There are two envelopes. Both contain a sum of money. One contains twice as much money as the other. The envelopes are identical and there’s no way of determining how much money is in either. You get to pick one envelope, open it, and keep the money inside it. You can switch envelopes as often as you like up to the point of opening one of them.
You pick up one of the envelopes and get ready to open it. But then you pause and consider whether you should switch envelopes and open the other one instead. You decide to analyze the choice mathematically.
Call the sum of money in your envelope X. The sum of money in the other envelope is therefore either 2X or X/2. There’s an equal probability that the other envelope will have either a higher or lower amount than your envelope.
So you figure out the expected value of the other envelope. That’s a standard mathematical operation - you add up all the possible values multiplied by the chance of them occurring. So the expected value of the other envelope is .5 (2X) + .5 (X/2). And that tells you the expected value of the other envelope is 1.25X. The value of your envelope is X. So you should switch envelopes and open the other envelope.
Already you might be thinking there’s something suspect about this result. What’s worse is that once you’ve switched envelopes, you can pause again to verify your results, and you’ll find that you can use the same logic to prove that your original envelope must have a higher expected value than the one you’re now holding so you should switch back. And so on.
Obviously there’s a flaw somewhere in this train of logic. But I can’t figure it out.
Keith Devlin wrote this explanation of what the flaw was but unfortunately I can’t follow the technical language of his explanation.
The expected value of either envelope is 1.25X because if you were to randomly choose any envelope on average you would accrue 1.25X per envelope. This should not be surprising. Proper treatment of your above example is that the expected value of the envelop you are holding is 1.25X, and by you logic, so is the alternate envelope. Therefore no matter which envelope you choose, there is the same expected value. Note that the expected value refers to the fact that over many trails in the end you would on average make 1.25X per trail.
Your problem is that before you switch envelopes, you need to first realize that the same logic shows that the first envelope has the same expected value. Choosing either gives the same expected value.
Ah, there’s a difference between the OP’s scenario and the one described in his link. In the OP, you have to decide whether to switch before seeing what’s in your envelope. In the link, though, you get a look at the check in your envelope first.
The Problem is that you are doing all of your calculations in terms of X the amount of money that you have in the envelope you are holding. As you say, if have the smaller amount of money in your envelope then the gain of switching will be X, while if you have the larger amount the loss by switching will be X/2. However, if you have the smaller amount in your envelope X will be smaller and so a gain of X won’t be that large. Meanwhile if you have the larger amount, then X will be larger and so the loss will be greater. This all evens out so that it doesn’t matter what you decide to do.
Getting back to the link, suppose you open the envelope and see $200. Now you could be playing one of 2 games. Either a $100/$200 game or a $200/$400 game. In this case you could make the argument that you should switch because it is more damaging to make a mistake in a $200/$400 game than it is in a $100/$200 game. But this assumes that the probabilities of the 2 games are equally likely, which there is no reason to believe. In fact whoever set up the game had some way of choosing how much money to put in the envelopes, and what the optimal choice is will depend on how this decision was made.
Not following you on this. It was set up so one envelope contains twice as much as the other. So if I define the contents of one envelope as X then the contents of the other has to be either twice as much or half as much as X.
The key is that there’s some upper bound on how much can possibly be in the envelopes (what that upper bound is depends on who offered them to you, but there’s still an upper bound). Let’s say that the upper bound is $100. If the envelope you’re holding is more than $50, then, then you can’t win by switching envelopes, and the expectation of your earnings if you switch is just X/2. And this possibility is something that you have to take into account in your expected value calculation.
The false idea that creates the paradox is that every possible value has an equal chance to be in the envelope. That is, the probability distribution is uniform (every point has the same probability) and includes the entirety of the positive integers. But, as the author of that article claims, such a distribution is not possible. The only possible value for the probability at each point is 0, and although this is always the case for a continuous variable case, the probability for the value to be in any finite interval also must be 0. Thus, you cannot have the probability of every single value be exactly the same.
