It depends on what the OP’s game is, which isn’t clear to me. Perhaps their game is “A die is rolled. Two envelopes are then filled, one with the number on the die, the other with twice that number. Then one of those envelopes is handed to you at random”. It hasn’t been specified, so far as I know, what the range of possible envelope values is meant to be. If the OP is willing to say “The game I’m interested in always has one envelope with $1 and one envelope with $2”, then that would settle it. But I haven’t seen that yet. Indeed, what they’ve said indicates they consider both {$1, $2} and {$0.50, 1} to be possibilities in their game. Which is fine; that is a game that exists. And in that game, there are more than 2 possible values for envelope contents. Sure, why not?
So, OP: you want us to talk about expected values. That means averages over some bunch of possibilities (for example, the expected value of a die roll is 3.5, because that’s the average over all six possibilities). But we can’t tell you what those averages are unless you tell us what all the possibilities are. So what are all the possibilities you want to consider?
That’s the same as saying during 1 roll of a die there is only 1 value involved. We can still ask about the probability distribution over all runs of the game.
What disqualifies “A die is rolled. Two envelopes are then filled, one with the number on the die, the other with twice that number” from being a game that matches the OP’s scenario?
If the OP wants to stipulate that there’s always one envelope with $1 and one envelope with $2, let them say so. And if not, let us accept whatever possibility space they want us to look at.
So, OP, tell us. Give us a list of all the possible envelope pairs. Is it just {1, 2}? Is it just {1, 2) and {0.5, 1}? Is it everything from {1, 2} up through {1024, 2048}? Is it all adjacent powers of 2? What is it?
Sure, you can ask about the probability distribution over all runs of the game if you want.
But it would be wrong to include numbers, at each step, that are not part of that trial of the game. This will give you incorrect results.
This would be like including dice roll as X and including X+1 and X-1 every time. Doing this introduces the number 0 and the number 7, neither of which appear on the die.
Suppose I set up a game where I roll a die, and then stuff envelopes, one with the die value, the other with twice the die value. Then I hand an envelope to you at random. I do this once and only once; I’m not going to play this game ever again.
Supposing your envelope has $3 in it, what’s the expected value of the other envelope?
The fact that the game is only played once doesn’t matter. In 1 single trial of a game, there’s no distribution to speak of; there’s just what things are. But when we talk about probability, we imagine a distribution of possibilities, regardless of what actually happened in this one instance. Probability isn’t about what actually happens; it’s about conceptual distributions of possibilities.
They’re unknown only in the sense that they take different values in different runs of the game. Which is precisely the thing you want us to ignore by invoking “1 trial, so the only possibilities are the possibilities which are present in this one trial!”.
In the game where we stuff envelopes based on a die roll, one with the value on the and the other with twice the value on the die, then hand one envelope to Alice and the other to Bob, it actually is true that, out of all the times Alice gets $2, on average, Bob gets $2.50. It’s not a lie that E[Bob’s envelope | Alice’s envelope = $2] = 1.25 * $2. It really is true: 1/12 of the time, Alice gets $2 and Bob gets $1, 1/12 of the time, Alice gets $2 and Bob gets $4, and the rest of the time, Alice doesn’t get $2.
(X + Y)/2 is the average of the actual values in both envelopes. That’s fine. It’s not what’s usually meant by “the expected value of Y”, which is the average of the value in the Y-envelope over all possibilities, not just the one that actually happened.
Two dice are rolled; I hand one to Mr. X and one to Mr. Y. What’s the expected value of Y’s die? It’s 3.5. Even if Mr. X actually got a 1, and Mr. Y actually got a 5, so that the average between the two of them was actually 3, that doesn’t change the fact that the expected value of the second of two die rolls is 3.5. Expected values aren’t about what actually happens; they’re entirely disconnected.
So, yeah, (X + Y)/2 and (2X + 0.5X)/2 are different things. But (X + Y)/2 is not what “expected value of Y” means. In some setups, it’ll happen to be the case that (X + Y)/2 always equals the expected value of Y anyway. Which is fine. But it’s not because expected values are always yielded by calculations involving only numbers on the table somewhere.
I also discussed a modified version where you had an envelope and didn’t open it, based on the idea that the specific amount in that envelope didn’t matter because the train of reasoning was consistent for any amount. But at this point I see a need to simplify. So let’s focus on only the open envelope scenario.
I understand the M+2M/2 argument that some people are making. But 1) it doesn’t resolve the paradox and 2) it doesn’t seem particularly useful in itself. It’s based on having information you don’t have. There’s no scenario in which you’re told the total amount involved so any plan based on that amount isn’t going to work.
