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...

that hurt my head

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

This was posted on Oprano a few weeks ago.

Regards,

Lee

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

Regards,

Alex

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

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

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

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.

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

exactly.

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

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.

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

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)

it's 50% before and after :2 cents:

This is a 100% useless... jk

:winkwink:

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..

this is the complete description..

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?

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:

ah no.. not really

Sorry for the repost :/

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

LOL
now I'm sorry for the repost :1orglaugh

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

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.

This 'riddle' is older them Methusalem himself and is nothing more then statistics. Get back to school :321GFY

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

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.

and who are you?

and what kind of education do you have?

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

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

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

LOL

that could explain why you are wrong

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.

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?

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.

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.

now that gives me an idea to go back in school...:1orglaugh :1orglaugh

haha this is bullshit.

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..

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

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

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!!!

that makes no sense.

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!!

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.