Box Switching:
One of my all-time favorite problems:
I have two boxes. Each has some positive amount of money in it, but I will give you no information about the possible dollar amounts other than the fact that one box has exactly twice the amount of money in it as the other. You randomly select one of the two boxes, open it, and find $100 inside. I now give you the option of keeping the $100 or switching boxes with me and keeping whatever's inside the other box. Which should you choose?
If you'd like a related problem that might help you think about this differently, click below.
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However, knowing I would be suspicious of your willingness to trade, perhaps you are engaging in reverse psychology. But, again, you don't have to offer, so this would be pointless, unless you just like playing head games and don't mind losing the $100. Answer -- don't take the trade.
If you are blinded, then perhaps it is always better to take the exchange. I mean, $100 is not that much, and there is no upper bound on the contents of the boxes. There is a lower bound ($0), so I have more room to move up than room to move down. Giving the limitations on the money supply, the upper bound is not quite infinite, and perhaps you are not that rich, but surely you have a lot more than $100 to offer, and I can afford to risk the $100.
Take the other box.
When you randomly select, your expected return is 1.5x, where x is the amount of money in lesser box. (The total amount of money is x+2x; you get, on average, 50% of that, or 1.5x.)
Once you see the money, you don't know if you've received x or 2x, but it doesn't matter, because the frequency of getting either is .5. So you'll get, on average, .5x or x, or .75x.
If you switch, though, you may get 2x or you may get .5x. On average, it's 1.25x. So, you gain .5x by switching.
After the $100 is revealed, you know that x=100 or x=50. You also know that your expected return is 200+50= $250, divided by two, or $125 if you switch and $100 if you don't.
First scenario: You get $100. Switching means that you have a 50% probability of gaining $100 more (changing a $100 box for a $200 box) and a 50% probability of losing $50 of what you have (exchaning a $100 box for a $50 box.
Second scenario: You don't have any knowledge, but you're still asked to switch, with a 50% chance of getting 2x your current amount of money, and a 50% chance of getting only .5x your money.
In other words, in each case the potential loss is only half the potential gain. Considered from a strictly mathematical standpoint, you should switch, since the possible gain is double the possible loss.
But ultimately, in both scenarios, the solution is the same.: Keep what you have, because in both cases, the fact that the other guy is asking you to switch boxes means he got box with the lesser amount, he knows it, and he's trying to fool you.
Or maybe I just think human beings are naturally cheating, lying bastards.
On the flip side, it makes no sense to always switch as a matter of policy. You've got a 50% chance of getting the "right" box on the first pick, so you can't have a better than 50% chance of getting the "right" box on the second pick. So the EV calculations can't be the right way to think about this.
A sure thing of $100 seems better to me than a 50% chance of only $50. Guess I'm "risk-averse."
You make the first switch, because EV=$125, right? But then the other guy asks you to switch again. But you don't have any information different that before the first switch! So a the second switch has an EV=1.25*($125)!
And so on...
So there's got to be something wrong with the EV calculation.
You stick with the $100 box. This is simply because 1/2 of $100 is $50, while twice $100 is $200. Since United States Currency has a $100 bill and a $50 bill, it is more likely that you're going to get a $50 bill. This does not hold if you open the box and find 5 20s. If you find 5 20s you should switch boxes, because the person likely went to the ATM, and only has 20s, and thus cannot have put $50 in a box.
Let's call the larger amount "Big" and the smaller amount "Small".
Bayes' Rule:
Prob(Big|$100) = Prob($100|Big)*Prob(Big) / [ Prob($100|Big)*Prob(Big) + Prob($100|Small)*Prob(Small) ]
Now, we know that the prior probabilities Prob(Big) and Prob(Small) are each 0.5. But we don't know the conditional probabilities Prob($100|Big) and Prob($100|Small). I think this is where Aaron Bergman's point comes in.
Anyway, this is making my head hurt.
I might have this whole analysis wrong - somebody please tell me.
This is different from the Monty Hall problem, where you get new information when he opens the door you didn't pick.
For the second question, the answer is, do whatever you feel like -- you have no more information about the contents with the box in your hand than you did before, so it doesn't matter.
Of course, this is ignoring the potential psychological aspect, which is why this is theory, eh?
This dilemma was memorably played out in the Princess Bride when the villain asked the hero to choose which glass of wine to drink when of the two glasses had been poisoned. The hero's only safe bet: secretly poison both glasses and spend the previous five years slowly building up an immunity to poison. I like that answer better than the EV analysis.
There are two probabilistics steps in the problem. First, you chose how much to put in the boxes.
When I open the first and find X dollars, all I know is that there is a 50% chance that the expected value of randomly picking a box is 0.75X and a 50% chance that the expected value is 1.5X.
Any amount of switching doesn't change those expected values or the randomness of whether my eventual selection is the high box or the low. (For instance, if you opened the box I didn't choose and it had $100 in it, I would have the exact same information and the decision to switch would be just as inconsequential.)
The only time you should take the choice for sure is if your box had only one penny in it when you opened it because that is the smallest amount of positive value possible and the other box would have to have two pennies.
The princess bride scenario was a true "game". The initiator knows which has the iocaine powder, they both have everything to lose (and supposedly much to gain, although she was not that hot), and the initiator essentially has no active role except the requirement to drink at the end. Of course they can play head games with each other...