Okay, I think I see what you’re saying but I’m not seeing how it’s relevant here. The envelope scenario doesn’t seem comparable to a classroom of students. It seems to be the equivalent of my test score on X and Y. Ignoring what other students got, if I know I got twice as much on my second test as I got on my first test and I know I got a 40 on my first test, then I don’t have a range of marks of my second test - I know I got an 80. Y is equal to 2X.
And if I was told my mark on the first test was a 40 and the mark of my second test was either exactly half or exactly twice my first mark, then again I know that it’s either 20 or 80. Y is either X/2 or 2X.
This is the trickier scenario. At least, it may be. Before I go any further, can you clarify one further question? The answer to this question isn’t strictly relevant to the expected value calculation in this scenario, but it will help make the discussion clearer anyway.
Prior to opening the envelope and seeing the $1 in it, what are all the possibilities for what the envelope contents are? I know, I know, it ultimately doesn’t matter what the possibilities without a $1 envelope are after you’ve opened your envelope and discovered that there is a $1 envelope, but just give me something to go on anyway, for concreteness’s sake. What, in general, are all the possible pairs of envelope values?
This type of problem is more word problem than a math problem. The key is finding the correct math problem to describe the words.
It does in that this is the correct math to go with the words. You only know that there are two envelopes with values M and 2M. You don’t know which you are holding. The other formula does not accurately represent the situation.
That is the key. It shouldn’t be useful. There are two envelopes. One with a higher value and one with a lower value. There shouldn’t be a formula to help you choose one. If you find a formula that tells you to pick one over the other, it is clearly being applied incorrectly.
Hypothetically, I suppose the envelope could contain any amount.
There are amounts that would render the problem solved in one easy step. If I opened the first envelope and found a penny inside, I’d know that the other envelope can’t contain a smaller amount so it must contain two cents. And if the first envelope contained more than one third of the money in the universe, I’d know that the other envelope couldn’t contain an amount twice as large so it must contain half the amount I have. And in a slightly less trivial case, if the first envelope contains an amount of money that can’t be divided by two then I knwo the other envelope can’t contain half of its amount so it must contain twice as much.
But let’s make a reasonable assumption that the first envelope contains an amount of money that could reasonably be halved or doubled. If you’re trying to work out a general procedure you shouldn’t assume a special case.
How so? Consider the situation I described in post 149:
What is inaccurate about this?
I agree. This is essentially the paradox. There’s an established formula that normally would produce an accurate answer. But in this situation it produces a false answer. What is it about this situation that causes this formula to produce a false answer?
It’s not really enough to say you should know the formula didn’t work here because it produced a false answer. That only works if you already know what the answer is supposed to be and can see the answer you got is wrong.
But you might be in a similar situation except that the answer isn’t obvious beforehand. In such a case you might use the formula and get a false answer but not realize it’s a false answer. There must be something in the situation itself that tells you whether or not the formula will work before you actually produce an answer.
Ok. Good. And I’ll go ahead and assume that all these possibilities are initially equally probable.
In that case: Your calculation is entirely correct. The expected value of Y, given that X contains $1, is indeed $1.25. That is, E[Y | X = $1] = $1.25. (My previous objections to your calculations, which are no longer relevant, were in the context of thinking you were looking at scenario 2) and trying to calculate the overall expected value of Y, which I now understand is not what you’re doing. You are indeed looking at scenario 1) and trying to calculate the expected value of Y given that X = $1)
Out of all the times you play this game and get a $1 envelope, the average value of the other envelope is indeed $1.25, for exactly the reason your calculations showed; out of all the times you get a $1 envelope, half of them have the other envelope being $0.50 and half of them have the other envelope being $2.
So what? Why does this seem paradoxical?
It seems paradoxical because what’s special about your envelope vs. the other envelope that would make the other envelope seem better? And if we can run this calculation no matter what X is, shouldn’t we just switch envelopes even without looking (and then switch back, which is absurd)?
Well, the answer, depending on the setup, is one of two things:
The setup is one where there’s a lowest possible envelope value. In this case, prior to opening the X envelope, you have no reason to prefer Y to X; the reasoning that prefers Y to X only works when X isn’t the highest value (exactly one bajillion dollars, say), which you aren’t assured of until you open X. After you open X, you gain new information, and it’s perfectly reasonable that the new information (which is asymmetric; it tells you something about X that it doesn’t tell you about Y) would lead you to conclude something asymmetric. It really is true that, for any value < one bajillion dollars, out of all the times when one envelope has that value, the other envelope has on average a higher value. But there’s nothing wrong with this. It’s just true. It’s counter-balanced by the fact that when one envelope has the highest value (one bajillion dollars), the other envelope is guaranteed to have a much lower value.
