A Math Puzzle

[Gogf]

You are given three boxes, and told that under one box lies a $100,000 bill, and under the other two nothing. You are told to choose a box by the person who has placed the bill, who after you have chosen a box, procedeces to eliminate another box, therefore showing that there is no money under the eliminated box. You are now given the option to stay with the box you have picked, or change to another one. What should you do?

I remember this puzzle because of BasketCase's recent thread, but unlike that, this one has a mathematical answer. Is it more likely that you will get the money from the box you chose, or the other? Why?

 

[punkbass2000]

Spoiler :

I can't explain in full, but I think I would choose the box I hadn't already picked. I had a 1/3 chance of picking the "right" one initially. If I did, then I lose. If I picked one of the wrong initially, then the placer has to indicate to me what the other wrong one is, in which case the other one is the "right" one. There is a 2/3 chance of this occurring.

EDIT: Changed to spoiler

 

[Gogf]

Good way to explain it. Correct.

Actually I thought it would have taken people a little longer to figure it out though .

 

[punkbass2000]

Apologies.

EDIT: Check out the Puzzle Quiz #2 thread for more puzzles similar to this (and many that aren't that similar, but are still fun )

 

[newfangle]

There is an article written by some guy, I think his name is John Allen Paul or something in some book he wrote.

Anyways, he proves that you should stick with the book you chose first.

EDIT: The book is called Innumeracy, I'm digging for more info.
EDIT: The author is John Allen Paulos.

 

[Gogf]

There is an article written by some guy, I think his name is John Allen Paul or something in some book he wrote.

Anyways, he proves that you should stick with the book you chose first.

And why is that?

 

[newfangle]

You'd have to read the book. I can't seem to find the individual article on the net. Anyways, propability is far from my speciality, so I could't tell you why.

Aren't I helpful?

 

[Syterion]

That's false. Everybody has heard of that puzzle(no offense, Gogf ), and it has been repeatedly shown to be 2/3 winning when changed, and 1/3 winning when not changed. I can't see how anything could possibly be refuted.

Incidentally, I believe I posted that puzzle in the first puzzle thread.

 

[newfangle]

That's the fallacy Syterion. It is not a 2/3 chance of winning if you don't switch. The problem is significantly more complex than that.

 

[Gogf]

Actually it is. Let me demonstrate:

You choose box A. If it's in box A and you change, you lose. If it's in box B, box C is elimated, and you move to box B. You win. If it's in box C, box B is eliminated, you move to box C, and win. How can that not make sense?

 

[newfangle]

It wouldn't be a famous puzzle if it was that straight forward.

I read the book 5 years ago, but it was something that always stuck in my head. You could either venture to your library if it interests you that much, or ignore me and continue to think you are correct. The choice is yours.

 

[Syterion]

How is it more complex? It's probability. How many events possible that win/total events possible. You have 1/3 chance of picking the right one. This is the only way you could lose when switching. Therefore 2/3 is right, no question.

 

[Syterion]

The reason it is famous is because common sense tells you it is 1/2 when you switch, for you have one correct and one not correct. But this is false, and the incorrectness of most people's intuition makes it a famous puzzle.

 

[Taliesin]

It is demonstrable that it makes no difference if you switch or not.
Assume you pick A. (It makes no difference, the same conditions apply to all three.)
Now where is the payoff?
If A pays off:

B is removed-- switch is not beneficial
C is removed-- switch is not beneficial
​

If B pays off:

C is removed-- switch is beneficial​

If C pays off:

B is removed-- switch is beneficial​

There are four possible cases, and two of them favour a switch. 2/4 = 50% odds. It's a coin toss.

 

[punkbass2000]

Why is whether B or C is picked the relevant factor in the number of outcomes? Like, why is this a combination vs. permutation equation? I pick A, if it is right and I switch, I lose. If it is not right and I switch, I win. That he can pick either B or C when I pick A is not important, AFAICT.

 

[Syterion]

It is demonstrable that it makes no difference if you switch or not.
Assume you pick A. (It makes no difference, the same conditions apply to all three.)
Now where is the payoff?
If A pays off:

B is removed-- switch is not beneficial
C is removed-- switch is not beneficial​

​

If B pays off:

C is removed-- switch is beneficial​

If C pays off:

B is removed-- switch is beneficial​

There are four possible cases, and two of them favour a switch. 2/4 = 50% odds. It's a coin toss.

The bold is where you are wrong. It doesn't matter which was removed, because the decision of the switch is the event, not which is switched to, because the one that is not eliminated has the prize. Look at it this way:

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?

 

[Taliesin]

The bold is where you are wrong. It doesn't matter which was removed, because the decision of the switch is the event, not which is switched to, because the one that is not eliminated has the prize. Look at it this way:

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?

True. Okay, you've convinced me. Actually, I had already realised I was wrong, when I went to get a die to do an empirical study and realised the second die roll when A landed was superfluous.

 

[Perfection]

Ah, the Monty Hall Paradox!

http://en.wikipedia.org/wiki/Monty_Hall_problem

 

[JohnRM]

You are given three boxes, and told that under one box lies a $100,000 bill, and under the other two nothing. You are told to choose a box by the person who has placed the bill, who after you have chosen a box, procedeces to eliminate another box, therefore showing that there is no money under the eliminated box. You are now given the option to stay with the box you have picked, or change to another one. What should you do?

I remember this puzzle because of BasketCase's recent thread, but unlike that, this one has a mathematical answer. Is it more likely that you will get the money from the box you chose, or the other? Why?

I would have to call bullcrap due to the fact that I know that a one-hundred thousand dollar bill does not exist and therefore not participate in this excersize.

 

[Perfection]

I would have to call bullcrap due to the fact that I know that a one-hundred thousand dollar bill does not exist and therefore not participate in this excersize.

Actually they do exist

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