This is much simpler. The initiator has the option to make the offer to trade with the chooser. If he wants to win, and knows the contents of the boxes, then he will only offer if it is of benefit to him. If he is blinded, it is again a very different game from the Princess Bride scenario...
"Uniform distribution" just means there's an equal chance of every possible outcome - like rolling a fair die, where you have an 1/6th chance of rolling anything from one through six. A "uniform distribution" to choose any random number of dollars is impossible - there's no such thing as a 1/infinity chance of something happening. It's important here, because the "50/50" chance of getting $200 vs. $50 is only true if the split is equally likely after you get your information - but it's not, because it wasn't equally likely before you got the information. This is the like the screwy behavior you get with the Monty Hall problem.
How about this? If you get an odd number of dollars (or cents, if he allows stray change in the box), then you know you have to switch. If you get an even number, it can go either way. That certainly should affect your decision in the second problem, and before you get the $100 it affects the "distribution" in the first problem as well.
I only meant it was like the Monty Hall problem in the sense that the "uniform distribution" is a bad way to think about it. Should have been more explicit about that.
In the blinded case as Aaron Bergman pointed out there's too little information in the problem to make a choice. The choice of distribution is surprisingly deep for those who haven't taken a proper
sadisticsstatistics course.Huh? The original problem specifies that we are NOT told the distribution. Thus, the original problem asks what we should do, given that we DON'T know the distribution. It cannot possibly be a correct answer to say that what we should do depends upon information that the problem says we don't get.
Consider -- What if there problem had stated: "I have 2 boxes, and they are (with equal probability) either $50 and $100, or $100 and $200, but I'm not going to tell you which. You take a box and it has $100 in it. Should you switch?"
The answer to this question is NOT "It depends upon whether the other box has $50 or $200." That is part of the information that, by the call of the question, we are not given. For the same reason, the answer to the original question is not "The answer depends on the particular distribution you choose." The answer might reasonably be: "The question does not give us sufficient information to provide a conclusive answer," but the answer to the question actually asked (in which it is given that we are not told the distribution) is not "the answer depends upon the distribution." The correct answer may be (and in my opinion is) "There is not enough information to conclude with confidence which choice leads to a higher expectation value, for reasons that would take a while to explain" but that is different from "The answer depends on the particular distribution you choose."
Indeed, the claim "The answer depends on the particular distribution you choose" is LESS right than "The answer depends upon what is in the other box." If you are allowed to say that the answer is "based on" information you do not have, then surely the fact of what is in the other box is a more important fact (leads to correct choice 100% of the time) than the fact of what the original distribution is (leads to correct choice 50% of the time).
Bottom line -- if the solution is allowed to be expressed in terms of "The answer depends on [some fact that is withheld by the nature of the question]," then why is "The answer depends on the particular distribution you choose" a better answer than "The answer depends upon what is in the other box."?
[NOTE -- This is one of my all-time favorite puzzles as well. I can't let the "answer" be a one- or two-liner.]
You guys are hopeless.
The correct answer is I tell you if you don't show me how much money is in the second box I'll hit you so hard in the mouth it will wake up your dentist. I then decide with certainty.
Come on.
Now, a clever man would put the smaller amount into his own box, because he would know that only a great fool would reach for what he was given. I am not a great fool, so I can clearly not choose the box in front of him. But Kevin must have known we are not great fools, he would have counted on it, so I can clearly not choose the box in front of me . . .
Dan
That said, your point about the stakes being different in the two "games" is spot on. Which leads me to think that Slipperman's comment is the one that ultimately ought to be dispositive here - if someone gives you a box with $100, take it.
There may be a genuine mathematical way to reason this out, but I'm not sure it's been put persuasively here yet. It's out of my bailiwick, so I defer to the better brains in the e-room, or as the Dread Pirate Roberts might've said, "Aaas Yooouuuu Wiiiiish."
Bull shit. Replace $100 with $100,000,000. Most of the population will hold the $100,000,000 if it was their box, and they would switch to get the $100,000,000 if the other person's box was opened. Yeah the information you have is the same, but the decision to switch can change depending on what's at stake.
The assumption that you should trade depends on constant marginal utility of money (almost always false) and on no risk-aversion (rare).
Cullen,
Is it true that by the time they started the game, neither knew? Perhaps my memory is not so good. But why then the analysis of whether Roberts/Wesley would be the type to put the iocaine in his own glass or in his opponents (see Dan Schmutter's reference)?
The EV analysis even works in the blinded case. It doesn't matter what is in your box, and whether you know what is in your box. Or can you show a difference in the EV calculation between the two cases?
This is really a Monty Hall problem, with the difference being there's no additional information. If there was a third box (two boxes of $N, one of $2N), the MH logic would say that you should switch after the third box (one of the two you did not already have in your hands) was revealsed to be $N.
Without the additional information of the third box, there's no basis for deciding to switch.
(The Wikipedia article claims that the best strategy for an N-door Monty situation gets you a (mathematically proven) (N-1)/N chance of winning; for two doors that's 1/2.)
Actually I'm starting to think you are right and I am wrong. The blinded case is still "x", and the EV of the trade is 1.25x. So no trades beyond the first. Perhaps that is true. I'll think about it some more...
But my original answer is still undoubtedly correct: in the non-blinded case (at least not blinded with respect to the box offerer), the offerer has absolutely nothing to gain by letting you exchange if he knows he has the box with the greater value (2x). Therefore you should not exchange.
