, please. I know it's hard to grasp, just get over it.

**ytcracker** · 12-17-2004, 04:46 PM

Originally posted by Fake Nick
this is where it all comes down too ,

you have 50% chance of your 66% rule being in effect and 50% chance of it not being in effect on your ONE try thus it makes no difference to change or not on ONE try

thats the math behind my reasoning and I know it is correct

nick
with that logic

i pick 6 lottery numbers
your logic is saying that there are two outcomes = win or lose
therefore i can either do one or the other
win or lose 50/50

i think get what you are saying - in one shot you could get ***lucky*** and beat the odds - but its the *illusion* of 50/50

the actual mathematical probability is 66%
it is irrefutable

**Libertine** · 12-17-2004, 04:49 PM

Originally posted by Fake Nick
no its not , there is not math involved in playing once on the lottery

the 66% chance of winning when changing is based on many many tries not on ONE try

on one try it will always be 50% chance of winning

Originally posted by Fake Nick
this is where it all comes down too ,

you have 50% chance of your 66% rule being in effect and 50% chance of it not being in effect on your ONE try thus it makes no difference to change or not on ONE try

thats the math behind my reasoning and I know it is correct

Please tell me this is some sort of sick joke and you don't actually believe this

**SilverTab** · 12-17-2004, 04:49 PM

Originally posted by Fake Nick

luck is an important factor in such a guessing game and should always be considered when calculating your odds

ok this is exactly like me telling you:
When you roll a regular dice, you have 1/6 chances of getting a 6

and you replying:
no, I have more chance, because 6 is my lucky number

**alexg** · 12-17-2004, 04:49 PM

Originally posted by Fake Nick
well , I still dont think that is accurate

let us say

A /

B /

C /

I chose C /

>> B / gets eliminated

so now it is down to A / and C /

you are now saying because it wasn't B / and I choose C / automatically the best option and most winning opportunity will always be A / ?

that just doenst make sense to me

you will have a 50 % chance of it being true and a 50% chance of it not being true

luck is an important factor in such a guessing game and should always be considered when calculating your odds

I cant understand how in ONCE case I would be better of chosing A / because it isn't B / and I choose C / initially

it would be if you dont count in luck , but who is the fool who does'nt count in luck in a guessing game ? ?

you are counting in luck, that's why there is still 33.333% that you will not win..

you agree that your chance increases when you do multiple tries..then you also agree that your chance is greater when doing it once... it is the theoretical CHANCE of winning...
of course you can see a better scale of results when you do this a million times

**Fake Nick** · 12-17-2004, 04:50 PM

Originally posted by ytcracker
nick
with that logic

not at all your 6 numbers are 6 times 1 out of 42 you have a lucky number , there are many many more possiblities

here there are two ,

on this once case the 66% will win or the 33% will win

even thoug that the statistics will always say 66% will win ,

in one case they are equal

**ytcracker** · 12-17-2004, 04:51 PM

i think youre just not understanding the theories behind the mathematics

if i was going to argue with a heart surgeon about how to operate on a patient i would be arguing out of total ignorance

even if i had some kind of opinion
i dont know what the fuck im talking about

**X37375787** · 12-17-2004, 04:52 PM

Should I run 100,000,000 iterations of the same program to prove the point?

Why are people so stubborn? Probability theory is not that hard, really. Especially not when it's been proven, black on white, how the outcome will be - the probabilities are set.

You can toss a coin 10 times, and the probability distribution may just be 30% heads, 70% tails. But with a significant number of tries, you will get closer and closer to the 50/50 distribution for the experiment.

**SilverTab** · 12-17-2004, 04:53 PM

Originally posted by ytcracker
i think youre just not understanding the theories behind the mathematics

if i was going to argue with a heart surgeon about how to operate on a patient i would be arguing out of total ignorance

even if i had some kind of opinion
i dont know what the fuck im talking about

exactly....

thing is...it doesn't take a brain surgeon to see that one LOL

when we proved it with countless graphs and examples....

**X37375787** · 12-17-2004, 04:53 PM

And "luck" certainly ain't no variable in probability theory. At least not the last time I checked.

**Fake Nick** · 12-17-2004, 04:53 PM

Originally posted by ytcracker
i think youre just not understanding the theories behind the mathematics

if i was going to argue with a heart surgeon about how to operate on a patient i would be arguing out of total ignorance

even if i had some kind of opinion
i dont know what the fuck im talking about

you are not understanding the fact that statistics are never based on one case

**Fake Nick** · 12-17-2004, 04:55 PM

if you play this game once you have absolutely no reason to change, the statistics never help you when playing once

**ytcracker** · 12-17-2004, 04:55 PM

Originally posted by Fake Nick
not at all your 6 numbers are 6 times 1 out of 42 you have a lucky number , there are many many more possiblities

here there are two ,

on this once case the 66% will win or the 33% will win

even thoug that the statistics will always say 66% will win ,

in one case they are equal

man nick you cannot possibly be this ignorant

thats like saying
theres a 5% chance for it to snow today and a 95% chance it wont

however it will either snow or it wont so 50% 50%

sure it might snow
but the chance of it snowing is 5%
thats the math
thats the point

**eGawd** · 12-17-2004, 04:55 PM

Let me explain it to you:

If the host says "STC is the Greatest" you have 2 choices. If you choes to say 'Yes' you will have 100% chance of dying of natural causes while choosing 'No' leaves you with the obvious alternative. Not responding is stupid like imaginary numbers.

**Fake Nick** · 12-17-2004, 04:56 PM

Originally posted by ytcracker
man nick you cannot possibly be this ignorant

thats like saying
theres a 5% chance for it to snow today and a 95% chance it wont

however it will either snow or it wont so 50% 50%

sure it might snow
but the chance of it snowing is 5%
thats the math
thats the point

that is taking it to the extreme,

if you play once statistics never help and I can prove that with stats

Interesting Math Riddle...

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