Yes, thank you to @Pasta for a sufficiently detailed exposition of the infinite series problem and the realted unequal distribution problem. With a noble assist by the back and forth between @Jay_Z and @Chronos. It all makes sense now.
And, as @Danger_Man points out just above (which @Chronos also said awhile ago) there are simple and complicated versions of the problem with little agreement on who is talking about exactly what.
Quite so.
If the two values within are x and 2x and both envelopes have the same text and you’re not allowed to open them, then it doesn’t matter which one you open.
The ‘incorrect expectation value calculations’ arise from not realising that the two values are x and 2x; but people think there are somehow three values; x, 2x and x/2!
I guess it comes down to, are the envelopes provided in pairs, having been sorted into sets of two that satisfy the written conditions before being distributed, or are they essentially sets (perhaps even open sets) of three or more, where only the first envelope is provided initially (and presumably not at random), and then the second envelope is provided on request, either at random (but from a prior pool of envelopes that will comply with the written conditions) or by a thinking agent?
Consider…
Scenario one:
Scenario two:
And I’m sure there are more scenarios you could come up with, depending on what is assumed about G and the conditions under which they provide envelopes.
Great, except that this isn’t true. It’s really not all that subtle. People have repeatedly tried to explain this but it appears that people’s “intuition” just overcomes the actual logic.
It depends on whether the second envelope is provided with the first, or whether it is provided only when requested from an unknown pool of other envelopes.
FWIW, as I read the OP it’s the first scenario (both envelopes provided together, which I described in Scenario 1 above and in my much earlier solution to the problem).
Both envelopes bear the statement “The other envelope is either half or twice this one”. If either one of those statements is true, the other one is, also.
Of course. But the question remains, how does the Game master arrange the problem to make it so?
True, you will get different distributions from “pick one value, flip a coin, and either double it or halve it to get the other value” or from “pick one value, double it to get the other value, and then shuffle the two envelopes”. The core point, though, remains “There are many possible distributions, and strategy in the game depends on first knowing or guessing what distribution you have”.
Puts a five in one envelope and a ten in the other?
My post above ignored some simpler cases, but maybe I’ll add a few words there (not saying anything new, but maybe saying something differently.)
If we literally know that the envelopes have to contain $5 and $10, and they are shuffled around, then the text on the back is incorrect or, perhaps more subtly, being interpreted incorrectly. The envelopes say “the other contains either twice or half this one’s amount” – semantically okay – but what they should say (or how they should be read) is, “if this is the $10 envelope, the other contains half; if this is the $5 envelope, the other contains twice”. The first terse statement is incomplete when it comes to usefulness in calculating the expectation value for switching. Saying “the expectation value of switching is 1.25 times what I’m currently holding” is wrong here because “what I’m currently holding” is a random quantity that is correlated with the other envelope. I do not truly have a chance to double my value if I have the high value, and I do not truly have a chance to half my value if I have the low value, and I have to average over both possibilities when calculating my return on switching.
If this fixed-two-value case is settled, then we can try to make a three-value case: $5, $10, $20. This doesn’t remove the subtlety in any way, though. All the same arguments hold, and I have to average over outcomes for all three possible values of “what I currently hold” when calculating the value of a switch. Whatever random value we currently hold is correlated with whatever could be in the other envelopes, and we can’t pretend it isn’t when calculating expectation values.
Going to four, or five, or any discrete or continuous infinite set of possible values still doesn’t remove this essential feature. You always have in the problem some notion of a distribution of initial numbers the envelopes are populated (or populatable) with.
This is where my previous post picks up the story, concluding that no matter what the underlying distribution is, a calculation that correctly includes that distribution by calculating the sum (or integral) of the value of switching for each possible holding, with each possibility weighted by the probability of having that holding and the probability that that holding is the smaller or larger envelope, then the EV for switching is always calculated to be the same as the EV for staying pat.
Briefly back to the three-value case (or similar): if you try to reason that the statement “the other contains half or twice this amount” has to be able to be true for some set up of two presented shuffled envelopes, then when you say “If I happen to hold 10, the other has to hold 5 or 20, and both are completely allowed” you have to logically step to saying “…which means I could also be holding 20, so my calculation should include that case, and in that case the other has to hold 10 or 40 and both are completely allowed (and still equally likely)”; and at this point you have to logically say “…which means I could also be holding 40 just as well, which means the other has 20 or 80…”, etc. Thus, applying the notion that the other could truly be double or half with equal probability is identical to saying that values all the way to infinity (and all the way to zero) are allowed, which is an ill-defined probability distribution. It cannot be the actual distribution of possible values in envelopes. The resolution is simply that for any realizable distribution of possible values for the envelopes, the probability that the other envelope is double or half is not in general equal, and the probabilities of the two options (higher or lower) depends on which value I’m currently holding, and which value I’m holding is random based on the distribution that was used to fill the envelopes. But critically, it doesn’t matter what that distribution is – a proper calculation that accounts for it will necessarily yield zero expected return on switching.