If everyone is blinded (or even just the offerer/initiator), then maybe the change is worthwhile...
You just showed why it is NOT a Monty Hall problem!! There is NO ADDITIONAL INFORMATION!!! Completely different. Let's Make a Deal -- lets not mention the poor fellow again...
Expected value makes sense in a business environment where you are doing more or less the same kind of thing over and over with the goal of amassing money. However, when you are doing something only once, I would think that an individual's own current wealth, risk aversion, etc., would do a lot more work than the expected value.
Over multiple iterations, this double-or-half game is incomplete as a model for human behavior also because it assumes a generous game provider. It is somewhat akin to a false dichotomy, in that it sets up an apparent logical conundrum that could not exist in nature. There is nothing in this game for the person providing the money under the boxes. Once we assume that the person providing the game wants to make money on it (after all, what point is expected value theory if we discount the most fundamental economic assumption), then that person has to either charge a fee or change the payout structure. Let's say they don't change the game, but charge a flat fee for each iteration ($100). With the knowledge that all people desire to make a profit over time, if the provider is in the business of doing this game, then we will be able to deduce a range of likely dollar amounts under the boxes. For instance, if we find $10,000 under the first box, we know that this outcome could only occur rarely if the flat fee is $100, and we will ignore the statement that "half of the time, the other box contains 1/2 the amoutn and half of the time it has 2X the amount" because even if strictly true as a mean average, it discounts the additional information that the dollar amount provides in the instant transaction.
It feels like I am still missing something, but that is my attempt to explain away the apparent logic-loop.
TheShadow is right. Knowing what's in the box you got does not provide you with any true information - other than what 2(x) and x/2 are.
The only real information you are provided is that you pick a box "randomly." That means you don't consider all the psycho-babble princess-bride nonesense.
Since you took the box randomly (which for purposes of the problem we assume can be done - let's say a coin toss), you've already made your best play and your worst. Keep what you've got or trade. It doesn't matter in any event.
For probability wizzes, this is essentially the point raised by Martingales. you can wiki it for more info.
It's an important example of the importance of priors. The "trick" is that people, given the statement of no knowledge of the distribution, assume that the distribution must be uniform.
So maybe I will go back, and say this is like Monty Hall after all -- he is getting advertising revenue to compensate his game!
You are being asked to make a choice which is more likely 50 and 100 or 100 and 200. That judgement depends on the wealth of the person asking the question and the amount of money available. I'd switch. Kevin seems kind of loose with the money.
OK, Symmetry, so what are the probabilities? 75/25? 60/40? We know it has to be one or the other.
Silly frequentist. ;)
As I've made clear, put this PhD probabilist down with Mr. Bergman.
You don't have a distribution here. You have 2 boxes. Not 100 boxes, not 100 possibilities within each box. This problem is bounded by the statement that one is twice the other. I repeat, there is no distribution to assume being normal or not normal. You're not smarter because you say that people assume a normal distribution. While this may be true, it does not apply here. If X is the larger value, than X/2 is the smaller. No distribution. Get it?
The "distribution" plays out like Mike Keenan said:
You are being asked to make a choice which is more likely 50 and 100 or 100 and 200.
And you can't assume that the two scenarios are 50/50. So that is why you can't do the EV...
You are correct if we assume no finite limit to the amount of money that can be put in a box. But in the "real world" there obviously has to be such a limit. Doesn't this negate your argument since a "natural" probability distribution for the natural numbers 1,2,3, ..., N where N is finite would be P(x)=1/N; 1<=x<=N?
I'm much more comfortable with the Bayesian argument!
http://mathworld.wolfram.com/ProbabilityMeasure.html
http://mathworld.wolfram.com/Measure.html
Theorem: There does not exist a uniform probability measure over the measurable space of natural numbers.
Proof: Assume there exists a uniform probability measure over the set of natural numbers N.
A uniform probability measure has the property that m(X) = k for all X such that X is a member of N. By the definition of a measure, k is a non-negative real number.
If k > 0, then there exists an integer j such that jk > 1. Define Y to be the set of the first j integers. Then, m(Y) = jk (union of disjoint subsets). m(N) >= m(Y) > 1. Since m(N) = 1, we conclude that k cannot be greater than zero.
If k = 0, then m(N) = 0 (union of disjoint subsets). Since m(N) = 1, we conclude that k cannot be zero.
Therefore, we have a contradiction, and we conclude that no such distribution exists.
It's the same problem as asking whether the sentence "this sentence is false" is false. You're given a logically impossible premise, so there is no answer.
When you find out that there is $100 in one of the boxes, you know slightly more about the situation than you did before (namely, that the other box contains either $50 or $200.) That's only relevant if you're thinking in terms of your own diminishing marginal value of money. If you're thinking economically, substitute appropriately discounted values and do the math with those instead of with the dollar figures. (And heck, if you're thinking economically, it's only one more step to work in things like the value of your pride in guessing correctly or shame in being wrong, reputational effects of being seen as a risk-taker or as conservative, and the like. Have fun with that.)
Other posters are correct to note that you should switch boxes if the likelihood is 50% that your benefactor put $100 and $200 in the boxes rather than $50 and $100. Indeed, you should switch if you believe the likelihood that the other box contains $200 is greater than one-third, since the expected monetary value of switching when the chance is exactly 1/3 is equal to $100.
