You have in front of you two envelopes with some undetermined amount of money in each. You are allowed to look at the contents of only one of those envelopes and you can only keep one of those envelopes. (You don’t have to keep the one you look at.) You take one envelope and open it and determine the amount of cash in it. Now knowing this amount, is there something you can do to determine whether you should switch or not that gives you a better than even odds that you will take the higher of the two envelopes?
Sure, ask.
No one’s going to tell you.
You need to have some prior knowledge of the distribution of amounts. No, you can’t just say “equally likely to be any amount”, because that’s not a valid distribution.
But the thing is, in the real world, you always have some information about the distribution. For starters, you know that neither envelope contains more than 100 billion dollars, because nobody has that much money to put in them. And if you have some idea of who set up the envelopes and what their motivations are, then you can constrain it further (for instance, if it’s a buddy who set up a bar bet, and you know that he’s solidly lower-middle-class, then you can be pretty confident that neither contains more than $50 or so).
Once you’ve established that prior, then you can start going about a calculation to base your decision on.
Assume you don’t know who set up the envelopes. Otherwise if it’s your bar buddy, the amount in the envelopes isn’t very indeterminate at this point since you’ve already limited the possible amounts.
Can you put the other envelope inside the one you open?
If it’s your bar buddy then both envelopes contain negative amounts of money in the form of IOU’s. The only way to win is not to play.
Nope. You can only leave with one envelope.
The envelope that you open has a limited amount of money, and the unopened one has an unlimited amount, so if any amount is just as likely as any other, then the second envelope has more possibilities that are higher than the first.
But that’s just nuts.
Or you might better argue that, if there is a minimum amount of money possible, (say a dollar) then you should always take the envelope if it has more than a dollar. In the long run such strategy wins because you will never pick the envelope that has the minimum when it occurs.
Interesting paradox relating to what the OP is asking.
You’re handed two sealed envelopes. You’re told one of them contains twice as much money as the other (and both contain an amount that’s divisible by two). There’s no way of determining the amount inside other than by opening it.
You pick one of the envelopes at random. You’re then offered a choice: you can keep the money that’s in the envelope you have or you can exchange it for the money in the second envelope. But once you open an envelope you’re committed to it. What do you do?
According to one line of reasoning, you should switch. You picked your first envelope at random so there is an equal probability of it containing the larger or smaller amount of money. And the second envelope also has an equal opportunity of containing the larger or smaller amount of money.
Let’s say the amount of money in the envelope you have is X. That means the amount of money in the other envelope is either .5X or 2X. As noted, there’s a 50% chance it’s .5X and a 50% chance it’s 2X. This means the expected value of the sealed envelope is 1.25X. In other words the amount of money in the second envelope is statistically more than the amount of money in the first envelope you picked.
Applying this logic you switch envelopes to improve your odds. But before you tear open the second envelope, you’re offered a chance to switch back. And you now realize the same line of reasoning is still in effect. If the amount of money in the second envelope is Y then the expected value of the first envelope is 1.25Y.
In other words, regardless of which envelope you’re holding in your hand, you should always switch to the other one.
Since you don’t know the amount in the second envelope, it can be thought of as simultaneously less than, equal to, and greater than the first envelope.
Schrödinger’s cash.
The amounts are still limited, and you can draw some information, because you specified that they contain cash, not checks or other financial instruments. Size and thickness of the envelopes puts an upper bound on its possible contents. High-denomination bills are also withdrawn from circulation, and often worth more to collectors than face value; one could argue that this takes them out of the category of “cash”, but an easier argument would simply be that the probability of the envelope containing high-denomination notes is low because there are few of them available. So, neither envelope is likely to contain notes of greater denomination than $100.
Now, that doesn’t tighten the constraints much, but in principle, it provides information. You could, for example, devise a decision tree that selects a keep/switch decision based on the relative thickness of the envelopes and the average of the denominations of notes in the inspected envelope.
Example Nodes:
Inspected envelope contains only low-denomination notes; uninspected envelope is thicker–>Switch to uninspected envelope. (The denominations in the uninspected envelope can’t be smaller, and there appear to be more of them.)
Inspected envelope contains only low-denomination notes; both envelopes are approximately the same thickness–>Switch. (Approximately same number of notes, chance of increasing denomination.)