Thus if you say that each envelope has a 50% chance of being higher and 50% chance of being lower, you would be incorrect. You have to determine what the actual prior distribution is, or guess at it, and use that distribution to calculate the expected gain. This would provide guidance as to whether or not to switch after seeing one of the values. If you don’t see either value, you sum over all the possible values, and while I’m not going to do the work to show it, it would come out that the two envelopes have the same expected value.
I don’t think it’s necessary that there’s an upper bound, just that you define your probability distribution such that it makes sense. Giving every possible value on the positive real line the exact same probability does not make sense, but would if you put in an upper bound. However, the amount could be exponentially distributed, with no upper bound. See Exponential distribution - Wikipedia
I’m not seeing your point here at all. The math works the same regardless of how big the number you plug in is. Basic arithmetic doesn’t change.
A word here on expected value. Suppose you find a gambling machine. You can put any amount of money in and the machine will flip a fair coin. If it comes up heads the machine will double your money. If it comes up tails the machine keeps half of your money. So every time you play, you either double or half your bet.
The game is weighted in your favor. Let’s say you put in dollar bills. You have an even chance of winning or losing. But when you win, you win a dollar and when you lose you only lose fifty cents. So the expected value of each bet is $1.25. You should play this game all day long. And the math remains the same regardless of whether you play with dollars or dimes or hundred dollar bills.
I don’t see that as the key. Just take whatever the total amount of money the guy who filled the envelopes can afford as the upper bound. Divide it in three. Put two thirds in one envelope and one third in the other envelope. You now have the set-up for the game - two envelopes with an unknown amount of money in them, one of which contains twice as much as the other.
I’m not seeing this. You have two identical envelopes and you know they have different amounts inside. If you pick one envelope at random, how is there not a 50/50 chance of whether you have the higher or lower amount?
Like I said, I don’t think I made a good point anymore, but to explain what the point was:
I thought you didn’t know how much you had in your envelope. Say that in reality, one envelope has five dollars and the other has 10. Then if X is the amount in your envelope, then if the other envelope contains 2X, then X = 5, while if the other envelope contains .5X, then X = 10.
It’s been awhile, but I’m pretty sure this is false. The definition of expected value is the average value over many trials, which is not the same as the value I am holding in the current trial. So while X = the value of the first envelope I picked during this specific trial, The average of the value of the first I envelop I pick over many trials would be 1.25X.
The definition of expected value as an average across all trials allows me to have a different expected value than the current value I am holding.
My previous comments were referring more to the situation where you see the value in the envelope, and I hastily said something about the unseen situation that was potentially not accurate.
The issue is that you are assuming that the envelope you are holding has some specific value. It doesn’t until you open it: it has a probability distribution. The other envelope has the exact same probability distribution. If you open the envelope, you can use this probability distribution to determine the likelihood the other is higher or lower, but until you do that the envelopes have the exact same amount.
Ok, what I quoted had very little to do with what I said.
Anyway, the issue is that if you assume a specific unknown value in the envelope, the amount in the envelope is not a number: it is a random variable. You can’t just do calculations with a random variable as if it were a fixed number and get answers like you expect. I’ll try to work up something a bit more detailed.
In my original misunderstanding of the scenario, X was defined as the amount of money in the particular envelope you’re holding in that particular trial. That amount is not going to change from trial to trial–that envelope will always have X dollars in it. You may not always pick it, but whether you pick it or not, it’s got X dollars in it. X is the same in each trial. But you don’t know what X is, and you don’ t know whether the other envelope has 2X or .5X dollars in it.
The envelope I have has X in it. The other envelope has 2X with chance .5 and X/2 with chance .5, so its expected value is 1.25X.
This is incorrect, as it assumes that X is a fixed number that can be worked with like any other fixed number, but it’s not. If you open the envelope to fix X and then use a proper prior distribution, you can work out the chance that switching will be beneficial. The prior distribution that every possible positive value is equally likely is not a proper distribution: the probability that the value falls within any interval of finite length is zero.
If you don’t know what X is for sure, you can’t work with X directly because it is a random variable with a probability distribution. You can work with its expected value, E(X), but you can’t use it in the same way as the above. It is not true that the other envelope has 2*E(X) with chance .5; that would assume that you know for sure that your envelope has X, which you don’t.