Everyone who isn’t already familiar with the idea owes it to themselves to read up on the law of total expectation. It’s far, far more relevant than Bayes’ theorem to this problem…
I fully understand what you are saying and don’t disagree with any of your statements regarding expected value conditioned on X=some value.
But that is not the situation in the OP. I realize that Little Nemo just posted that he/she has abandoned the OP in favor of a game that matches exactly what you are talking about, but I was unaware that the game had changed. I did not get that memo.
Yes, I read it and found it interesting. I think I followed everything you said although it would be easy to convince me of anything when you start juggling infinities.
I didn’t really see it as abandoning the OP as it was narrowing it down to avoid distractions. I said in the OP that the first envelope had a specific value X. But when people started talking about how X was a bunch of different possibilities, I decided to pin it down by metaphorically opening the envelope and saying “Okay, X is not a range of values. X is one dollar.”
I didn’t ask “what is the value of Y?” I asked “what are the possible values of Y?” Obviously not all possible values of Y are the actual value of Y.
So I agree that the set of values (2X,.5X) contains the value Y for sure. But I’m not sure what you mean about the set of values (2X,.5X) also contains a value that is not even allowed in this game. How is it allowed or not allowed?
You said there are 2 envelopes, and X is in the first envelope. So the other envelope has either 2X or .5X
When you open the envelopes you will find that either 2X or .5X was a value that appeared in at least 1 envelope. You will also find that either 2X or .5X was a value that did not appear in any envelope.
Question:
Are you unable to see that the 3 different values X, 2X and .5X do not all appear in envelopes when you open them during this 1 trial of the game?
1 of those values is not allowed in this trial of the game because your game specifically states that there are only 2 values that can appear in the envelopes.
We don’t know in advance whether 2X or .5X is the bad value, that’s why there is an “or”, it is one OR the other, and it just so happens that the one it is not doesn’t appear in ANY envelope in this trial of the game.
Isn’t this just a two-door version of the Monty Hall problem?
Also, to simplify the scenario so you can’t metagame the envelope giver, why not have the amount of money in the form of a check instead of actual cash?
(The even/odd problem still does sort of hold, though, if you open your envelope and get, say, $1.23)
It’s not that the calculation is wrong, it’s that he has incorrectly defined the problem:
There are two envelopes, one has A, the other has B.
The relationship between A and B is utterly unnecessary for this thought experiment.
You take one in your hand and call it X.
So you have a 50% chance that X = A and a 50% chance that X = B.
Where is the third value coming from?
I don’t want to sound snarky but these are the basics of probability you guys are questioning here.
Little Nemo, back to the question of the set of values (2X,.5X), you agreed Y is in the set but then you hesitated on the question about whether a value in the set does not appear in the game at all.
Can you answer the question above?
Or just answer this, assuming the OP and the 1 current trial of the game, and envelope 1 has $10 and envelope 2 has $20:
- Which envelope has the value X?
- Which envelope has the value 2X?
- Which envelope has the value .5X?
If you’re saying one envelope has $10 and one envelope has $20 then you’re not assuming the OP.
Or just answer this, assuming the actual OP and the 1 current trial of the game, and envelope 1 has $10 and envelope 2 has half or twice the amount of envelope 1:
- What are the values in the two envelopes?
If you can express the contents of both envelopes now in just two values, you’ll have impressed me.
I am assuming the rules of the OP and instantiating 1 example game for illustrative purposes because you seemed to have difficulty when I list the values as X, 2X and .5X
How about this?
In the OP, after you open the envelopes, do you see X, 2X and .5X in an envelope?
RaftPeople, et al, supposing the open envelope has $10 in it and the closed envelope has $5 in it, what would you say the expected value of the closed envelope given that the envelope has $10 in it is?
Supposing the open envelope has $10 in it and the closed envelope has $20 in it, what would you say the expected value of the closed envelope given that the open envelope has $10 in it is?
I feel Little Nemo’s snark above was justified.
I understand your point about the conditional expected value. Assuming infinite possible values it works, I get that.
The OP did not say the “conditional expected value”, just the “expected value”, which is just the simple average of the 2 values available in the 1 trial in the OP, X and Y.
The reason I’ve focused on X, 2X, .5X is to show Little Nemo why that does not produce the same number as (M+2M)/2.
Which means (2X+.5X)/2 is telling Little Nemo something different about the situation than (M+2M)/2
Maybe it would clear things up for Little Nemo if you (Indistinguishable) described in basic simple language what (2X+.5X)/2 represents in that situation and why it’s a mistake to use that for envelope choice guidance. I know you probably have already done this in one of your posts, but most of them are pretty detailed and involve formulas, I’m saying just recap it in simple real world terms without formulas.
