Blaster Master: The assumption that there is a 50% chance that you’re playing one game or the other is built into the original paradox (which is how Little Nemo’s expected value analysis gives a result of 1.25X). My point is that even with that assumption, there is no paradox, as long as you talk about X consistently.
If the likelihood of the two games is not equal, all it does is change the conditional probabilities. If I knew that my friend didn’t have $200, and I opened an envelope and saw $100, the probability of the other envelop containing $200 given that mine contained $100 is 0, and the analysis would correctly suggest that you not switch envelopes.
To add some clarity, to my point, in the generalized case, where we don’t open the envelope, we don’t know anything about the distributions before hand except that we either have V or 2V with equal probability, so when we say X, we can’t say the other one has either half as much or twice as much, as if X is V, the other envelop MUST have 2V, and if X is 2V the other envelope MUST have V, so the expected gain is zero.
In the $100 case, it’s a bit more interesting becase opening the envelope doesn’t clarify between the two cases, whether we switch depends entirely upon our belief about the likelihood of each distribution at the beginning of the game, as the probability of seeing $100 in our envelop is exactly equal regardless of whether we’re playing 50/100 or 100/200. So, if before we thought they were equally likely, then we can use expected value to say we should switch; if, however, we think there’s > 80% chance we’re playing 50/100, we shouldn’t switch.
I don’t see how you get that value though; how is it built in? All we know is that one envelope contains V and the other contains 2V, but the combinations that fit that description are infinite… 1/2, 2/4, 50/100, etc. If we assume all are equally likely, fine, but practically speaking, we know that they’re not. When we open the evevelope and see what’s inside, we eliminate all but two of those possibilities, but unless those two possibilities were equally likely before hand, they won’t be equally likely afterward. And if all values are equally likely, then no matter which one we open, and what it’s contents are, we should switch.
This is where the second part comes in, we have some initial idea, based upon what show we’re on, about what some likely and unlikely values are for what’s in the envelopes. If I have some knowledge about their typical prizes, then I already have an idea of what distributions are likely and unlikely before I open the envelopes, and seeing the value in the envelope I opened doesn’t change how likely either distribution is relative to the other. Instead, I know where it is relative to the expected prize, and know that if it’s relatively close, I probably have 2V and shouldn’t change, and if it’s relatively much less, I probably have V and should change, the expected value of either envelop is still (2V+V)/2.
Apples and oranges. Your scenario is completely different than the scenario in the OP.
In your scenario, there are two envelopes. They have, respectively, either $10 and $5, or $20 and $10. You pick one. In your description, you always pick the one with $10. The expected value is $12.50.
But the apparent paradox only arises if you try to use the expected value logic to argue that you should switch back to the original envelope. If you switch to the 1.25Y envelope, the alternative is not to switch back to another envelope that suddenly has an expected value of 1.5625Y – it’s to switch to an envelope that contains Y, which you clearly wouldn’t do.
In other words, I don’t really see any paradox in saying “if you don’t open the envelope, don’t switch, if you do open the envelope, always switch*.” Those are two different games.
Based on your expectation of what the possible prizes could be, of course.
Fair enough, though I do want to clarify (not for you, but for anyone reading) that the reason not to switch if you’re playing the “don’t open” version, isn’t because it hurts you, but because the expected value is the same, so it just doesn’t matter.
Either way, like the link in the OP indicates, the apparent paradox rises in conflating the probabilities, much like with the Monty Hall problem.
Since OP has been correctly answered, miscellaneous comments are in order.
(1) No one’s mentioned the odd/even effect. Opening one evelope and seeing 5 cents I know the other envelope must have the double amount, rather than half, since U.S. mints no half-pennies.
Thus the benefactor will never put an odd amount into either envelope, lest our response be too obvious. On seeing 10 cents, I know the other envelope has 20 cents, since 5 would be an odd amount. Therefore the benefactor will never put an amount that’s not a multiple of four, and seeing 20 cents I know the other has 40 cents since 10 cents isn’t a multiple of 4.
Continuing in this way, almost all amounts are forbidden. Either both envelopes contain zero (:eek:) orOn Wednesday they came to hang the prisoner … and he wasn’t expecting it!.
(2) For the opened envelope case, with the odd/even effect ignored, best strategy is to switch whenever X < Threshold. Determining Threshold is left as an exercise.
(3)
I’m going to call the Fraud Hotline on this. Oh, we know Buck Godot is really Warren Buffet posting under a false name and would happily pay $10 billion if he lost it, but what about $10 trillion? Or better yet, for those of us who’ve always dreamed of winning the lottery in a big way, 10 sextillion tons of rhodium?
Just in case it’s not clear to everyone, this is not the same as the “unexpected hanging paradox” because the benefactor can know which envelope is which and the recipient can know the benefactor knows so if the amounts are $5 and $10, he can give the recipient the $10 envelope who cannot assume $5 is not in the other one since he will never be given that first.
— or he can go and find an old half cent piece put $50.01 in one envelope and $25 plus the half-cent in the other and then take a picture of your surprised face when you open that envelope that you were absolutely sure had the larger amount.
I’m not following the people who say it makes a difference what X is and you can’t express the content of the other envelop in terms of X.
Let’s say we do the simpler version where we open the first envelope. I open it and find $100 inside. I now know that the other envelope contains either $50 or $200 with an expected value of $125.
