Interesting Math Riddle... - GoFuckYourself.com

alexg · 12-15-2004, 04:50 PM

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

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

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

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

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

why?

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

Steen2 · 12-15-2004, 04:51 PM

that hurt my head

kirupai · 12-15-2004, 04:51 PM

What if he chooses the same box, he still has the 50%

European Lee · 12-15-2004, 04:52 PM

This was posted on Oprano a few weeks ago.

Regards,

Lee

alexg · 12-15-2004, 04:55 PM

Quote:

Originally posted by European Lee
This was posted on Oprano a few weeks ago.

Regards,

Lee

I don't visit that board, and this is GFY...

Regards,

Alex

Manowar · 12-15-2004, 04:59 PM

you got me..... the percentage is higher but i dont know if i'd switch, because its only higher now there are 2 options left

SilverTab · 12-15-2004, 05:00 PM

Quote:

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

Top Jimmy · 12-15-2004, 05:02 PM

Switching doesn't increase his odds, removing one box did.

I wouldn't switch. I always go with my gut instinct.

Antonio · 12-15-2004, 05:04 PM

After one of the boxes is eliminated you are left with 2 boxes, so your chances of getting it right are 50% NO MATTER which box you choose.

xclusive · 12-15-2004, 05:05 PM

There is no reason to switch it's now 50 50 chance either way...

sean416 · 12-15-2004, 05:05 PM

Quote:

Originally posted by Top Jimmy
Switching doesn't increase his odds, removing one box did.

I wouldn't switch. I always go with my gut instinct.

exactly.

He could elect to change his box, then chose the same one and he'd be guessing at 50%.

StuartD · 12-15-2004, 05:07 PM

It's a brain teaser to make you think the odds of it being his original choice is lower than the odds of it being the other box.

it isn't.

Arousal Design · 12-15-2004, 05:07 PM

1 of 3 boxes...

picks one. one's opened that is empty leaving two left.

the one he picked and the one that's left... either of which contain the money. what difference does it make if he switches he's still picking one of two boxes left...

sounds like 50/50 proposition to me? =D

alexg · 12-15-2004, 05:08 PM

Quote:

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

well I also said something like that when I first heard it...
but it's not very accurate... i mean, you already chose it at the beginning, and had a 33.333% chance of winning, and you stay with the same choice.. nothing changed...you still have 33.33%
how can it be different?

SilverTab · 12-15-2004, 05:11 PM

Quote:

Originally posted by alexg
well I also said something like that when I first heard it...
but it's not very accurate... i mean, you already chose it at the beginning, and had a 33.333% chance of winning, and you stay with the same choice.. nothing changed...you still have 33.33%
how can it be different?

you had 33.33% when there was 3 boxes!

now that he opened 1....there are 2...no matter which one you pick you have 50/50!

(unless of course you switched to the one he just opened...then you are stupid)

SetTheWorldonFire · 12-15-2004, 05:15 PM

it's 50% before and after

This is a 100% useless... jk

StuartD · 12-15-2004, 05:16 PM

Quote:

Originally posted by alexg
well I also said something like that when I first heard it...
but it's not very accurate... i mean, you already chose it at the beginning, and had a 33.333% chance of winning, and you stay with the same choice.. nothing changed...you still have 33.33%
how can it be different?

Because one was removed from the equation dumbass

alexg · 12-15-2004, 05:26 PM

Quote:

Originally posted by MaskedMan
Because one was removed from the equation dumbass

dumbass?

I was indeed wrong...

It's not 50%, It's 66.66%!

read here: http://256.com/gray/teasers/#deal

your intuition doesn't always work... this is pure statistics..

alexg · 12-15-2004, 05:28 PM

Quote:

Originally posted by alexg
dumbass?

I was indeed wrong...

It's not 50%, It's 66.66%!

read here: http://256.com/gray/teasers/#deal

your intuition doesn't always work... this is pure statistics..

this is the complete description..

Baal · 12-15-2004, 05:32 PM

It's actually a 66.66666...% chance of the other box winning.

Why? Because the information changed. Essentially, you gave the participant a choice of choosing ONE box (which would be the choice if he stuck with the original box) or choosing TWO boxes, one of which can't win and that one would be revealed as a loser (at which point the participant would switch their choice from the original to the other one).

Run the stats, and switching will always reveal a greater chance of a win.