That was great. Thank you.
I’ll say that this snip really shined the spotlight on it for me.
The statement on the envelopes is the truth, but not the whole truth. And the part the (naïve) reader unwittingly invents to supply the unnoticed missing part is what leads them astray.
The gamesmanship of nearly every word puzzle / paradox / brain teaser problem is to include sentences that seem both complete, unambiguous, and grammatical, but which on closer examination either leave out critical info or can be reinterpreted to mean something rather different.
This puzzle lives and dies on the hidden-in-plain-sight-ambiguity trick.
In a way this error is mathematically akin to the super-naïve approach to probability that says “It’ll either ran tomorrow at my house or it won’t. Since there are two possibilities that means the odds must be 50/50.” Well … not really if e.g. you live in an arid climate or in a rain forest.
This puzzle just requires more sophistication to notice the same kind of error: assuming a distribution not actually in evidence.
I’m just not understanding why the wording is misleading or incomplete. One envelope has to contain twice what’s in the other. Which means the remaining envelope has to contain half what’s in the first. You just don’t know which is which.
One envelope can’t contain literally any value you can think of, with the other being either twice or half that amount. It can only contain twice or half of the other, which is an unknown but fixed amount.
See what I outlined as scenario 2:
But note that I do hold to this description of the OP:
ETA: Really, this problem is much more about set theory (and how new evidence may inform our understanding of the set and its boundaries) than logic in general or expected value specifically.
Right. Per the OP, there are two envelopes in front of you.
But I’m also not following how this changes with the second envelope being offered in some different manner, unless the hypothetical is tortured beyond recognition (e.g., what if the guy providing the envelopes deliberately only offers a second envelope with half every time, etc.).
No, just that, while the Player (P) only ever gets access to two envelopes, it is possible that the Guy (G) providing the envelopes has access to at least two, in addition to the first envelope already in P’s possession, of which G provides one additional, presumably at random, upon P’s request. So it’s not just that the second envelope could hypothetically be half or double, it’s that there exists in fact (unknown to P) an envelope that is half the first envelope, and an envelope that is double the first envelope. A set of three, rather than of two, but of which P will only ever have access to two.
Even so, for G to be able to make a random choice to provide a second envelope from a set of three, it must be the case that the original envelope already in P’s possession is half of one and double the other, otherwise there is a possibility that the statement on the envelopes would not be correct when the second envelope is provided to P. If, for example, P starts with the smallest of three envelopes, and then by chance G gives P the largest envelope as their second upon request, then in fact the second envelope is not double the first, it is four times the first.
To me, this turns it into a completely different hypothetical.
I agree. It doesn’t match the scenario as described in the OP, but it explains how a slightly different framing changes the problem significantly.
To be sure we’re talking about the same scenario:
Two envelopes, one with $5 and one with $10. I shuffle them and hand you one. Both envelopes say “The other envelope either holds twice this one or half this one.” All fine, as you say.
It’s misleading, though, because which of those two cases is true depends precisely on what is in your current envelope, yet a naive expectation value calculation would have you act as if that dependence wasn’t present. It suggests that you can say “I have some value x, so the other has 2x or 0.5x, and thus switching has value 1.25x.” But the correct calculation has to include the fact that the two options for the other envelope are correlated with what value I have in my current envelope. So, the correct calculation in this simple case is “If I have $5 now then switching gains me $5. If I have $10 now then switching loses me $5. I don’t know which case it is so I need to average over them (weighting by the probability of each case, which is equal here). So, I have an equal chance of gaining $5 or losing $5, so the EV of switching is zero.”
Adding smoke-and-mirrors by having the initial values not be so cut and dry (i.e., not just $5 and $10 but some random determination scheme) makes it easy to slip into thinking the 2x and 0.5x options are somehow suddenly not correlated with one’s current holding. But that’s not correct. The correlation is essential and cannot be removed.
Yes, the misunderstanding depends on an illogical interpretation. The description of the envelopes is entirely accurate and the only info required. The “paradox” is just bad math.
I think that another lesson of this problem is that, for any calculation ever, there are multiple ways to do the calculation. And all valid calculations will yield the same answer. So when yo have a choice of doing an easy calculation that you know is valid, or a difficult calculation that’s so complicated that it’s hard to be sure if it’s valid, just do the easy calculation.
To whit, in this case: If you have two identically-labeled envelopes and can’t look in either of them, then it doesn’t matter which of the two you pick. It’s really easy to say that and stop there. If you do some other calculation that gives a different answer, then you know it’s wrong. You could try to figure out precisely how it went wrong… or you could just not do the calculation that way.
This same principle often shows up with perpetual-motion machines. Conservation of energy is easy. So when someone shows you a purported perpetual-motion machine and asks why it won’t work, the simple answer is just to say “It violates conservation of energy”.