6 ms·

Isn't the expected value of both envelopes, prior to picking either, 0.5 * X, where X is the sum of the envelopes' value? So one way to argue this is a fallaci

Isn't the expected value of both envelopes, prior to picking either, 0.5 * X, where X is the sum of the envelopes' value? So one way to argue this is a fallacious problem is that you can't take an unknown (the actual value of the envelope that you picked) and pretend it's a known. The value of the envelope you are holding, prior to opening it, is 0.5 * X. The value of the other envelope is also (0.5 * (.66 * X)) + (0.5 * (.33 * X)) == 0.5 * X. This may be too ill-expressed -- but there's something wrong with treating the value of an unopened envelope as anything other than completely probabilistic. Put differently, don't you have to say 0.5 chance I picked the bigger envelope (call that A). In which case, the other envelope is 1/2A. So if I switch, I lose 1/2A. 0.5 chance I picked the smaller envelope (A/2). So if I switch, I gain 1/2A. Which means if I switch, half the time, I gain 1/2A and half the time I lose 1/2A, for an expected gain of 0 from switching.

sunir 16y ago
That is the right answer. The flaw is that the "A"s in the formula are different actual values depending on which envelope you have chosen, but the analysis has them as the same. In other words the paradoxical analysis is done relatively, which is wrong. You have to analyze probability from an absolute frame of reference.
- InclinedPlane 16y ago
  Exactly, A doesn't equal A for all parts of the formula, so it's silly to add it up.
- miloshh 16y ago
  Well, of course you're right, as is yanowitz. But there is something unsatisfying about this explanation. Intuitively it is obvious, but why is the relative reasoning incorrect? There is no formal distinction between absolute and relative quantities, and no theorem that says that expected values can only be taken from absolute quantities. There are just random variables, and these have some distributions, and they can be independent or not. Nothing prevents you from taking an expectation of a random variable that is a ratio of two other random variables. Another angle - I could say that our definition of expected value, based on weighted arithmetic average, is completely arbitrary, and instead define my own expected value G[X], based on the geometric average. Suddenly, the relative approach becomes correct: sqrt(2 * 0.5) = 1, so the expected relative improvement from switching is 1. What the hell is going on?
  - yanowitz 16y ago
    You're combining two worlds and then adding up probabilities. But there's no world where all three possible values are in play at the same time. There's only two, and you either picked high or low. It's a 50/50 shot.
    - miloshh 16y ago
      Which three worlds? After you have picked one envelope, there are two possible relative outcomes - 2 and 0.5. The average is 1.25 - there is nothing wrong with the math, the only problem is that taking an average of two relative quantities is intuitively not useful for decision making. Putting the intuition on a formal grounding is not something you succeeded with.
  - scotty79 16y ago
    > Intuitively it is obvious, but why is the relative reasoning incorrect? Because in the reasoning symbol A is used to denote "expected value of a random variable representing the amount in the envelope you picked" but later on "expected value of a random variable representing the amount in the envelope you picked provided that you picked envelope with more money" and "expected value of a random variable representing the amount in the envelope you picked provided that you picked envelope with less money". Error comes from splitting reasoning for two cases and failing to factor in the condition on which you split in your further calculation of the cases. If you solve some equation and you have to split your reasoning in two (or more) cases, the while reasoning those cases you have to remember the condition that you assumed for given case and factor it in (possibly toss away some solutions). I'd like to see some day less one dimensional way of writing down mathematic reasoning so one can see how information flows through through the course of a proof and errors like this would show up more easily.
    - miloshh 16y ago
      OK, now I see there are two slightly different possible formalizations of the problem: 1. You are told that the two envelopes contain amounts A and 2A, but you aren't told what A is. After you pick one envelope, you are allowed to open it, and then you're given the choice to switch. Here the optimal move depends on the distribution of A, and if you don't know it, you can't do much other than pick randomly. After some googling, this is the more common formalization, and it is analyzed in several math papers and blogs. 2. (The version I was assuming.) You are told that the envelopes have, say, $100 and $200. You pick one and you aren't allowed to open it yet. Now you're given the option to switch one last time. There is no problem with undefined priors and weird conditional probabilities in this version. However, the freaking paradox still holds! The expected value you get by switching is $150, no question about that. But the expected relative gain you get by switching is 1.25, there's also no question about that! This is the real paradox to me. Taking an expectation of a relative quantity is intuitively wrong, but why exactly?
techiferous 16y ago
That is the clearest, most succinct explanation I've heard. Thanks.