What would you pick... one box with a 33.3333...% chance of Winning, or TWO boxes, each with a combined chance of 66.66666...% chance of winning?

who · 12-15-2004, 05:33 PM

Quote:

Originally posted by alexg
well I also said something like that when I first heard it...
but it's not very accurate... i mean, you already chose it at the beginning, and had a 33.333% chance of winning, and you stay with the same choice.. nothing changed...you still have 33.33%
how can it be different?

Well, are you also saying that if the gameshow host then proceeded to pile on a million empty boxes onto the table, that his chances have suddenly decreased to 0.000001% and even if he doesn't switch boxes, he's doomed?

YOu silly monkey.

BlueDesignStudios · 12-15-2004, 05:34 PM

Quote:

Originally posted by xclusive
There is no reason to switch it's now 50 50 chance either way...

ah no.. not really

Baal · 12-15-2004, 05:36 PM

Quote:

Originally posted by Baal
It's actually a 66.66666...% chance of the other box winning.

Why? Because the information changed. Essentially, you gave the participant a choice of choosing ONE box (which would be the choice if he stuck with the original box) or choosing TWO boxes, one of which can't win and that one would be revealed as a loser (at which point the participant would switch their choice from the original to the other one).

Run the stats, and switching will always reveal a greater chance of a win.

What would you pick... one box with a 33.3333...% chance of Winning, or TWO boxes, with a combined chance of 66.66666...% chance of winning?

Sorry for the repost :/

alexg · 12-15-2004, 05:37 PM

Quote:

Originally posted by Baal
It's actually a 66.66666...% chance of the other box winning.

Why? Because the information changed. Essentially, you gave the participant a choice of choosing ONE box (which would be the choice if he stuck with the original box) or choosing TWO boxes, one of which can't win and that one would be revealed as a loser (at which point the participant would switch their choice from the original to the other one).

Run the stats, and switching will always reveal a greater chance of a win.

What would you pick... one box with a 33.3333...% chance of Winning, or TWO boxes, each with a combined chance of 66.66666...% chance of winning?

yes... I have corrected myself above that post...

alexg · 12-15-2004, 05:37 PM

Quote:

Originally posted by Baal
Sorry for the repost :/

LOL
now I'm sorry for the repost

Alky · 12-15-2004, 05:43 PM

say the boxes are:
A B C

say i choose A, and box C is removed ... why should i move to B?

I could have easily selected B in the begining and C was removed... Then I'd go back to A using your theory.

My point? same odds.

alexg · 12-15-2004, 05:45 PM

Quote:

Originally posted by Alky
say the boxes are:
A B C

say i choose A, and box C is removed ... why should i move to B?

I could have easily selected B in the begining and C was removed... Then I'd go back to A using your theory.

My point? same odds.

have you read this before posting?
http://256.com/gray/teasers/lets_deal.html

Baal · 12-15-2004, 05:55 PM

Quote:

Originally posted by Alky
say the boxes are:
A B C

say i choose A, and box C is removed ... why should i move to B?

I could have easily selected B in the begining and C was removed... Then I'd go back to A using your theory.

My point? same odds.

Check out "alexg"'s link.

Short answer? In your example, would you rather choose just A, or BOTH B and C, knowing that one of the losing choices (between B and C) would be revealed? In revealing a losing choice, you are gaining information. A still has a 33.333...% chance of winning. I'd go with the other 66.66...% switch, personally.

Nembrionic · 12-15-2004, 06:13 PM

This 'riddle' is older them Methusalem himself and is nothing more then statistics. Get back to school

azguy · 12-15-2004, 07:09 PM

((A+B)/C + ((n)^2 / C-B+A) + 1) * (n)^2 - C - (A + B^2) = A. You're right, he should switch.

sickkittens · 12-15-2004, 07:12 PM

It makes as much sense as the fact that it makes no sense and makes sense at the same time....if that makes any sense.

alexg · 12-16-2004, 04:22 PM

Quote:

Originally posted by Nembrionic
This 'riddle' is older them Methusalem himself and is nothing more then statistics. Get back to school

and who are you?

and what kind of education do you have?

Kevsh · 12-16-2004, 04:24 PM

Quote:

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

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

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

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

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

why?

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

wrong, he has a 50% chance either way - there's only 2 boxes left .

Kevsh · 12-16-2004, 04:25 PM

<---- i swear i didnt see other answers before i posted. honest. really.

alexg · 12-16-2004, 04:26 PM

Quote:

Originally posted by Kevsh
wrong, he has a 50% chance either way - there's only 2 boxes left .

have you read this thread thoroughly before replying? please do so...

alexg · 12-16-2004, 04:27 PM

Quote:

Originally posted by Kevsh
<---- i swear i didnt see other answers before i posted. honest. really.

