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-   -   Interesting Math Riddle... (https://gfy.com/showthread.php?t=403411)

alexg 12-15-2004 04:50 PM

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...

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 :2 cents:

This is a 100% useless... jk

:winkwink:

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. :2 cents:

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 :1orglaugh

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 :321GFY

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 :321GFY
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

buddyjuf 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.


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