A Math Puzzle

[JohnRM]

Actually they do exist

http://en.wikipedia.org/wiki/Large_denomination_bills_in_U.S._currency

There are none of these bills that exist as legal tender.

 

[Perfection]

There are none of these bills that exist as legal tender.

So, they still exist.

Also they could be old Taiwanese dollars

 

[BasketCase]

The following experiment should make clear what's going on with the three boxes:

Shuffle a deck of cards and draw one at random without looking at it. Now have a friend go through the deck and turn face-up 50 cards that are not the Ace of Spades. Your friend now holds card #51, and you hold card #52. Either #51 or #52 must be the Ace. It will almost always turn out to be #51 (the card you're NOT holding) that's the Ace.

Initially, your chance of getting the Ace of Spades was 1 in 52, and the act of turning over a bunch of non-Aces does not change this chance. Hence, go for the box you didn't originally pick.

 

[Jorge]

You have 1/3 chance to pick correctly initially. When you do so, you lose when you switch. When you pick wrong first (2/3 chance), and switch, you win. Simple enough, no?

Wrong. You had 1/3 chance initially. But after opening one box and seeing that there is nothing inside, the problem changes. The chances of having pick wrong first is now condicionated to the fact that one box is known to be empty, and is hence 1/2, not 2/3.

 

[Mise]

I prefer to think of it this way:

Say you have a large number of boxes N. The probability that you have picked the correct box the first time p -> 0 as N -> infinity . If someone then eliminates ALL the other boxes, bar yours and one other, and all those boxes are empty, the prob. that the other box is the correct box -> 1 .

I mean, if someone did that with you, had 100 boxes say, and said for you to pick one, and then revealed all the others except one box, wouldn't that set of alarm bells in your head? Wouldn't that look just a little bit suspicious?

 

[BasketCase]

Somebody go get three boxes and a co-conspirator to do the experiment a hundred times or so, and post the results.

 

[Mise]

Somebody go get three boxes and a co-conspirator to do the experiment a hundred times or so, and post the results.

I tried it with 10 playing cards (find the Queen of Hearts) with my mum over the holidays and it worked every single time (about 25 times or so!). I'm trying to perfect the game so I can make lots of money from it.

 

[Scuffer]

Wrong. You had 1/3 chance initially. But after opening one box and seeing that there is nothing inside, the problem changes. The chances of having pick wrong first is now condicionated to the fact that one box is known to be empty, and is hence 1/2, not 2/3.

You already know that at least one of the other two boxes is empty already, so showing you that one of those boxes is empty changes nothing whatsoever.

When you pick the first box, you know you have a 1/3 chance of being right, and 2/3 of being wrong. It remains the same when the box everyone knew was going to be empty, was shown to be empty.

 

[Jorge]

You already know that at least one of the other two boxes is empty already, so showing you that one of those boxes is empty changes nothing whatsoever.

When you pick the first box, you know you have a 1/3 chance of being right, and 2/3 of being wrong. It remains the same when the box everyone knew was going to be empty, was shown to be empty.

The box that is first opened also had a chance of 1/3 of not being empty at the begining.

Solve this problem: we have 3 boxes (A, B, C), two empty and one with the bill, and 3 persons (1, 2 and 3). Person 1 chose box A, person 2 choses B and person 3 choses C.

At first the three persons have 1/3 of getting the bill, right?

Now we open box C and is empty. Person 3's chances are now 0. And acording to you this paradox would happen:

Person 1 believes he has 1/3 and also that person 2 has 2/3 and wants to swicht.

Person 2 also believes he has 1/3 and that person 1 has 2/3, so he also wants to change.

How is it possible that person 1 has 1/3 (in his mind) but 2/3 in person's 2 mind? Your logic is wrong.

To prove it, you can make the experiment. You can repeat the thing 999 times (for example, to make it divisible by 3). Aprox. 333 times, the bill will be in box A, 333 in box B and 333 times in box C (i.e., 1/3 each)

Now you asure that box C is empty. Out of 999 repetitions, only 666 have the box C empty. And out of that 666, 333 times the bill will be in box A and another 333 times it will be in box B (that is, 1/2 for each box).

 

[Jorge]

OK, forget what I wrote, I have just read again the problem and realised that the man that put the bill does not open a random box, but one he already knew was empty.

 

[Mise]

The box that is first opened also had a chance of 1/3 of not being empty at the begining.