But you have <i>no idea</i> what this probability is, and the problem doesn't tell you any information that helps to determine it. Perhaps there's some information in the real world that would help - you know that your benefactor is a cheapskate, or you read something in his body language as you consider your decision - but that's not in the problem.
Nice summary, not too redundant, but still the question persists -- if you don't know the probality of 50/100 vs 100/200, what can you do? Can you do any calculation at all? If not, and if you are not answering it in terms of (very simple) game theory, then the question is really not worth answering -- thanks alot, Kevan! You prob/stat whizzes need to let me know what to do in an uncertain world!
Here's why the EV analysis doesn't work:
Suppose you pick the first box. You know that there's money in it, but not how much. Then if the EV analysis were true, the expected value of the second box would be 1.25 the value of the first box REGARDLESS OF WHETHER YOU OPEN IT. But the same would be true if you chose the second box.
If the game is dishonest, you should keep the $100, because it's more likely you chose the right box. If you chose the wrong box, why would he ask you to switch.
Box X and Box Y
Only two boxes I note.
"Each has some positive amount of money in it, but I will give you no information about the possible dollar amounts other than the fact that one box has exactly twice the amount of money in it as the other."
X and Y values are positive integers
X=2Y or X=1/2Y
"You randomly select one of the two boxes, open it, and find $100 inside."
X=100
Solve for Y
Y=200 or Y=50
"I now give you the option of keeping the $100 or switching boxes with me and keeping whatever's inside the other box. Which should you choose?"
Expected Value of Y:
1. number of possible values for Y - 2
2. probability for each value is thus %50
3. Solve for expected value:
E(Y)= 1/2*200 + 1/2*50
E(Y)= 125
Now - the value of Y is not ever going to be $125 - however this Expected value thingy is telling me to take the second box.
You are choosing between two outcomes. One outcome yields 1/3rd of the total money in the boxes, the other 2/3rds. The expected value of choosing box A is 1/2 the total money -- the same as box B.
2) This is not the Monty Hall case. With MH you first choose 1 of 3 doors assumed equally likey to have the best outcome. Let's choose A. They then show that one door you didn't choose was a loser--let's say door B. But you already knew that either B or C was a loser, since there can be only one winner. (Indeed the winner could be A.) There is no new relevant information offered.
Do you want to change your choice to the 3rd door or keep your first choice? Yes you should change your choice from A to C. But you are not changing your view based on new information. Rather you have been given a different opportunity than before. Instead of choosing 1 of 3 doors, you are allowed to choose between door A and the better of door B or C. You get the outcome of 2 doors instead of one. Monty Hall is not about new information. Wikipedia gives the right answer for the wrong reason.
If the first box you opened had an odd number in it (say, $5.01), then you know it can't be the box that contains the 2x higher amount (since an odd number can't be 2x anything.) Since the box you opened has an even number, keep it. It has a higher probability of being the "2x" value, since it is even.
The second example does not give this information, since you haven't opened the box.
Why not? Probabilities are just a lack of information. The "reality" is that the likelihood of 50 and 100 is either 1 or 0. But that doesn't help. And if there is no additional information, then both scenarios are 50% likely.
Both boxes cannot hold odd amounts (to the penny or the dollar) because one box holds "exactly" twice as much as the first). This reduces the number of possibilities so that knowing one box hold $100 does affect your choice.
Before the first box is opened, the possibilities are [box 1/box 2]:
odd/even, and box 2 holds twice the amount in box 1
even/odd, and box 1 holds twice the amount in box 2
even/even, and box 1 holds twice the amount in box 2
even/even, and box 2 holds twice the amount in box 1
When box 1 is revealed to hold an even amount, we are left with these three possibilities [box 1/box 2]:
even/odd, and box 1 holds twice the amount in box 2
even/even, and box 1 holds twice the amount in box 2
even/even, and box 2 holds twice the amount in box 1
It is twice as likely that box 1 holds the greater amount. Keep the $100.
I don't have a PhD in statistics, but I'm fairly comfortable saying that "Probabilities are just a lack of information" is just not true. Known probabilities are actually a wealth of information. Uncertainty is a completely different art...
Can someone with a PhD in statistics explain further?
The paradox, of course, is that switching in the second scenario is ridiculous, since you haven't learned anything more. You could keep switching forever! And yet, if you open it, the weird thing is that you're going to want to switch... 100% of the time. So then, why did you even need to open it? But how can that be?
Well, by my reasoning, you want to trade if you know the dollar value of your box, but otherwise it doesn't matter.
I'm basing this on the idea that if the numbers are entirely unknowable and can be anything, the potential gain or loss are both equal: infinity. But if you know the value of the box, then the potential gain or loss is fixed, and the gain is twice that of the loss. This means that fixing the value does actually change the facts.
To illustrate, let's say you know the boxes aren't going to be worth more than a million dollars. In that case, whether you switch obviously depends on what your number is. High number = stay, low number = switch. Knowing the number, then, makes all the difference.
If you don't know, however, you may in fact have a box worth a million dollars. In that case, you're automatically going to lose $500,000. Which seems to illustrate something important: If you have anything from 500k to 1mil, you're going to lose a LOT of money. But if you have from 1 cent to 499k, you could actually still lose half your money, while the gain even if you double is only modest.
All of this is basically why, if you don't know the values, you don't know whether you should switch. getting rid of the limit, you could win a million dollars, you could lose 2 million, you could win 4 million, or you could lose 8 million. If you know the value, however, then there are only two options, and the one is more good than the other one is bad.
