[Guest]

there are two currently closed boxes. it is known that one box has 3 times the amount of the other. you open a box and see 9 dollars in it. is it in your benefit to switch? obviously it should be as you either get 27 dollars or you get 3 dollars, so on average you'll make a profit. now for the weird part. you pick a box, but are not allowed to open it. is it still in your benefit to switch?

[Guest]

edit: nevermind, read it wrong.

Edited June 24, 2009 by James8421

[Guest]

I think you should just keep what you pick. I don't see any advantage with switching when you can't see what you pick first.

Heres the possibilities with a switch:

pick 9$ switch to 27$= 18$+

pick 27$ switch to 9$= 18$-

pick 3$ switch to 9$= 6$+

pick 9$ switch to 3$= 6$-

[Guest]

I think its is always good to switch.......

Though chance for making profit or loss is always same (1/2 for both profit and loss)...........

The amount for profit is 2x, much larger than the amount of loss 2x/3 (considering x is the amount in first box).

[Guest]

james, that's what i would think too, but note that the problem doesn't state an upper limit as to how much money is in each box, that is if you pick 27 its equally likely you'll get 81. to better demonstrate, let's say there are 3 boxes, you pick the middle box, but don't know how much is in it, you are told one of the boxes has 1/3, and the other has 3x as much. would it be in your benefit to switch?

however, in respect to the original problem, since you don't know what's in each box, how can it be in your benefit to switch, that is what would be the difference between switching and initially picking the other box?

[Guest]

If you switch and the new box has higher amount, you win 3x; you gain 2x

but, if you switch and the box has lower amount, you win x/3; you loose 2/3x

Net gain by switching is (2x - 2x/3) = 4/3x which is always positive

... so you should switch

[Guest]

deegee, wouldn't it be equally beneficial to switch back?

[Guest]

Now you dont want to be testing and questioning your luck so much!

Well, if you are switching back, you are starting from a different amount than when you started with the first box.

What i wrote was the gain (expected) in the long run. However, the expected maximum winning amount after the first switch is 8x/3 and not 3x (which is if you are lucky by not switching). The upper bound of expected winning has changed.

And, if you switch back, it would be the same as not switching.

Thats what I think (no concrete proof behind all this). Of course, I may be far from being correct.

[Guest]

deegee, firstly, how are you starting from a different amount, you didn't know the amount in the first place!

secondly, by deciding to switch you have essentially chosen a box.

so, once again you can either gain 3x or lose 1/3 by switching again, it seems to me.

[Guest]

there are two currently closed boxes. it is known that one box has 3 times the amount of the other. you open a box and see 9 dollars in it. is it in your benefit to switch? obviously it should be as you either get 27 dollars or you get 3 dollars, so on average you'll make a profit. now for the weird part. you pick a box, but are not allowed to open it. is it still in your benefit to switch?

I don't see how it would be either advantageous or disadvantageous to switch. The way this is worded it boils down to:

box A contains 3^x (call this the loser) and box B contains 3^x+1 (the winner) dollars.

All other things being equal, the probability of getting box A is 0.5, and box B is 0.5. By showing you the box you choose, i don't see how the that gives you any more information about the boxes, other than the 2 possibilities for x. I can't think of any (off the top of my head) system that can show an x at which it would be more advantagous to switch, because the function is exponential.

Same goes for the second part. Your probability of getting a particular X doesn't change if you switch or not, in this case, ESPECIALLY since there is NO information about the contents.

[Guest]

tpaxatb,

for the first part i would have to disagree. by knowing the contents, you essentially cut the problem into asking, is it a good idea to risk losing 1/3 in order to gain 3x the amount?

for the second part i would tend to agree, i can't see how it would be advantages to switch, since you don't know the contents, but i can't see a clear way to prove this is the case, especially given mine and deegee's earlier arguments.

Edited June 24, 2009 by phillip1882

[Guest]

james, that's what i would think too, but note that the problem doesn't state an upper limit as to how much money is in each box, that is if you pick 27 its equally likely you'll get 81. to better demonstrate, let's say there are 3 boxes, you pick the middle box, but don't know how much is in it, you are told one of the boxes has 1/3, and the other has 3x as much. would it be in your benefit to switch?

however, in respect to the original problem, since you don't know what's in each box, how can it be in your benefit to switch, that is what would be the difference between switching and initially picking the other box?