Inspected envelope contains only low-denomination notes: uninspected envelope is thinner–>Judgment call. (I’d go with Switch for this node, because the opportunity cost of losing part of an envelope-sized stack of small bills is low enough that I’d risk it for a chance at a single larger bill.)
And so on. As you populate your decision tree, mark the nodes with indeterminate or “judgement call” choices. See how many are the result of inspecting which envelope (assuming the envelopes are of discernibly different thicknesses). If inspecting one envelope leads to more fuzzy nodes than the other, use that information to determine which envelope to inspect. From there, just follow the tree.
You could be more rigorous about defining your tree than I was with my examples; perhaps you could establish a value scale for denominations using ratio of (equal or smaller):larger denominations in normal circulation, and use the average of that value for more quantifiable comparisons.
How much of an edge this would give you, I don’t know. Maybe not enough to be worth the effort. It’s something you could theoretically do, though.
Well, the one who sets up the envelopes has to be careful with what amounts he puts inside…In order to ensure that you don’t give extra information when somebody opens one of them, it is a condition “sine qua non” that both amounts are divisible by 4.
Because if you open an envelope and find an amount divisible by 2, but such that, when divided by 2, it gives you an odd number, then (given that we are told that both envelopes contain amounts divisible by 2) it immediately follows that that particular envelope is the one that has less money inside.
For instance, if you open an envelope and see that it has $50 inside, you immediately know that that envelope is NOT the “big” one. Why? Because 50/2=25, an odd number, and we are told that both envelopes contain an amount of dollars that is even. That means that the only possibility is the other one: This envelope contains $50 and the other $100.
Just a little aside! 
Nitpick: the IOU’s I’m familiar with are not negative money, they’re unsecured promissary notes which might or might not be negotiable into money based on the character of the person who wrote them.
The buddy’s bar tab might possibly represent negative money, in that it’s the notice of money which he owes the bar. (Of course, just because you take an envelope containing the tab, that doesn’t really mean you’ve assumed his debt or obligation.)
Interesting observation that hadn’t occurred to me. But wait. If both amounts have to be divisible by 4 and the one you open is divisible by 4 but not by 8, then you know the other one must be the larger, else it would not be divisible by 4. So both must actually be divisible by 8. But wait again. If they both have to be divisible by 8 and you open one not divisible by 16, then the other one must be larger, else it would be divisible by 4 and not 8. So they must both be divisible by 16. But wait…
Aaaah, but therein lies the rub – You tell the patsy… I mean, the subject that both amounts are divisible by 2, and (without telling him anything else) make sure that the smallest one is divisible by 4.
Thus, you might have, for instance, $16 in one envelope and $32 in the other. No matter which one he opens, there is no chance that he will get the extra information that will allow him to decide that the other one must be the envelope with the higher pay-off. Because he has only been told that “both amounts are divisible by 2”.
I’m taking the one that has dollar bills in it. You just said cash, not in what form. Unless the first one I open has some Euros or British pounds in it I’m taking the US dollars. I don’t trust you enough to take the other envelope and get some Zimbabwe bills.
Honestly though, unless it was a really small amount, like under $10, I’m keeping it, free money is free money.
Randall Monroe shared this problem here. It’s a fascinating problem that does actually have a solution even when you know nothing about the distribution of numbers.
Briefly, the answer is
Choose a random real number between 1 and 0. The amount in the envelope you open is x. If your random number is less than 1/(1+e^-x), keep the envelope you opened. Otherwise, switch for the other one.
See the linked post for more details.
The puzzle as stated by Monroe is much more rigorously information-free. It posits real numbers, rather than the much more limited set of possible numbers represented by cash denominations. The numbers are presumably written on identical slips of paper, as opposed to being represented by tangible cash, which removes an option for distinguishing between the envelopes. (It also has the player randomly selecting an envelope for inspection, but that doesn’t matter with indistinguishable envelopes.)
The solution from the abstract scenario should apply here as well, giving a universal strategy for better-than-even odds. However, I think applying the decision tree first, then falling back to the strategy Monroe posted if the tree hits a fuzzy node, would give better odds than the purely mathematical approach.
Yup, I agree. The real world of money and physical envelopes means you can assume some type of distribution in this specific problem. But given what I think was the OP’s intent, that seemed to be a little bit of fighting the hypothetical. This problem is interesting because even in the case where there’s absolutely no way to guess at a distribution, there actually is still a way to solve it.