But I want to go back to the point that, if you really have *no information* about the likelihood of two choices, then each choice is 50% likely. We're not talking about the spin of quarks here, so a priori the odds that each box contains a certain dollar amount are either 1 or 0. The probability we assign just reflects our level of uncertainty about what the boxes contain. But like CB, if someone can give a theoretical explanation of why I'm wrong, I'd love to see it.
So we bound the distribution and call it uniform. Or we specify that it is 50/50. We eliminate odd-even trickery. We blind the other guy so his motivations don't enter into it. We take whatever action is necessary to deny you the pathetic pseudo-erudite cop-out of "gee, we can't know."
I still think the EV analysis has to be wrong because you have a 50% chance of picking the "right" box on the first try, and switching--whether you peek inside or not--can't change that.
Can someone who has been wasting pixels on proofs and jargon explain either 1) why the EV analysis is wrong, or 2) why I'm wrong?
www.j-bradford-delong.net/movable_type/archives/001395.html
My previous proof is wrong.
Independent of the initial box chosen, a switch would be made!
Yes, it does have to be one or the other, which tells us that the probabilities add up to 1. However, we don't know what they are specifically without information not given in the problem.
Actually, I should ammend that. The problem says "I will give you no information about the possible dollar amounts" rather than "You have no information about the possible dollar amounts." If it was the former then we'd be well and truly stuck, but since its the later we can bring in information like the fact that theres no way to put 10 trillion dollars in a box, how much money people and this person specifically tend to use to prove points, etc. From these you can try to work out probabilities based on the state of your knowledge about the universe.
Random:You don't have a distribution here. You have 2 boxes.
I disagree. As long as those two boxes could, based on the current state of our knowledge, contain more than one possible value we have a probability distribution across those values.
Regardless of "distributions", if a switch is called for even when you don't know what you have, then the original choice is moot. If offered the choice again, you would switch again. This can't be helpful.
So if the expected value analysis doesn't hold up, then what does this tell us about the probabilities to start with.....
>I think part of what's feeding the battling intuitions here is that, in fact, 100/200 is *not* equally likely as 50/100, because in general in life higher amounts are less frequent than lower amounts.<
Well, if there is a limited amount of money in the world, I think that's necessarily true. What that would mean (at the very least) would be that if you had a box for more than half of that total value, you are going to lose a FORTUNE if you switch. Meanwhile, your only possibility of winning would be if you had less than half of the world's wealth in your box. And even then, you could still lose. Thus, in the real world, your potential losings may actually be greater than your potential winnings (that is, if there's a top limit, and you don't know what's in your box).
I think this explains pretty well why if you don't know the values it's not necessarily true that doublings are more good than halvings are bad. Because when you're at the outer reaches of the potential money, you always get halved.
That's theoretical, though, and merely explains why the math isn't as simple as it appears. Nevertheless, if my box had 5 dollars, or 100, or a million, or 100 million, I'd always switch (if he has 100 million, isn't it just as possible he has 200 million?)
The problem I see with traditional EV is that there's nothing to suggest that the person running the puzzle is as likely to put $200 in a box as he is to put $50 in a box.
Suppose I specify that they are equally probable. EV still can't be right. What's the mathematical reason for that?
Going in, you have a 50% chance of choosing the box with the higher value, and 50% chance of picking the lower value. I presume that if you played this game 1000 times, 50% of the first picks chose the higher value box and 50% chose the lower value box. If that is so, then switching 100% of the time will result in ending up with the higher value box 50% of the time and the lower value box 50% of the time. If that is right, it shouldn't matter whether you switch or not. What am I missing here?
If the chooser keeps the original box there is a 50% chance that she is getting x, and a 50% chance that she is getting 1/2 x. The expected value of what she holds is .75x. (It doesn't matter whether she gets to look inside the box or not, because that tells her nothing about whether she has x or 1/2 x). If the chooser switches, the situation is the same, a 50% chance of x and a 50% chance of 1/2x, again equaling .75 x. So switching gets you nowhere; the expected payoff is the same either way.
Nice point, very elegant. You are not as simple as you would like to admit...
Actually, if you specify that then suddenly the EV does become valid. Before you opened the box you didn't know that this was one of those Xs for which P(2X)=P(.5X), but once you've opened it you do knowand can apply hte EV accordinly. Since this can't be true for all X, that means that switching is probably a bad idea for every other number.
Discussion of Expected Values is now concluded.
something about the distribution of money amounts, regardless of what is told to you. Then whether you should switch depends on what you believe.
http://brainyplanet.com/index.php/Envelope%20Solution
Unfortunately this seems to be on a slow server that often returns blank or truncated pages.
The two boxes have a smaller amount 'X' and a larger amount '2X' in them. Assuming your choice is random, your expected value on choosing (and though it doesn't effect the analysis opening) is (X+2X)/2=1.5X
If you have chosen the box with the smaller amount you get X. If you have chosen the box with the larger amount you get 2X.
Now the switching comes in. If you initially had chosen the box with the smaller amount you will now receive 2X. If you initially had chosen the box with the larger amount you will now receive X. The expected value post-switch is 1.5X which is the same as the pre-switch scenario.
Therefore, on an expected value analysis you have no reason to switch.