LOL

that could explain why you are wrong

12-16-2004, 04:29 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.

Nembrionic · 12-16-2004, 05:58 PM

Quote:

Originally posted by alexg
and who are you?

and what kind of education do you have?

My name is clearly visible on the left.
I have the second highest level of highschool education you can get over here in Holland and did college, computer science, which involved a lot of math.

And?

power182 · 12-16-2004, 07:14 PM

Quote:

Originally posted by alexg
have you read this thread thoroughly before replying? please do so...

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

power182 · 12-16-2004, 07:22 PM

Quote:

Originally posted by alexg
have you read this before posting?
http://256.com/gray/teasers/lets_deal.html

Random is the keyword here. As history shows, game shows are anything but random. When making a choice one must calculate not only the odds but the enviroment as well; as I said bad example.

reynold · 12-16-2004, 07:38 PM

now that gives me an idea to go back in school...

smit · 12-16-2004, 07:52 PM

haha this is bullshit.

BlueDesignStudios · 12-16-2004, 07:56 PM

I cannot believe that people are still 'debating' this..

though the results appear to go against logic.. when you think about it / look at the solution it's correct..

smit · 12-16-2004, 07:59 PM

the links posted are dead, show me another link with the reasoning behind it

Odin88 · 12-16-2004, 08:03 PM

Stupid. It is obvious from that point in time he has a 50% chance regardless, so switching does not increase his probability of winning.

yuvalus · 12-16-2004, 08:13 PM

dude it's 5 am here and i'm back from a party but i can't sleep; still i got an answer for you so-called riddle

I WOULD SWITCH THE BOXES

why?

if you will change to the other box, you will have 2/3 chances! you'll duoble your first chance!

the chance of you choosing the right box when you have 3 boxes is 33%
the chance of you choosing the wrong box out of 3 boxes is 66%

thus, when you have only 2 boxes (your first choice and what is left) if you change your selection your chances of winning are HIGHER!!!

smit · 12-16-2004, 08:16 PM

that makes no sense.

chodadog · 12-16-2004, 08:19 PM

Quote:

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

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

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

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

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

why?

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

I like to call this Retard-o Math. It's a 50 percent change no matter what. There's no reason to change the box. He's just as likely to fuck up either way.

yuvalus · 12-16-2004, 08:25 PM

Quote:

Originally posted by chodadog
I like to call this Retard-o Math. It's a 50 percent change no matter what. There's no reason to change the box. He's just as likely to fuck up either way.

dude don't be an asshole

when you had you first selection there was 33.3% you were right (1 box out of 3)
and 66.6% you were wrong

now, when there are 2 boxes, if you change you slection, ther is 66.6% you are right (100-33.3) and 33.3% you're wrong (100-66.6%)

That's it!!

chodadog · 12-16-2004, 08:29 PM

Let me explain to you what is wrong with this. Let's say it was 100 boxes. Picks a box and then 98 are opened. By your logic, that means he still has a 1 in 100 chance of having the right box if he keeps his selection, and a 1 in 2 chance of winning if he changes his box.

It's not possible because the probability has to sort of add up. You can't have 1 in 2 and 1 in 100 as all the possibilities. You can have 1 in 100 and 99 in 100. You can have 1 in 2 and 1 in 2.

Go get 100 boxes, try it out with a friend and see how many times you get it right and how many times you get it wrong when you perform the test 100 times without changing your choice. You really think you're only going to win 1 time?

You people suck at math.

12-15-2004, 04:50 PM	#1
alexg IL4L.com Join Date: Aug 2003 Location: Israel - ICQ: 162136565 Posts: 11,287	Interesting Math Riddle... Well, not really a riddle, but a rather interesting and not so easy to understand fact.... Let's say there's a show on TV where the participant has to chose 1 out of 3 closed boxes. In one of them there are $10k, and the other 2 are empty. He makes a choice, and before opening the box the host of the show opens one of the two other boxes which is empty (there must be at least one, because a total of 2 boxes out of the 3 are empty). now, should the participant change his choice to the other last box that the host didn't open? or should he stay with his choice? most would think it doesn't matter at all... but the correct answer is that he should switch... why? at the beginning the participant was to chose one of 3 boxes... one of the three contained the cash, therefore he had a 33.333...% of chosing the right one... now, after the participant has opened one of the empty boxes, if he changes his choice... then he has to chose 1 out of 2 boxes, and has a 50% chance to get it right... but if he stays with his old choice, he still has a 33.333% chance of course... __________________ Find fuck buddies in your area! Last edited by alexg; 12-15-2004 at 04:51 PM..