Oh, I thought we were dealing with just those dollar amounts.

Either way I still don't see any advantage in switching no matter the dollar amounts. Since you can't see what you pick, by switching you are basically second geussing yourself, and it would be the same as if you picked the box you switch for in the first place. Make sense?

[Guest]

tpaxatb,

for the first part i would have to disagree. by knowing the contents, you essentially cut the problem into asking, is it a good idea to risk losing 1/3 in order to gain 3x the amount?

for the second part i would tend to agree, i can't see how it would be advantages to switch, since you don't know the contents, but i can't see a clear way to prove this is the case, especially given mine and deegee's earlier arguments.

Is there a way that you can show your thought processes for part I? In other words, how does knowing the value of the case (and therefore the value of x) help you in any way? Can you explain exactly WHAT the additional information is that you have garnered from being shown the value in the case you have selected?

I cannot see how showing you the value of x differs in any way than not showing you. By that logic (because you know that case(win) holds 3 times case(lose)), you should always pick a case, then switch, regardless of whether you are shown the amount in the case.

Assign lose to x and win to x+1...the set of possible values that determines what is in the case.

From the outset, probability P(win) = P(lose) = .5

If you never switch, your probability of picking P(win) will always be 50%.

If you always switch, you have a 50-50 shot of picking P(win). But, you also had a 50-50 shot of picking P(lose). You'll switch to P(lose) from P(win) 50% of the time. In addition, you swtich to P(win) from P(lose) 50% of the time. This makes the overall odds of getting to P(win) 50%.

Which is why I said it's a coin flip...i don't see the advantage, nor do I see a disadvantage, to switching cases, unless you can show me something that i'm missing here...

[Guest]

Hehe I love these paradoxes.

Its obviously 50-50. The argument being that any logic you apply which makes a switch favourable can be used to switch back. However, looking at what's inside does not change anything, which implies that for the original scenario where you knew you'd picked $9 it would also not matter if you switched or not. Although the OP agrees that it is 50-50 it compares the money that would be won or lost to reason that a switch is favourable. But this CANNOT be done. And here's an explanation why.

Since you picked the box at random the box you chose is 50-50 at being the high one. You see that you picked $9. So there are 2 possibilities:

The boxes contain $3 and $9 OR $9 and $27. But which is it? The big flaw is to assume that it is 50-50 for it being the $3 or the $27 you'll switch to. No probability can be placed on the two possibilities. You cannot take an unknown variable that can take 2 values and assume it to be 50-50 either way. It reminds me of a really silly argument I heard: we do not know if God exists so therefore there is a 50% chance that he does. This logic makes no sense whatsoever.

The only information we know is that there is a 50% chance that we will get the better deal if we switch. However, if you wish to evaluate your chances by multiplying by the money won or lost it can't be done.

[Guest]

psychic mind,

okay let's add coin tossing, to make the problem clearer. you pick a box open it, you see it has mine dollars. now i tell you that i'll flip a coin. if it lands on head i'll put 3 dollars in the other box, tails then 27 dollars. you however cannot see my flip. would it be in your benefit to switch? how would this be any different than the original problem?

now imagine i flip a coin. if it lands on head ill put 9 dollars in one box and 27 in the other. if tails, i'll put 3 dollars in one box and 9 dollars in the other. you pick a box, however are not allowed to look inside. would it be in your benefit to switch?

it seems to me the answer should be no, but once again i don't see a clear way to prove it.

[Guest]

to better understand my difficulty with the second problem, lets instead say i flip x coins.

if the number of heads i get are even, i'll put 3^x in box A, and 3^(x+1) in box B.

if the number of heads i get are odd, i'll reverse the two.

you pick one of the boxes, say B. would it be in your benefit to switch to box A without looking inside?

argument for: since you have 50/50 chances of getting 1/3 or 3x as much then it should be in your benefit to switch.

argument against: it would be equally beneficial to switch back.