On a marginal value analysis you may have a good reason to switch depending on different values of X. (For a $1 revealed amount I always switch because $2 in hand is worth more to me than the 50 cent loss even though my EV is the same in both cases. For a $1 million revealed amount I might not switch because the $500,000 'loss' would make a big difference to me. For a $100 million revealed amount I would probablly switch because the $50 million loss wouldn't feel as much to me as the gain I could get by getting $100 million. But in all cases the EV of switching or not switching remains the same.)
Suppose a dog has 4 puppies and each puppy is a 50/50 independent coin flip of being a boy or girl. Is it most
likely that the litter will be 2 boys and 2 girls?
Her answer would be, "No, it is most likely that the litter will be 3 of one sex, 1 of the other."
This is both correct and incorrect. If the event space is (all four same sex, 3 of one 1 of the other, 2 and 2) then she is correct since the probabilities of these events are, respectively, (1/8, 4/8, 3/8) and 4/8 is the highest of this list.
On the other hand, if the event space is (4 male, 3 male 1 female, 2 and 2, 1 male 3 female, 4 female) then the probabilities are (1/16, 4/16, 6/16, 4/16, 1/16) and thus
the most likely is 2 and 2.
Of course, she never gives you the event space, which, like the distribution in this problem, is necessary to answer the question.
Given that the first statement was essentially "You can't decide" disguised in language which says "I'm smarter than you," it can hardly be a shock that the discussion went on.
All the yammering about uniform distributions and natural numbers and whatnot is basically people who know too much missing the point. Sort of like what would happen if you asked a law professor "Can I legally do X?" and got back a 180 page law review article discussion deontological vs. welfarist thories of justice.
My (pre-law) training was in physics, and I can assure you that if Choset puts up a puzzle about the uncertainty principle, my respone will not include the words "commutation relations" however relevant they may be.
HowieKevan poses us is,Deal... or no deal?
It wasn't disguised in language to me at all. It was the most parsimonious language possible, and completely accurate.
Sort of like what would happen if you asked a law professor "Can I legally do X?" and got back a 180 page law review article discussion deontological vs. welfarist thories of justice.
Except it was hardly 180 pages. More like asking a law professor some question and he responds succinctly by using the precise Latin legalese for the situation, which unfortunately does not always illuminate without a longer response. Ask a math question, get a math answer.
Naturally, of course, to spell out the answer in more detail requires using more words. But there you go.
What an odd comment...
1) If you bound the distribution, the paradox goes away.
If we know that the most that can be in a box is $100, and we get a box that has $60 in it, we don't switch. If we get a box that has $30 in it, we do switch.
The paradox is that we *always* switch, no matter what's in the first box. That makes no sense because opening the first box seems to give us no information. In this case, however, opening the first box does give us information, as expected.
2) You're correct that the EV analysis is wrong. The reason, however, is that the given assumption in the problem is impossible.
You open the first box, and there is $100 in it. The boxes are either $100/$50 (scenario A) or $100/$200 (scenario B).
Before opening the box, these two scenarios are both equal to zero. Otherwise, the total probability of all the possible scenarios would be infinite, and that's impossible.
After opening the first box, the probability of these two scenarios appears to be equal, and this conclusion leads some people to believe that the probabilities are both 50%.
In reality, the probabilities are undefined. Bayes' theorem gives us the probability of scenario A:
Pr{A|$100 in box 1} = Pr{$100 in box 1|A} x Pr{A} / (Pr{A} + Pr{B}) = 0/0
Therefore, Pr{A|$100 in box 1} = 0/0. This value is *undefined*, not 50%. If the numerator and denominator were both very small numbers approaching zero, then 0+/0+ could equal 50%, but 0/0 does not.
This result comes about because the problem statement itself is a contradiction. It is the reason that EV analysis fails and that your intuition is correct.
Sorry if you don't like "proofs" and "jargon," but they're the only way you know you're right rather than just arguing intuitively about questions that have a correct answer.
So at first, I have no idea whatsoever how much money I'm going to get. It could be one dollar or it could be a trillion dollars. Ok, probably not a trillion (but again, if the first box is half a trillion, why not?).
Then I open my box. Blam! It's a million dollars. I'm thinking, "Sweet! That's huge! I'm getting at least 500k here." So now I have the question of whether to switch.
So how do I possibly not switch? Not knowing anything else, I don't see how it can be anything other than 50/50 for 500k or 2 mil.
Maybe the problem is that it wouldn't actually be 100 dollars. If there were truly no limits, the number would be something with like a million zeros on it, right? If the number were truly picked randomly.
If there are no limits, though, I still say you have to switch, if you're given a specific number.
On the other hand, contrary to BrainyPlanet there is a well-defined version of the question which exhibits the same paradoxical behavior. Suppose that there is a 1/2^n chance that the envelopes contain the values $3^n and $3^(n+1). This is a good probability distribution since the sume of 1/2^n is 1.
Suppose you open one envelope and find $2^k. You can easily calculate that the expected value for switching is (11/9)2^k. This is clearly larger than 2^k, so you should always switch no matter what you find in the envelope!
Ah, but then what happens if you don't open the envelope? Here the answer is a lot trickier and there's room for debate. But what I would say is that you now compute the expected value for staying the same, and the expected value for switching. Both of these calculations turn out to be infinite! So there's no way to say that switching is better since both switching and staying the same have infinite expected value.