12-15-2004, 04:51 PM	#2
Steen2 Confirmed User Join Date: Feb 2004 Location: Vancouver, Canada Posts: 7,662	that hurt my head __________________ ICQ: 2262.73945

12-15-2004, 04:52 PM	#4
European Lee Confirmed User Join Date: Dec 2002 Location: Daytona Beach Posts: 7,133	This was posted on Oprano a few weeks ago. Regards, Lee __________________ YOUR TRAFFIC IS SAFE WITH US - CONDOM CASH PUTTING THE GAY ADULT INDUSTRY ON YOUR PC!! AFFORDABLE 2257 COMPLIANT GAY + BI CONTENT

12-15-2004, 05:02 PM	#8
Top Jimmy Confirmed User Join Date: Dec 2004 Location: Down your street Posts: 2,513	Switching doesn't increase his odds, removing one box did. I wouldn't switch. I always go with my gut instinct. __________________ Don't reply. Let it die.

12-15-2004, 05:05 PM	#10
xclusive Too lazy to set a custom title Join Date: Apr 2004 Location: Buffalo, NY Posts: 35,218	There is no reason to switch it's now 50 50 chance either way... __________________ I support MediumPimpin.com / Shemp's Outlawtgp.com /

12-15-2004, 04:51 PM	#3
kirupai Confirmed User Join Date: Aug 2003 Location: California Posts: 894	What if he chooses the same box, he still has the 50%

12-15-2004, 04:59 PM	#6
Manowar jellyfish Join Date: Dec 2003 Posts: 71,528	you got me..... the percentage is higher but i dont know if i'd switch, because its only higher now there are 2 options left

12-15-2004, 05:04 PM	#9
Antonio Too lazy to set a custom title Join Date: Oct 2001 Location: Spartaaaaaaaaa Posts: 14,136	After one of the boxes is eliminated you are left with 2 boxes, so your chances of getting it right are 50% NO MATTER which box you choose.

12-15-2004, 05:07 PM	#12
StuartD Sofa King Band Join Date: Jul 2002 Location: Outside the box Posts: 29,903	It's a brain teaser to make you think the odds of it being his original choice is lower than the odds of it being the other box. it isn't. __________________ This is me on facebook This is me on twitter

12-15-2004, 05:07 PM	#13
Arousal Design Confirmed User Join Date: Jun 2002 Location: Connecticut. Posts: 1,358	1 of 3 boxes... picks one. one's opened that is empty leaving two left. the one he picked and the one that's left... either of which contain the money. what difference does it make if he switches he's still picking one of two boxes left... sounds like 50/50 proposition to me? =D __________________ Make Major Bank With Major Personality - NaughtyBank! Featuring NaughtyAllie and NaughtyJulie!

12-15-2004, 05:15 PM	#16
SetTheWorldonFire Confirmed User Industry Role: Join Date: Feb 2002 Location: California Posts: 7,444	it's 50% before and after This is a 100% useless... jk __________________ www.STWOFDesign.com hit me up on icq 154206276 or Skype: JaimeGizzle

12-15-2004, 05:32 PM	#20
Baal Confirmed User Join Date: Jan 2002 Location: SoCal Posts: 696	It's actually a 66.66666...% chance of the other box winning. Why? Because the information changed. Essentially, you gave the participant a choice of choosing ONE box (which would be the choice if he stuck with the original box) or choosing TWO boxes, one of which can't win and that one would be revealed as a loser (at which point the participant would switch their choice from the original to the other one). Run the stats, and switching will always reveal a greater chance of a win. What would you pick... one box with a 33.3333...% chance of Winning, or TWO boxes, each with a combined chance of 66.66666...% chance of winning?

12-15-2004, 05:43 PM	#26
Alky Confirmed User Join Date: Apr 2002 Location: Houston Posts: 5,651	say the boxes are: A B C say i choose A, and box C is removed ... why should i move to B? I could have easily selected B in the begining and C was removed... Then I'd go back to A using your theory. My point? same odds.