[Guest]

psychic mind,

okay let's add coin tossing, to make the problem clearer. you pick a box open it, you see it has mine dollars. now i tell you that i'll flip a coin. if it lands on head i'll put 3 dollars in the other box, tails then 27 dollars. you however cannot see my flip. would it be in your benefit to switch? how would this be any different than the original problem?

now imagine i flip a coin. if it lands on head ill put 9 dollars in one box and 27 in the other. if tails, i'll put 3 dollars in one box and 9 dollars in the other. you pick a box, however are not allowed to look inside. would it be in your benefit to switch?

it seems to me the answer should be no, but once again i don't see a clear way to prove it.

There really is'nt anything to prove. I think of it more like a human nature kind a thing. If you get to look, and are asked to switch, and are told what the outcomes are, yea people will probably switch because like in the OP, on average you should(but not neccesarily) 'win' more than you 'lose'. Now when you don't get to see the box you pick first, that just makes you realize that no matter what, the odds are the same no matter what you do, therefore it would'nt exactly be an advantage to switch. When you are unable to look, your odds are the same either way you look at it. Say the 2 boxes are 9$ and 27$...you have a 50/50 chance at 'winning' and a 50/50 chance at 'losing' right from the start....now if you switch, you still have the same odds of win vs lose. Flipping coins won't change anything really, because in the end you'll still have 2 unknown dollar amounts in 2 boxes of which you have to choose from. Matter of fact, you could just tell the person playing the 2 dollar amounts that are in the boxes, and it will still be a 50/50 either way. Still would'nt see an advantage to switch even after I know what is inside the 2 boxes(without being able to see what one I picked first of course).

I think that no matter what you do it is all the same.

You could switch in both scenarios(being able to look, and not), and you could keep your first pick in both scenarios, and have the same chances of having less or more money either way.

Test it out....

Set up say 10 sets of boxes....grab two people....put one in a different room (so he is unaware of the boxes the first guy picks)....have the first guy go thru and pick his boxes, look in them, then switch.....now have the second guy come in and pick his boxes, look in them, and keep it....Who would you bet on to have more money after 10 rounds? I would'nt bet on either.

Now do the same without letting them see the box they pick...I think it would be the same outcome.

There is one huge uncontrollable factor in this whole problem....what box you end up being stuck with..

By the way you said it would be an advantage to switch every time when you can see the box right? I can make it an advantage not to switch.....Pick a box in your head...now pick the other one and say you don't want to switch.

[Guest]

psychic mind,

okay let's add coin tossing, to make the problem clearer. you pick a box open it, you see it has mine dollars. now i tell you that i'll flip a coin. if it lands on head i'll put 3 dollars in the other box, tails then 27 dollars. you however cannot see my flip. would it be in your benefit to switch? how would this be any different than the original problem?

now imagine i flip a coin. if it lands on head ill put 9 dollars in one box and 27 in the other. if tails, i'll put 3 dollars in one box and 9 dollars in the other. you pick a box, however are not allowed to look inside. would it be in your benefit to switch?

it seems to me the answer should be no, but once again i don't see a clear way to prove it.

You seem to be missing the point. Flipping the coin assumes that it is 50-50 but it might not be. So look at it this way. I set up the boxes so they contain $3 and $9 but you don't know this. You are told that you picked the $9 box so in this example you would always loose by switching, but of course you wouldn't know that. Now if i continued to set it up like this and you continued to pick the $9 first you would always loose. So what I am saying is we dont know what the probability is to switch and win. So with what you know it wouldn't matter what you did.

[Guest]

Well for the second case where you dont know the amount in the first box, I ran a simulation on excel for 65000 tries (using random numbers to determine which box contains what and which is the first chosen box). Turns out that in both cases (whether you switch or not), the expected winning amount is the same.

Edited June 25, 2009 by DeeGee

[Prof. Templeton]

1/2(x) + 1/2(3x) = 2x

where x = the smaller of the two amounts. For either box, at any time.

This reminds me of the CapitolOne Visa commercial. "What's in your wallet" Those that know Gardner will know what I'm talking about.