It doesn’t matter that from 1 cent up through one bajillion dollars is a very large range. We can illustrate the same thing with a narrow range of possibilities. E.g., if the only possibilities in the world are {0.25, 0.5}, {0.5, 1}, {1, 2}, and {2, 4}, each in both possible orders, then it really is true that, over the cases where the first envelope contains a 1, the second envelope’s average value is 1.25. That’s not false. It’s true. You can read it right off the list; there’s 8 cases total, 2 of which start with 1 [(1, 0.5) and (1, 2)], and over those two cases, the second value on average is (0.5 + 2)/2 = 1.25. That’s right. It’s exactly right. There’s nothing wrong with it. It’s true.
And it’s not paradoxical. It doesn’t imply that we should blindly switch envelopes back and forth infinitely often: we’ll only want to switch when we open an envelope and see that it doesn’t contain the highest value (4); if we open an envelope and see a 4, we won’t want to switch. So it’s all just dandy. It is indeed true that out of the times when we see $1, the average value of the other envelope is $1.25, and there’s nothing wrong with this.
That’s the setup where there is a highest/lowest possible value overall. That’s simple. Slightly trickier is:
The same basic setup, but with no initial limits on how high or low envelope values can get. I’ll return to this later, but what happens here is essentially the trickiness about infinite sums I began talking about earlier in the thread. It’s not actually as scary as it might look; it’s not that complicated, it’s not hard to grasp if you want to. But I’ll talk about it later.
Okay, it seems we’re on the same page about why the paradox exists.
Not sure I can accept this an a general explanation. It seems like a special case - the equivalent of Friday in the Surprise Hanging Paradox. The issue, as I see it, is whether the logic that’s appliable on Friday is also applicable on any day of the week.
So I’m looking forward to what you say in regards to this case.
You aren’t using the “established formula” correctly. You are using a random variable X as a constant in a probability calculation, when the probabilities depend on different values for X. You can’t get a value for the expected value of Y without using the expected value of X, not X itself. X is one of (a,b) (let’s assume a > b), and Y is the other one, and both a and b are fixed before you choose and can’t change. If you want to use X = a, then you also have to use Y = b, and the expected value of a constant is a constant. If you want to take both possibly outcomes into account, you can’t say that X = a in both of them, since the other outcome isn’t X = a, Y>a, it’s X = b, Y = a. Instead you are making X fixed, but then making Y and therefore what was in both the envelopes before you started uncertain. It is uncertain to you, but it isn’t uncertain to the person who put the money in the envelopes. It isn’t uncertain to physical reality. Your calculation is along the lines of “if the envelopes contained 1 and 2, you would have a higher expected value than if the envelopes contained 1 and 0.5”. It has nothing to do with your choice, it has to do with how the predetermined amounts of money in the envelopes is distributed.
Little Nemo is calculating E[Y | X = 1] perfectly correctly to be 1.25. There is no mistake there. What do you think E[Y | X = 1] is?
Consider, BarryB, a game where I stuff envelopes with money: 25% of the time, I stick 25 cents in one and 50 cents in the other. 25% of the time, I stick 50 cents in 1 and 1 dollar in the other. 25% of the time, I stick a dollar in one and two dollars in the other. And 25% of the time, I stick two dollars in one and four dollars in the other. Then I flip a coin to pick an envelope to give you and keep the other envelope for myself.
Out of all the times when you get an envelope with one dollar in it, what’s the average value of my envelope? It’s $1.25.
That doesn’t mean the average value of your envelope and my envelope, in any particular round of the game, is $1.25. But the average value of your envelope and my envelope is not what’s meant by “The expected value of my envelope”.
Put another way: BarryB, you and many others seem to be interested in the calculations “The expected value of Y, given that one envelope contains a dollar and the other envelope contains two dollars” (which is $1.50) and “The expected value of Y, given that one envelope contains a dollar and the other envelope contains 50 cents)” (which is 75 cents). Fine, those are things, but they’re not the relevant expected values in the situation Little Nemo describes, because the situation they describe isn’t one where what you know is that “one envelope contains a dollar and the other envelope contains two dollars”, and it isn’t one where what you know is that “one envelope contains a dollar and the other envelope contains 50 cents”. The situation they describe is one where what you know is that “the X envelope contains a dollar”. And “the expected value of Y, given that the X envelope contains a dollar” is indeed $1.25.