12-15-2004, 06:13 PM	#29
Nembrionic Confirmed User Join Date: Feb 2003 Posts: 2,424	This 'riddle' is older them Methusalem himself and is nothing more then statistics. Get back to school

12-15-2004, 07:09 PM	#30
azguy Confirmed User Join Date: Nov 2004 Location: Scottsdale, AZ Posts: 5,167	((A+B)/C + ((n)^2 / C-B+A) + 1) * (n)^2 - C - (A + B^2) = A. You're right, he should switch.

12-15-2004, 07:12 PM	#31
sickkittens I am a meat popsicle. Join Date: Jul 2002 Posts: 25,100	It makes as much sense as the fact that it makes no sense and makes sense at the same time....if that makes any sense. __________________ HIGHEST PAYOUTS FOR NO-CONSOLE TOURS IN THE ENTIRE INDUSTRY! THIS SIG CAN BE YOURS FOR $200 - ICQ: 78881543

12-16-2004, 04:25 PM	#34
Kevsh Confirmed User Join Date: Dec 2004 Location: TO Posts: 8,619	<---- i swear i didnt see other answers before i posted. honest. really.

12-16-2004, 04:29 PM	#37
buddyjuf Guest Posts: n/a	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.

12-16-2004, 07:38 PM	#41
reynold Too lazy to set a custom title Join Date: Oct 2002 Location: Global Traveler Posts: 51,271	now that gives me an idea to go back in school...

12-16-2004, 07:52 PM	#42
smit Confirmed User Join Date: Nov 2003 Posts: 387	haha this is bullshit. __________________ SIG TOO BIG! Maximum 120x60 button and no more than 3 text lines of DEFAULT SIZE and COLOR. Unless your sig is for a GFY top banner sponsor, then you may use a 624x80 instead of a 120x60.

12-16-2004, 07:56 PM	#43
BlueDesignStudios Confirmed User Industry Role: Join Date: Feb 2003 Location: Australia Posts: 9,492	I cannot believe that people are still 'debating' this.. though the results appear to go against logic.. when you think about it / look at the solution it's correct.. __________________ Blue Design Studios - Adult Design Specialists! Email me for a free quote: [email protected]

12-16-2004, 07:59 PM	#44
smit Confirmed User Join Date: Nov 2003 Posts: 387	the links posted are dead, show me another link with the reasoning behind it __________________ SIG TOO BIG! Maximum 120x60 button and no more than 3 text lines of DEFAULT SIZE and COLOR. Unless your sig is for a GFY top banner sponsor, then you may use a 624x80 instead of a 120x60.

12-16-2004, 08:03 PM	#45
Odin88 Confirmed User Join Date: Jul 2003 Location: au Posts: 3,267	Stupid. It is obvious from that point in time he has a 50% chance regardless, so switching does not increase his probability of winning.

12-16-2004, 08:13 PM	#46
yuvalus Confirmed User Join Date: Oct 2003 Posts: 1,906	dude it's 5 am here and i'm back from a party but i can't sleep; still i got an answer for you so-called riddle I WOULD SWITCH THE BOXES why? if you will change to the other box, you will have 2/3 chances! you'll duoble your first chance! the chance of you choosing the right box when you have 3 boxes is 33% the chance of you choosing the wrong box out of 3 boxes is 66% thus, when you have only 2 boxes (your first choice and what is left) if you change your selection your chances of winning are HIGHER!!! __________________ Mail me: [email protected]

12-16-2004, 08:16 PM	#47
smit Confirmed User Join Date: Nov 2003 Posts: 387	that makes no sense. __________________ SIG TOO BIG! Maximum 120x60 button and no more than 3 text lines of DEFAULT SIZE and COLOR. Unless your sig is for a GFY top banner sponsor, then you may use a 624x80 instead of a 120x60.

12-16-2004, 08:29 PM	#50
chodadog Confirmed User Join Date: Apr 2002 Posts: 9,736	Let me explain to you what is wrong with this. Let's say it was 100 boxes. Picks a box and then 98 are opened. By your logic, that means he still has a 1 in 100 chance of having the right box if he keeps his selection, and a 1 in 2 chance of winning if he changes his box. It's not possible because the probability has to sort of add up. You can't have 1 in 2 and 1 in 100 as all the possibilities. You can have 1 in 100 and 99 in 100. You can have 1 in 2 and 1 in 2. Go get 100 boxes, try it out with a friend and see how many times you get it right and how many times you get it wrong when you perform the test 100 times without changing your choice. You really think you're only going to win 1 time? You people suck at math. __________________ 26 + 6 = 1