## to switch or have it not matter?

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- phillip1882
**Posts:**145**Joined:**Fri Jun 14, 2013 9:11 pm UTC**Location:**geogia-
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### to switch or have it not matter?

here's a puzzle that's one of my personal favorites. you have two boxes. it is known that one of the boxes has 1/3 the dollar value of the other box. you open a box, see 9 dollars, and are allowed to switch. should you do so? or is it no better?

good luck have fun

- ahammel
- My Little Cabbage
**Posts:**2135**Joined:**Mon Jan 30, 2012 12:46 am UTC**Location:**Vancouver BC-
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### Re: to switch or have it not matter?

Naive answer, which I expect somebody will enjoy telling me why it's wrong:

**Spoiler:**

He/Him/His/Alex

God damn these electric sex pants!

- phlip
- Restorer of Worlds
**Posts:**7573**Joined:**Sat Sep 23, 2006 3:56 am UTC**Location:**Australia-
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### Re: to switch or have it not matter?

We've had this puzzle before... the short version is:

**Spoiler:**

Code: Select all

`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

### Re: to switch or have it not matter?

**Spoiler:**

In the beginning the Universe was created.

This has made a lot of people very angry and been widely regarded as a bad move.

--Douglas Adams

This has made a lot of people very angry and been widely regarded as a bad move.

--Douglas Adams

### Re: to switch or have it not matter?

phlip wrote:We've had this puzzle before... the short version is:Spoiler:

It's possible to construct a distribution such that P((X/3, X)) < 3*P((X, 3X)) for all X > 0, but I guess any such distribution has an undefined/infinite mean.

- phlip
- Restorer of Worlds
**Posts:**7573**Joined:**Sat Sep 23, 2006 3:56 am UTC**Location:**Australia-
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### Re: to switch or have it not matter?

Derek wrote:It's possible to construct a distribution such that P((X/3, X)) < 3*P((X, 3X)) for all X > 0, but I guess any such distribution has an undefined/infinite mean.

Well, sure, the St. Petersburg Paradox is always an option if you want to go into that realm...

Code: Select all

`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

### Re: to switch or have it not matter?

phillip1882 wrote:you have two boxes. it is known that one of the boxes has 1/3 the dollar value of the other box. you open a box, see 9 dollars

This is either inconsistent or it is underdefined. I see that from the fact that it is given that the amounts contained in both boxes are set beforehand, yet you have also given us here the value (presumably representing a constant) in a particular box that "I open", implying choice and/or volition on my part. However, you fail to specifically define how was the box chosen. If it was an act entirely dependent on our own choice (knowing that different people may make their choices differently), then you would not be able to give us the result of that choice specifically, in this case as a dollar amount. Therefore the choice of the box must have been done in some other way, or if there was an element of our own decision making in this step then it would be a part of our answer.

If it was chosen from the flip of a fair coin then it would have the same inconsistency - you wouldn't be able to give us a specific constant dollar amount, only an example amount or a parameter, e.g. n together with the condition that we cannot infer the possibility or impossibility of the other value being 1/3 the revealed value.

If, however, the initial choice was made in a way that I cannot be certain to have been at least fair and impartial,

**Spoiler:**

You've also not explicitly stated that we don't know anything about the value of the second box. It is only implied and I assumed this in the above.

Please clarify or rectify these aspects and then you can have a definitive solution.

### Re: to switch or have it not matter?

I argue you do have a bit of information - specifically, whether or not the money can be divided into thirds. If I open the first box, and the amount in there is $9.01, I can infer that this is the smaller of the two (since there is no way the other box contains 1/3 of this amount), so I always switch.

In the case that the money IS divisible by 3, you have a 1/3 chance that it is the smaller amount and coincidentally divisible by three (since 1/3 of all numbers are divisible by three), and therefore 2/3 chance that it is the larger of the two. In this case, the expectation value of switching is (2/3)*(1/3)+(1/3)*3=$1.22 (relative to a dollar)

So... I guess you still switch? I thought this was going to somehow go somewhere more interesting...

In the case that the money IS divisible by 3, you have a 1/3 chance that it is the smaller amount and coincidentally divisible by three (since 1/3 of all numbers are divisible by three), and therefore 2/3 chance that it is the larger of the two. In this case, the expectation value of switching is (2/3)*(1/3)+(1/3)*3=$1.22 (relative to a dollar)

So... I guess you still switch? I thought this was going to somehow go somewhere more interesting...

### Re: to switch or have it not matter?

So... I guess you still switch? I thought this was going to somehow go somewhere more interesting...

By this I mean my argument, not the general discussion

### Re: to switch or have it not matter?

thefargo wrote:So... I guess you still switch? I thought this was going to somehow go somewhere more interesting...

By this I mean my argument, not the general discussion

It becomes far more interesting (and more similar to the linked thread) if you don't know the value of either envelope prior to opening either. Knowing that one contains $9 and the other is equally likely to be $3 or $27 greatly simplifies the decision tree (though I'm struggling to figure out exactly why).

Borrowing from St. Pete's Paradox... let a fair coin be flipped until tails comes up. Call "k" the number of times heads showed up first. In the envelopes put 3^k and 3^(k+1). When you pick an envelope, you don't know if you have k or k+1, both are equally likely to have occurred. What is your expected prize if you switch? What is the expected prize if you don't switch?

Of course, this is pretty much just re-stating the problem in the other thread with powers of 3 instead of 2.

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