Everything I know about this problem I learned from a slightly drunken discussion with Kenny of antimeta who is a graduate student studying philosophy of probability. So any good observations in my comment should be attributed to him. Any mistakes should be contributed to either me, or the wine that I was drinking at the time.
For the reasons discussed by some prior commenters, if I open Box 1 and it contains $100, I will think to myself "if I change to Box 2 I will either gain $100 or lose $50. Being risk-neutral, the expected gain outweighs the expected loss. Therefore, I will change to Box 2."
BUT ... nothing in that reasoning depends on the specific value of $100. I would have reasoned "this box contains X, changing to the other box will lead to either a gain of X or a loss of 1/2 X, and therefore I will change," no matter what X was.
Nor does anything in this reasoning depend on it being YOU who chooses the first box. Suppose the situation were that I (the player) gets to choose one of the two boxes, open it, note the amount inside, then keep that box or change to the other one.
Based on the above reasoning, my optimal strategy would always be "pick one box, open it, and then, whatever is in it, reject it in favor of the other box." But if I'd opened the other box to begin with, by identical reasoning, I would have changed that box for the first one.
In any event, that seems to be the reasoning I would go through ... except it doesn't seem to make a whole lot of sense.
I actually find it fairly condescending when someone assumes that I won't understand them, and I expect that a lot of people feel the same way.
If you don't like rigorous math, skip over my posts, and don't take them so personally.
Yes, that is a better analogy. But if the answer is "you can't decide without more information" then there is no need to dress it up in Latin or anything else.
Nor does it make sense to stand on what lawyers would call "nice" distictions like the distribution when the average layman is just going to assume 50/50 and go from there. The interesting question is what happens in the 50/50 case, and why are our intuitions so contradictory?
I'm not opposed to parsimonious languge in and of itself, having used it in many discussions with physicists and lawyers, but it should hardly surprise anyone that non-specialists don't accept the parsimonious explanation only accessible to specialists. To return to the uncertainty principle example, I would hardly expect this sentence to end all discussion here at VC: "The paradox is resolved because you cannot have simultaneous eigenstates of non-commuting observables."
If the box the person picks (Box #1) holds $100, one cannot assume that there's a 50% of Box #2 holding $50, and a 50% chance of Box #2 holding $200. Just because there are only two choices for monetary amounts in Box #2 does not mean each choice is equally likely (nothing in the original question says that).
The point of all the math types here is the following: there is no way to completely speicify the process by which amounts are chosen for the boxes are such that the answer to the question "what is the probability that the other box is twice the amount I have now?" is always 50/50. For instance, if the process is to draw from a uniform distribution between $0 and $100, put that amount in one box and put twice that amount in the other box, then if you open a box and it has over $100 in it, the probability that the other box has $200 in it is zero, not 50%.
Just because you weren't told the procedure by which amounts were put in the boxes (other than one has twice the other) doesn't imply you have no beliefs about the matter. You must have some belief about where these amounts came from. And whatever these beliefs are, the probability of the other box having twice the amount of the box you opened is not 50%.
Think about it this way...
I give you the following information:
* a is either equal to zero or one.
* a is equal to zero.
* a is equal to one.
What is the probability that a = 0? It's not 50%. It just doesn't exist.
It is not correct to say "you can't decide without more information". Like I state in my previous post, you can't help but have beliefs regarding the process which determines how much money is put in the lower amount box. The answer is not "you can't decide" but you whether you should switch depends on what these beliefs are.
* a is equal to zero.
* a is equal to one.
What?
As for the reasoning behind my answers, I'll wait a bit, since the comments keep flowing.
I say the following "I have two boxes. In one box I put a certain amount of money. I then flipped a coin to decide whether the other box would contain double the amount of the first box or half the amount of the first box."
I hand you one box and ask you if you would like to switch. Isn't there a 50% probability that the second box contains half and a 50% probability that the second box contains double?
Why doesn't this result in the paradox?
Dan
I'm not taking it personally. As it happens, I don't mind math (having done graduate physics work before bailing out for law school) but I can't read your notation so I do skip most of it.
That said, I was and am interested in this problem because it exposes an obvious flaw in how we think about probability which is worth trying to understand (I'm still groping around a bit; Sebastian and Box Keeper seem to have gotten it right but I can't say exactly why).
But all the highly technical objections of "we don't know the distribution" strike me as analgous answering the question of "how long will the ball remain airborne?" with "Unless we know if we're on the moon or not, we can't tell what the acceleration of gravity is." Technically correct, not particularly helpful. Make a simple assumption that speficies the problem and explain.
Now replace my dollar values with variables. You've switched, and Kevan shuffles the 2 boxes (including the one you originally had) 3-card-monte-like so that you don't know which is which. You had x, and now you have either x/2 or 2x. Should you switch again? The EV of switching again is .25(x+x/2) + .25(x+2x) = 9x/8, compared to the 5x/4 you now have. (Alternatively, you are given the option to switch back to the box you originally had. Should you? No.) So you should switch once, then stay pat.
I don't see how this problem is any different than the one above, and it doesn't require assuming some sort of uniform probability distribution among all integers, just among 2 or 3 boxes.
You did not address the issue of whether the initiator was required to offer the choice, or whether he is trying to "win". In a game scenario, if he offers to exchange when he doesn't have to, then he is holding the "losing hand".
Of course, this comes down to the question not being specific along these lines.