The first box opened (not the one you picked) has a 100% chance of NOT containing the money. That was stipulated in the problem description -- the first box opened is ALWAYS empty.

edit: nevermind

 

[Scuffer]

1/3 in the box you chose. 1/3 in each of the others, so 2/3 in the ones you didn't. That doesn't change when you remove one of those two boxes, therefore it must be 2/3 in the remaining box. The reason it never becomes 1/2 is that the removal of the box only ever happens on the boxes you didn't chose, never the one you did.

In the problem you put forward, someone's box is removed, which is never the case in the original problem. I don't know what the solution is, but it is a different problem. Presumably, as you are dividing the box into 3 groups, not 2, there would be no incentive to change.

And yes, I have done it 1000 times (in excel admittedly) and the practice matches the theory. The problem with you analysis above:

To prove it, you can make the experiment. You can repeat the thing 999 times (for example, to make it divisible by 3). Aprox. 333 times, the bill will be in box A, 333 in box B and 333 times in box C (i.e., 1/3 each)

Now you asure that box C is empty. Out of 999 repetitions, only 666 have the box C empty. And out of that 666, 333 times the bill will be in box A and another 333 times it will be in box B (that is, 1/2 for each box).

is that you are disregarding all the times C is full. If C is full, then you should discard B instead. In that instance, 333 times A is full, and 666 times B or C is full. In other words 1/3 A, 2/3 not A.

edit] Last time I had this discussion, it was a drunken 3am outside a chipshop - this is much more civilised!

 

[Jorge]

The first box opened (not the one you picked) has a 100% chance of NOT containing the money. That was stipulated in the problem description -- the first box opened is ALWAYS empty.

edit: nevermind

As it's written it is not clear what procedure was taken to chose the first box. If it's random and by chance the box is empty, then the problem is different from the guy chosing a box he already kew for sure that was empty(and in the problem they don't say it very clearly).

 

[Timko]

I read this problem in a book about a year ago (I had already heard it though) and the book also said that the woman with the highest IQ in the world had a maths/problem column and answered this riddle. And then lots and lots of people sent letters in making very patronizing remarks about her being wrong. Obviously she was correct - as we have shown - but the book printed at least a dozen letters, brimming with superiority, mostly from PhD's!

Very funny to read people's letters, when they have an absolute conviction that they are right, but are utterly, utterly wrong.

 

[The Last Conformist]

This one fooled me the first time I came across it (about ten years ago). I still hate it for that.

As someone said, the reason it's famous is that it defeats most people's mathematical intuitions.

 

[Erik Mesoy]

The Curious Incident of the Dog in the Nighttime

covers this. I followed the proof. You should switch. Draw it as a tree diagram if you like.

Here's how to get it past your "mathematical intuition" - this is the way I think, not just a piece of logic, really...

The bill is in a box, I have 1/3 chance of picking the correct box.
If I picked correctly, switching means I lose. (1/3rd chance)
If I picked wrongly, switching means I win. (2/3rds chance)

See how switching gains you the correct box in a majority of cases?

 

[Timko]

The Curious Incident of the Dog in the Nighttime

covers this. I followed the proof. You should switch. Draw it as a tree diagram if you like.

I think that might have been the book I was thinking of.

 

[Rout]

I remember my maths teacher asking us this and we did'nt believe the answer (switch) at first. Apparently people have actually conducted simulated computer experiments with data supporting the 1/3 chance first pick, 2/3 chance switch explanation. So if you don't follow the logic of the maths (like me some days), theres proof anyhoo

 

[stratego]

Hello CFC, good to be back.
By now a lot of us has already agreed on the 1/3, 2/3 ratio.
Well consider this:
Let's say instead of having one player we have three players, and they happen to each choose a different door. After a while, the host eliminates one door so one of the players is out. Would both of the remaining players increase their chance of winning if they switch?

 

[superisis]

Can I put another perspective on the problem (ok, maybe even changing the prerequisits a tad)... if the three boxes look exactly alike and the 100,000 dollar guy picks a remaining box at random and it's empty, one would assume that you should stick with your box. As if you choose box A and:

box A has money in it, the chance of the box picked being empty = 100%

box B has money in it, the chance of the box picked being empty = 50%

if the box is chosen at random and turns out to be empty (it is more likely that the other box not chosen is empty too), thus your chance of winning is higher if you stick with your box. But it don't matter since probablility doesn't exist.