The conditional expected value, conditioned on the value of envelope 1, is the appropriate value to use, once one has opened envelope 1 and seen what is in it, knowing nothing else. That is not a mistake.
When the OP says the “expected value”, it is implicit that they want the relevant conditional expected value, just as when one says “Two cards are drawn from a deck. One is handed to you and one is handed to your opponent. You see that your card is a Jack of spades. What’s the probability that the opponent’s card is higher than your card?”, it is implicit that one wants the conditional probability P(the opponent has a higher card than you | you have a Jack of spades) = 12/51, rather than the unconditional probability P(the opponent has a higher card than you) = 1/2. This conditionalizing is implicitly understood in these problems to be the desired translation into the mathematical formalism of averages/expected value/probability. It is exactly appropriate to consider expected values conditioned on the value of envelope 1 after the information available to one is all of and no more than the value of envelope 1.
It is furthermore the case that the unconditional expected value can be obtained from the conditional expected value using the law of total expectation, so that E[Y] = E[E[Y | X]] = E[1.25 X] = 1.25 E. That seems paradoxical, but you have not given any account of it; your haranguing misses the point.
No one in this thread is under the confusion that (2X + .5X)/2 produces the same number as (M + 2M)/2, where X is defined as the value in envelope 1, Y is defined as the value in envelope 2, and M is defined as the minimum of the values in envelope 1 and envelope 2. Everyone can see that those expressions are two different random variables (in that their values are not always the same).
But (M + 2M)/2 is E[Y | M], which is not the relevant conditional expectation after opening envelope 1, as one is not in a position to know what M is. The relevant conditional expectation is E[Y | X], because one is, tautologically, in a position to know the value of X after opening envelope 1.
If you want, we can re-construct the paradox using only Y and M, though. We have that E[Y] = E[E[Y | M]] = E[(M + 2M)/2) = E[3/2 M] = 3/2 E[M]. But we also have that E[M] = E[M | Y] = E[(Y + Y/2)/2] = E[3/4 Y] = 3/4 E[Y]. So E[Y] is simultaneously equal to both 3/2 E[M] and 4/3 E[M]. Paradox!
I know, I’ve used the notation “E[…]” quite a lot through this post. Well, it’s convenient and a lot easier than typing out “expected value of…” and its variants all the time. If you have any questions about what any uses of the notation mean, just ask me. E[A] means the unconditional expected value of A. E[A | B] means f(B), where f is the function defined by f(b) = the expected value of A conditioned on B = b.
No envelope was opened in the OP.
In other words:
Given only that the lowest value in the two envelopes is m, what is the expected value of the second envelope? Well, it has a 50% chance of being m and a 50% chance of being 2m, so its expected value is (m + 2m)/2 = 3/2 m.
By the law of total expectation, this means the (unconditional) expected value of the second envelope is 3/2 times the expected value of the minimum of both envelopes.
But:
Given only that the second envelope contains y, what is the expected value of the minimum of both envelopes? Well, it has a 50% chance of being y and a 50% chance of being y/2, so its expected value is (y + y/2)/2 = 3/4 m.
By the law of total expectation, this means that expected value of the minimum of both envelopes is 3/4 times the expected value of the second envelope. Put another way, the expected value of the second envelope is 4/3 times the expected value of the minimum of both envelopes.
But this means 4/3 times the expected value of the minimum of both envelopes is equal to 3/2 times the expected value of the minimum of both envelopes. This seems odd, and is unaccounted for by the posters who mistakenly believe Little Nemo’s mistake is not looking at M. It is only actually accounted for by noting that the expected value of the minimum of both envelopes is actually infinite, and thus multiplying it by 4/3 and multiplying it by 3/2 can indeed both be taken to be infinite as well and thus the same as each other.
No, but I’m talking about the information you have before you open the envelopes.
My OP is the equivalent of “I’m playing roulette. What’s a better bet? Betting a corner at 8-1 or a sixline at 5-1?”
And your answer is the equivalent of “You can’t know which bet will pay better until you know where the ball lands.”
Fine, but… consider opening one and seeing what happens. If you open the first envelope and see k dollars, then the expected value of the second envelope, conditioned on that information, would be 1.25 k. And this works no matter what value you might see in the first envelope.
Now we can apply the law of total expectation to avoid actually having to open any envelopes: by applying the law of total expectation to the above, we find that the unconditional expected value of the second envelope is the unconditional expected value of 1.25 * (the value in envelope 1) = 1.25 * (the unconditional expected value of envelope 2).
That is the paradox. If you do not see how the law of total expectation applies to the paradox, you have not grasped what Little Nemo is getting at.