Or let’s say I open an envelope and find two dollars inside. I now know the other envelope contains either one dollar or four dollars with an expected value of $2.50.
Or let’s say I open the envelope and find $8,000,000 inside. I now know the other envelope contains either $4,000,000 or $16,000,000 with an expected value of $10,000,000.
And so on. X is a variable only in the sense that you can assume any amount is in the first envelope. The values of 2X and X/2 and the expected value are always fixed in relation to that.
We’re talking about how you are changing X when you switch envelopes. If you count X as the same, the scenario breaks down . The expected return is determined as 1.5X for envelope 2 before the first switch, and then is determined to be 1.5X for envelope 1 before the second switch. Those are the same expected return, so obviously it makes no difference which envelope you open.
For 1.5X in envelope 1 to be greater than the 1.5X in envelope 2, you have to be changing the value of X. What you are likely doing is assuming that, since you determined in the first switch that envelope 2 is better, that it must have more money in it. Thus, you are increasing X to 2X. This is an unwarranted assumption.
Think about what expected return actually means: it means that, if you repeat the same test over and over, you will average out to 1.5X. The whole point is that you are picking randomly, and that, by doing so, you will tend to pick the X envelope as many times as the 2X envelope. This has no indication on how well you will do in a single iteration.
The only way expected return makes sense is if one choice has a better expected return than the other, which, again, you’ve already determined is not the case. You are using the wrong tool to solve this problem.
For anyone who wants to get a better understanding of randomness and probability check out The Drunkard’s Walk by Leonard Mlodinow. He explains why people typically misunderstand these things. He explains this very thing in the book.
Not that the OP should be able to tell. There are at least three different inconsistent answes floating around, at least two of which have proponents who realize they disagree with each other.
When you say that your envelope contains X, so the other envelope must contain either 2X or X/2, you’re introducing a value that isn’t in the problem. You’ve now got three possible outcomes: X/2, X, and 2X. But the problem has only TWO outcomes: M and 2M. So, you’ve introduced a non-possible value and that’s what’s screwing up the calculation.
Expressed another way: suppose that one envelope contains $40, and the other contains $80. Now, YOU don’t know that, but that’s the reality. You pick one envelope, say it’s the $40 envelope (although you don’t know that). The probability of finding $20 in the other envelope is zero – it’s not (1/2)($20). Your calculation has introduced a non-possible, third value. That’s why it’s flawed.
Better approach is, assume that the envelopes contain M and 2M, some fixed amount. You pick an envelope:
(a) There’s a 50% chance that you picked M, and the other envelope contains 2M.
(b) There’s a 50% chance that you picked 2M, and the other envelope contains M.
First, what’s your expected value from the envelope you picked?
Answer: (1/2)(2M) + (1/2)M = 1.5M, as expected. (The expected value is halfway between the two outcomes; note that you can’t actually win that amount!)
What’s the expected value in the other envelope?
Answer: (1/2)2M + (1/2)M = 1.5M … the same.
What’s the probability of a gain by switching envelopes?
IF YOU PICKED M: then the other contains 2M, and switching gives a gain of M. Expected value = Probability X Outcome = (1/2)M
IF YOU PICKED 2M: the the other contains M, and switching gives a LOSS of M (or a gain of -M): Expected value = Probability X Outcome = (1/2)(-M)
Overall expected gain from switching: (1/2)M + (1/2)(-M) = 0, as one would think. There’s no advantage/disadvantage to switching.
Your arithmetic here is correct only if the $50 and $200 are each 50% likely. There’s no basis to know that. (If the point isn’t clear, you might suspect Mr. Benefactor is too much the tightwad to put $200 in. Or, contrariwise, too generous to put in just $50. Without further info, we may be unable to provide any better estimate than 50%-50%, but that doesn’t make that estimate correct.)
Strange as it may seem, reasoning for the unopened and opened envelopes must be slightly different; don’t confuse the cases.
I think I see what you’re saying here but this seems like a trivial issue. I can resolve this by using X as the unknown content of my first envelope and determine that the expected value of the other envelope is 1.25X. Then when I switch envelopes I’ll use Y as the unknown content of my new envelope and determine that the expected value of my original envelope is 1.25Y.
I follow what you’re saying here but it didn’t really clear up the paradox for me. What it seems you’ve done here is simply avoid the paradox.
The original paradox, as I saw it, was that you have what appears to be an easily solved problem. You use what should be routine mathematical operations to solve it. You arrive at a solution but you realize the solution is wrong.
But you only realize this because the problem was so simple. If the problem had been more complicated, the solution might not have been obvious and you would have worked your way through it to arrive at what would appear to be a real answer.
So the paradox is why mathematical operations that should have led to a correct answer did not.
Now I agree that your post did arrive at the correct answer. But you did so by using a different path. You essentially said “We can see that Method A gave us the wrong answer so let’s try Method B. And look, this gave us the correct answer.”
So what you’re saying is true but it doesn’t make it clear (at least to me) why Method A didn’t work.
I realize it probably appears that I’m just being stubbornly argumentative in rejecting everyone’s help. But that’s not the case. I’m genuinely confused by this paradox and I really am hoping that somebody can explain it in a manner that will give me that “ah hah” moment of enlightenment.