The two box paradox

[phillip1882]

there are two closed boxes. you are told one box has 3 times more than the other. you pick a box and open it. you see that is has 9 dollars in it. would it be in your benefit to switch? obviously the answer is yes, as you stand to gain more than you lose. now imagine after picking a box, you are not allowed to open it. would it still be in your benefit to switch?

 

argument for:

since you either still gain 3 times more or lose 1/3, the answer should be yes. for example, lets say there are three boxes, you pick the middle box, but don't know how much is inside, you are told that in one of the boxes you didn't pick holds 1/3 the amount and the other 3 times the amount of the box you picked. clearly its beneficial to switch to one of the two boxes.

 

argument against:

since you don't know how much is in each box, you have 50/50 chances of getting what you want. to demonstrate, it would be equally beneficial to switch back to the original box.

 

anyone see a clear way to prove which is better?

[Jway]

I'm thinking outside the box...er two boxes. Since I wasn't told I can't take both boxes, I'm doing just that. I'm picking one, taking the other (really taking both). If I'm told, "but, um, you see, you have to give one back," then I will. Both boxes will be given back, in good time, and I will specifically say to whoever cares. You are not allowed to open either box until I am 18.4 miles away from this destination.

 

IMO, this little exercise wasn't set up that well, but to answer within the integrity of what was being asked: I believe I honestly would be happy with either box I am allowed to open. I may stand to gain more from choosing (and opening and taking contents of) one box over the other, but in my view, if I could play a lottery where I either get some money, or more money, I would be going for "some money" and feel I would win no matter what I chose.

[Hasanuddin]

Hi phillip1882,

 

Doesn’t the question depend on several other things?

 

First, how much fun do you enjoy playing the game. If you’re having a really good time with friends you’d be swayed to draw out the game.

 

Next, how badly do you need the money? To me, either $3 or $27 I don’t really care, neither one of these amounts matter to me at this stage of my life. I'm right in the middle of refinancing... it's going through and freeing up a lot more than 27 dollars a month (PS, everyone, you homeowners, I help you taking advantage of the Obama refinancing deal... it's the "stimulus package," but aimed at normal homeowners who haven;t defaulted or "gotten into trouble," or just mild trouble.) Therefore since the dollar amounts are trivial. I’d take the next move just to satisfy curiosity.

 

However, remembering the recession of the early nineties where I was out of work for over sixteen months my answer would have been different. Picture being down to your last 95 dollar and going to the grocery store to buy $93 food just so I wouldn’t be tempted being frivolous… that was me, and I had no idea how another cent would come my way. Now if I was back in those shoes and I was invited to play the same game, then I would have cautiously taken the first box and been very grateful to you. Back then, if I were to buy rice, beans, spice, and a tiny meat, I could’a stretched that nine dollars out for over two weeks. I might have later wondered if I had picked the box, but I'd be content to be fed.

 

Basically it is a game/question that is viewed and answered differently depending on the perception of the person playing. I believe Maslow’s hierarchy of human needs would help with this question. If you haven’t read them then I’m sure you’ll enjoy them.

[maddog]

there are two closed boxes. you are told one box has 3 times more than the other. you pick a box and open it. you see that is has 9 dollars in it. would it be in your benefit to switch? obviously the answer is yes, as you stand to gain more than you lose. now imagine after picking a box, you are not allowed to open it. would it still be in your benefit to switch?

Told in the first method, you have opened the box and have $9 in your hand. At this

point it equal probability (50/50) that other box has $3 or $27. You have $9 now. This

is a Deal/No Deal mentality. Never exercised on the show would be to stand with the $9.

Of course if you like to wager, go for the other box. No Paradox.

... anyone see a clear way to prove which is better?

This is not something to prove with the information given. Game Theory, yes; rigorous

provable Mathematics, No. :)

maddog

[Erasmus00]

I was in the middle of typing up the interesting history of this paradox, which goes to the heart of frequentist v. bayesian probability when I realized that there was probably a wikipedia article. And there is: Two envelopes problem - Wikipedia, the free encyclopedia

[Boerseun]

Although Schrodinger will have it that the $9 in the envelope is both dead and alive until you open it...:shrug:

[lawcat]

The best thing is not to switch. The odds are the same. First , odds are 50-50, barring additional inputs, between the two boxes. Second, the odds are 50-50 that your pick is either larger, or smaller, between the two boxes. However, in switching the box you are wasting time in decisionmaking and anxiety of expectations. Since odds are even, efficiency says to stick with the original choice, barring additional inputs.

[CraigD]

The best thing is not to switch...

This is correct, but misses the point of the paradox – see the wikipedia article “two envelopes problem” Erasmus linked to for a decent encyclopedic description.

 

In short, the puzzle is to explain the paradox of how the ordinary, elementary, and usually trustworthy mathematical statistical technique of expected value seems to go wrong to produce a paradox in which it appears to instruct you that you should always change your selection. It’s a whodunit mystery, where the perpertrator of the crime is some step in reasoning or assumption.

 

If one knows the exact numeric outcomes [math]V_1,\,V_2,\,\dots\,,V_n[/math] (AKA payoff) and probabilities [math]p_1,\,p_2,\,\dots\,,p_n[/math] of each occurring for an event, the expected value of it is

[math]V_{\mbox{expect}} = V_1P_1+V_2P_2\,\dots\,+V_nP_n[/math].

For example, the expected value of a fair 6-sided die roll is

[math]1\frac16+2\frac16+3\frac16+4\frac16+5\frac16+6\frac16 = \frac{21}6 = 3.5[/math]

which can be confirmed by noting that, after many die roles, the average roll approaches 3.5.

 

The box/envelope question seems to lend itself to this same approach (if it doesn’t, the philosophical underpinnings of mathematical realism are in serious peril! :eek:) – but if stated in the usual way (see the wikipedia article), gives:

[math]V_{\mbox{expect}} = 3x\frac12+\frac{x}3\frac12 = \frac{5x}3[/math]

where [math]x[/math] is the amount in the first box you pick, and [math]\frac{10x}6[/math] is the expected value – what you should, on average, get, if you change you choice to the other box. So you should always change you choice, and change it as many times as you’re allowed, which is wrong, and just plain silly. :( (Note that the wikipedia article’s version states that one envelope has 2 times the payoff of the other, giving [math]V_{\mbox{expect}} = \frac{5x}4[/math], while post #1 has it as 3 times, but this doesn’t change the essence of the paradox)

 

To appreciate the paradox, the reader should try to understand what’s wrong here (trust me and wikipedia – the philosophical underpinnings of mathematical realism are not in jeopardy here ;)). Appreciated well, this paradox build intuitive and formalizable comprehension of what a value is in mathematics, and how they correspond to commonplace events like this problem’s.

 

I’ve what I think’s a pretty good, formally sound yet intuitively strong, explanation of this paradox, but don’t want to spoil it for people new to the paradox, so will wait to see what others come up with. Enjoy, all, and thanks, Phillip, for starting this thread :thumbs_up