[plasmid]

Phillip, could you clarify if there are any rules about what values the boxes may contain? Are they restricted to have integer dollar amounts, or are you allowed to have fractions of dollars and even fractions of cents? If you don't have any restrictions, then I propose that if you open the box and find $9, then there is NOT a 50% chance that the other box holds $3 and a 50% chance that the other box holds $27 as you might expect. In fact, there is a 75% chance that the other box holds $3 and a 25% chance that the other box holds $27. I hate to post and run without showing my logic, but hopefully I'll have some time this afternoon.

[Guest]

Phillip, could you clarify if there are any rules about what values the boxes may contain? Are they restricted to have integer dollar amounts, or are you allowed to have fractions of dollars and even fractions of cents? If you don't have any restrictions, then I propose that if you open the box and find $9, then there is NOT a 50% chance that the other box holds $3 and a 50% chance that the other box holds $27 as you might expect. In fact, there is a 75% chance that the other box holds $3 and a 25% chance that the other box holds $27. I hate to post and run without showing my logic, but hopefully I'll have some time this afternoon.

Box A contains 3 times box B. That means that one of the boxes contains 3^x and the other contains 3^x+1 for some value x. (it doesn't matter which is which, the relation itself that one box has 3 times the other box always holds true). If you are shown a particular value in a box, you can bound x by computing x = log(val)/log(3). Therefore the other box either contains 3^x-1 dollars or 3^x+1 dollars.

Example: you are shown ten dollars. That means that x is log(10)/log(3) which is approx 2.0959. That means the other box either contains 10*3 dollars (3^3.0959) or 10/3 dollars (3^1.0959). However, remember that these values were fixed at the start. Regardless of whether the other box contains $3.33 or $30, you have still selected either x or x+1 for some x. If the other box contains $3.33, you have selected box x+1 (x = 1.0959, x+1=2.0959). If the other box contains $30, you have selected box x (x = 2.0959, x+1=3.0959). Remember, it is the relation that counts, not the value of the other box.

Once again:

Call x = lose and x+1 = win. From the outset, your chances of picking win (i.e. picking the box that has x+1) is 50%.

There are four possible outcomes in your scenario:

Pick win and switch -> lose

Pick win and stay ->win

Pick lose and switch -> win

Pick lose and stay -> lose.

The probability of the other box being win when you picked lose is 100%

the probability of the other box being win when you picked win is 0%

the probability of the other box being loose when you picked win is 100%

the probability of the other box being loose when you picked lose is 0%.

In all cases, your probability of winning if you stay is 50%. There are two possible outcomes in this situation, each with a 50% shot.

If you switch, there are two possible outcomes. You have a 50% shot of one of the starting points (win/lose). When you start at win, there is a 100% chance of moving to lose.

When you start at lose, there is a 100% chance of moving to win. Since you have a 50% chance of starting at win or lose, it means that each outcome has an overall chance of 50%.

Put another way:

There are two sets of outcomes: Set a = {x, x-1} and {x, x+1}. You have a 50 percent chance of obtaining either set (picking a box). Regardless of the value of x, there is no more information about the set that has been chosen. In other words, showing you x gives you no information about x-1 or x+1. Since x-1 and x+1 are mutually exclusive, and you have a 50 percent chance of choosing set a over set be, for some x, there is a 50% chance that the other box contains x-1 or x+1.

(I tried to use as little math and as much prose as possible...does that help?)...

edit=changed formatting

Edited June 25, 2009 by tpaxatb

[Scudo]

Now I am new and therefore a "beginner", but it seems to me, with two boxes or three your equally likely to pickup any given amount. The intriguing part is the comment about switching boxes. If I switch, do I notice the weight change? If so I may notice something that others do not. If I switch, may I switch again? If I notice the original box was heavier can I switch back?

I don't see how it would be either advantageous or disadvantageous to switch. The way this is worded it boils down to:

box A contains 3^x (call this the loser) and box B contains 3^x+1 (the winner) dollars.