_delirium 16y ago
Under a frequentist interpretation of probability that's correct (and the paradox doesn't arise), but Bayesian decision theory does allow you to reason in the way the paradox proposes, starting in a situation where you've already picked an envelope, and estimating the utility of switching to the other one by multiplying the probability that you're in the high->low configuration with the value of a high->low switch, and likewise with a low->high switch. So you have 50% chance of halving your money, 50% of doubling, and thus estimated 1.25x return. Frequentist probability doesn't let you do that, because it doesn't let you say "there's a 50% likelihood I'm in this state, and 50% that I'm in the other". Instead, it says you must be in one or the other state (non-probabilistically), and the probabilities are only attached to a previous process that generated that outcome. But Bayesian probability does let you interpret the probabilities as belief in each current state, so you have a 50% belief you're in one state, and a 50% belief you're in the other, and the value of any decision is thus 0.5 * value_if_I_was_in_situation_A + 0.5 * value_if_I_was_in_situation_B. (If you don't allow that kind of computation, Bayesian decision theory has to be revised in some other cases as well.)
- deleted 16y ago
  [deleted]
- john7 16y ago
  No, Bayesian decision theory does not allow you to reason in that way. Suppose the amounts in the envelopes are A and B. A Bayesian would say: Prob(I have A) = 1/2 and Prob(I have B) = 1/2 The expected gain from switching is: Prob(I have A)x(what I'd gain from switching = B-A) + Prob(I have B)x(what I'd gain from switching = A-B) which is: 1/2 x (B-A) + 1/2 x (A-B) = 0
  - _delirium 16y ago
    Ah yes, this one can be solved that way; oops. My understanding of the literature is that variants are much more problematic, though, and require more complex restrictions on typical Bayesian probability frameworks to exorcise them. This is one good review, with a proposed solution (from a Bayesian perspective): http://books.google.com/books?id=14_ykEOAZ6AC&pg=PA49 http://books.google.com/books?id=14_ykEOAZ6AC&pg=PA49
afusiak 16y ago
Heard on the street?
robinhouston 16y ago
“You can't take an unknown and pretend it‘s a known.” is reasonable. However, the paradox still stands if you are allowed to open your chosen envelope and count the money inside before deciding whether to switch. The problem genuinely does have to do with the use of an impossible probability distribution. It’s not just a straightforward mistake (as in the wrong answer to the Monty Hall problem, say).
- yanowitz 16y ago
  Let's say you open the envelope. And there's $20. That means the set of two envelopes is either $20 and $10 or $20 and $40. In one case, the amount of money in play is $30, in the other, $60. Now, you have no idea which it is. And the fact that you're holding a $20 doesn't tell you anything. The $20 is fake information. It hasn't revealed any real info, as you knew you'd open something. The key is not to combine a world where the total envelope value is $30 and where the total envelope value is $60 which is what happens if you add up probabilities. Instead, there's a 50% chance you're holding 1/3 of the total value, 50% chance you are holding 2/3. So if you switch, half the time, you add 1/3, half the time, you lose 1/3. net expected gain from switching: 0.
- miloshh 16y ago
  "use of an impossible probability distribution" - this sounds interesting. Which exact distribution do you mean? To me it seems (though I might be wrong) that the distributions are perfectly valid. You have a pair of random variables (X, Y) that take values (100, 200) or (200, 100) with equal probability. Then E[X] = E[Y] = 150, there is no question about that. Also, E[X/Y] = 1.25, there is also no question. The only question is why E[X] is useful and E[X/Y] is not useful for our decision making - and I honestly don't know why.