Vizzini: I'm afraid so. I can't compete with you physically, and you're no match for my brains.
Man in black: You're that smart?
Vizzini: Let me put it this way: Have you ever heard of Plato, Aristotle, Socrates?
Man in black: Yes.
Vizzini: Morons!
Man in black: Really! In that case, I challenge you to a battle of wits.
Vizzini: For the princess? [Man in black nods] To the death? [Man in black nods again] I accept!
Man in black: Good, then pour the wine. [Vizzini pours the wine] Inhale this but do not touch.
Vizzini: [taking a vial from the man in black] I smell nothing.
Man in black: What you do not smell is Iocaine powder. It is odorless, tasteless, and dissolves instantly in liquid and is among the more deadly poisons known to man.
Vizzini: [shrugs with laughter] Hmmm.
Man in black: [turning his back, and adding the poison to one of the goblets] Alright, where is the poison? The battle of wits has begun. It ends when you decide and we both drink - and find out who is right, and who is dead.
Vizzini: But it's so simple. All I have to do is divine it from what I know of you. Are you the sort of man who would put the poison into his own goblet or his enemy's? Now, a clever man would put the poison into his own goblet because he would know that only a great fool would reach for what he was given. I am not a great fool so I can clearly not choose the wine in front of you...But you must have known I was not a great fool; you would have counted on it, so I can clearly not choose the wine in front of me.
Man in black: You've made your decision then?
Vizzini: [happily] Not remotely! Because Iocaine comes from Australia. As everyone knows, Australia is entirely peopled with criminals. And criminals are used to having people not trust them, as you are not trusted by me. So, I can clearly not choose the wine in front of you.
Man in black: Truly, you have a dizzying intellect.
Vizzini: Wait 'till I get going!! ...where was I?
Man in black: Australia.
Vizzini: Yes! Australia! And you must have suspected I would have known the powder's origin, so I can clearly not choose the wine in front of me.
Man in black: You're just stalling now.
Vizzini: You'd like to think that, wouldn't you! You've beaten my giant, which means you're exceptionally strong...so you could have put the poison in your own goblet trusting on your strength to save you, so I can clearly not choose the wine in front of you. But, you've also bested my Spaniard, which means you must have studied...and in studying you must have learned that man is mortal so you would have put the poison as far from yourself as possible, so I can clearly not choose the wine in front of me!
Man in black: You're trying to trick me into giving away something. It won't work.
Vizzini: It has worked! You've given everything away! I know where the poison is!
Man in black: Then make your choice.
Vizzini: I will, and I choose...[pointing behind the man in black] What in the world can that be?
Man in black: [turning around, while Vizzini switches goblets] What?! Where?! I don't see anything.
Vizzini: Oh, well, I...I could have sworn I saw something. No matter. [Vizzini laughs]
Man in black: What's so funny?
Vizzini: I...I'll tell you in a minute. First, lets drink, me from my glass and you from yours.
[They both drink]
Man in black: You guessed wrong.
Vizzini: You only think I guessed wrong! That's what's so funny! I switched glasses when your back was turned! Ha ha, you fool!! You fell victim to one of the classic blunders. The most famous is never get involved in a land war in Asia; and only slightly less well known is this: Never go in against a Sicilian, when death is on the line!
[Vizzini continues to laugh hysterically. Suddenly, he stops and falls right over. The Man in black removes the blindfold from the princess.]
Buttercup: Who are you?
Man in black: I'm no one to be trifled with. That is all you'll ever need know.
Buttercup: And to think, all that time it was your cup that was poisoned.
Man in black: They were both poisoned. I spent the last few years building up immunity to iocaine powder.
To clarify, the problem could read: Has Kevan put 50 and 100 into the boxes or has Kevan put 100 and 200 into the boxes.
But in my version of the problem they are. I've said that I determined which scenario to apply by a coin flip.
Dan
other than the fact that one box has exactly twice the amount of money in it as the other. You randomly select one of the two boxes, open it, and find $100 inside. I now give you the option of keeping the $100 or switching boxes with me and keeping whatever's inside the other box. Which should you choose?".The puzzle remains the same.
How does the puzzle remain the same? In the original you know something about the contents of the second box. In your version you know literally nothing.
Dan
1) I must offer you the switch either way. I'm not playing strategically at all.
2) Despite the unrealistic nature of this assumption, I am indeed assuming there is a limitless amount of money in the universe. If the box I gave you contains every dollar the US Mint has ever printed, the other box might have twice as much.
Mike Keenan: You are precisely right that the two scenarios (50 vs. 100 or 100 vs. 200) are not equally probable. But I believe that far from making my reasoning wrong, that is precisely the insight that makes my answer correct.
Of course I could be wrong.
I don't think you switch once and hold under my version. Each time you have a box, there is a 50% chance the other box contains double and a 50% chance the other box contains half, because it was determined with a coin flip. Why isn't the expected value analysis that leads to infinite switching valid here?
Dan
I like to. . . think outside the box.
If the guy gives you that option only after you've chosen, though, and it is his money, it seems to me like you're in the Princess Bride situation.
The possibility of additional switching is not part of the problem and can't be reasonably analyzed unless you restate the problem with the necessary info. If the third box is created only after you have chosen the first two, it inevitably allows for experimenter manipulation and the logical answer would be to stop at two. If there are two "third boxes" each placed before you depending on which second box you got, I think it's back to the EV problem.<