All other things being equal, the probability of getting box A is 0.5, and box B is 0.5. By showing you the box you choose, i don't see how the that gives you any more information about the boxes, other than the 2 possibilities for x. I can't think of any (off the top of my head) system that can show an x at which it would be more advantagous to switch, because the function is exponential.

Same goes for the second part. Your probability of getting a particular X doesn't change if you switch or not, in this case, ESPECIALLY since there is NO information about the contents.

[Guest]

there are two currently closed boxes. it is known that one box has 3 times the amount of the other. you open a box and see 9 dollars in it. is it in your benefit to switch? obviously it should be as you either get 27 dollars or you get 3 dollars, so on average you'll make a profit. now for the weird part. you pick a box, but are not allowed to open it. is it still in your benefit to switch?

This is quite an interesting puzzle/paradox. I think the question throws you off a bit though with the bit highlighted in bold.

Actually the paradox comes about because "obviously" it shouldn't benefit you at all to switch. You had a 50/50 chance of getting the better box to start with, you still have a 50/50 chance of getting the better box after a switch. The paradox is that, when you try doing the maths, you get an answer that actually says you are better to switch because "on average" you get more.

This is where the fault in thinking comes. A couple of people on this thread have attempted to show it makes no difference with big arguments about how it is a 50/50 chance, but these miss the flaw as well.

Here is the maths problem simplified:

Chance of getting better box when switching is 50/50 - this is obvious and does not need proving.

Expected win = 0.5 x 9 + 0.5 x 1 = 5, which is better than 3, so you should switch.

The error comes about when you say that, because you've seen the 3, this is now simply a probability question of the boxes containing 3 and 9 or 1 and 3. This is not the case.

In this particular instance these are the only two possibilities. In this particular instance you have a 50/50 chance of winning more if you switch, so it makes no difference.

If you want to work out an expected "average" win though, you need to have a general case. This is not a general case any more, as you have looked at a box. If you were to do this a second time and look at the box, you will probably see a different number and you will have a different specific case. In a specific case, an "average" win is nonsensical - you can't win 5, you can only win 1 or 9 if you switch - similarly you can't carry on repeating this until you average 5, as this is not a repeatable event - it is a specific case.

If you return to the general case, we can see that there are an infinite number of equal and opposite cases that are equally as likely. For example, the case where the boxes contain 3 and 9 and you pick the 3, and the case where the boxes contain 3 and 9 and you pick the 9 are equally as likely. In both cases if you switch you win/lose an equal amount of money - i.e. if these two cases happened in sequence and you swapped both times, you would end up in exactly the same situation as if you didn't swap either time.

It's easy to see that any one situation has such an equal and opposite that is equally as likely. As such, it follows that the average win is the same if you swap as if you don't swap.

Now if you were given constraints such as a maximum amount, the amounts need to be integers, etc. then that would change matters and make things more complicated. But without those constraints this is simply a 50/50 situation and you may as well stick with what you have.

[dyalDragon]

in essence, the way the OP is worded brings this down to a human nature problem i think...

if you open the box and see 3 dollars, switch... either way it goes you can't really buy anything with 3 dollars anyway

but if you open the first box and see 3000 dollars, the cost of switching has just gone up a lot in terms of what your options are (i.e. 3000 dollars is much more useful than 1000, but 9000 is even better!)

taking all this into account i think it depends on the amount in the box and the person making the decision, either way you are getting money out of the deal

as far as not being able to open the box, sure the expected gain (over multiple repetitions) is greater than the expected loss, but if you won't know what is in the box until you finish making your decision and can open it then it doesn't really matter to anyone except the person who put the money in the box in the first place

[bonanova]

Your chance of having selected the more valuable box is 50%.

Switch if you like, but don't expect to increase your winnings.

The boxes are known to contain $x and $3x.

The expectation value of a box chosen at random is $2x.

The expectation of switching is an increase or decrease of $x.

Switching therefore is not profitable.

If one box is known to contain $9, the expectation for x is 4.5;

the expectation for the other box is $4.50 or $13.50.

Conclusion:

The two cases of [$3, $9] and [$9, $27] cannot be equally likely.

Why that is so is the subject of another puzzle.