  - brazzy 16y ago
    If it were (100, 200) or (200, 100) with equal probability, you'd switch iff the first envelope contains 100. Nothing interesting about that scenario. The problem states that (and is only interesting because) one of the envelopes contains twice as much as the other, but not how much, so that the amount in the first envelope tells you nothing. This is actually impossible because there is no information about how the amount (let's say the bigger of the two) is distributed, which usually implies uniform distribution. But a uniform distribution is only possible if you assume an upper limit (otherwise, what's the expected value?). If there is an upper limit, the question again becomes quite easy (you switch if the first envelope contains less than half the upper limit).
btilly 16y ago
Isn't the expected value of both envelopes, prior to picking either, 0.5 X, where X is the sum of the envelopes' value?* Only if the value of X is determinate. As matters stand for the person picking, it is not. It is simply undefined. The solution to the "paradox" is simply that expected value doesn't always mean what we'd like it to mean.
- Confusion 16y ago
  Yanowitz defines X as the sum of the value of both envelopes. That amount is not undefined and not indeterminate. The amount inside an individual envelope isn't undefined either: it has a probability distribution.
  - btilly 16y ago
    Sorry, but no. No probability distribution has been defined for the amount inside an individual envelope, and you can't just pull one out of thin air. That's not how mathematics works. If you had a probability distribution for that, you could make all sorts of arguments from it. But we're not given one, and we're given no information from which we can imply the existence of one. Therefore there is no such distribution. Period. (Interestingly, assuming such a distribution gives useful information even when it is the wrong distribution, but that is a complicated variant on the problem.) Now you can argue until the cows come home about what probability "really" means, or what is "really" true in a made up problem. Plenty of philosophers are willing to argue that with you. But the way that mathematicians look at this is quite simple. You can only work with the information you are given, and you can't make up information that you weren't given. Arguing about the fine points of reality within a piece of fiction is, not to put too fine a point on it, just plain silly. After all it is a made up example. So don't do that. We are given certain information. We look in the one envelope if we wish, and we get a dollar value. Based on that we can draw inferences about what might be in the other envelope. We can create an expected value for said envelope. All that is fine. But our intuition about what the expected value has to say about what we should do is wrong. Nothing is wrong in the math - it is our intuition that is wrong here. Our intuition is based on what happens if you encounter large numbers of similar, but independent, events. However in this case we have a singular, and extremely unlikely to be repeated, event. Therefore the basis for our intuition is lacking here. And we are mislead.
ewjordan 16y ago
Your math is right, but there's more to the story here, specifically to understand why the switching argument is wrong - this is commonly misunderstood, and has more to do with the implied "fairness" of the putting-numbers-in-the-envelope process than the calculation of the odds afterwards. The super quick version: the number drawing process can't be fair, and that makes those .5/.5 hi/low probabilities wrong in general. The real impossibility is the idea that we can "pick a random number" such that all numbers are equally probable, with no constraints on the size of that number. If this isn't intuitively obvious, then just try to normalize a uniform probability distribution over the positive real line, you'll see the problem. In order to select the two amounts, the original problem statement implies that we need to randomly pick A (which we can take as the bigger envelope) in exactly such a way, so basically, the "paradox" is dead-on-arrival, because the true prior distribution that A was picked from must be different - for example, there might be an upper bound, say $100, in which case if we see that our envelope has any number over $50 on it, we are 100% sure that we hold the larger envelope. Otherwise, without some bounds or tapering of the distribution, we can't normalize it, and we can't use it to draw numbers from, so we have no numbers in envelopes to even discuss. tl;dr version: If you can write me an algorithm that "randomly selects a number" (let's even make it an integer to make it even simpler - of course I mean an Actual Integer, not those short bit sequences that we pretend are integers when we program), such that no integer is more likely to be chosen than any other, then I've got two envelopes that I'd like to show you. :) On another note, the truly insane might notice that in the same way that we can formally sum series like 1+2+3+... = -1/12 (http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%C2%B7%C2%B7%C2%B7 http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%C2%B7%...), and 1+1+1+... = -1/2 (http://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%C2%B7_%C2%B7_%C2%B7 http://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%C2%B7_...), we might consider the probabilistic analogues of these formulas, and "formally normalize" a PDF that is uniform over the positive integers, with a constant probability of -2 for every integer (whatever that means). Similar things could be tried over the real numbers, though it gets messier because the integral analogues of those formal sums tend to come out to zero (for instance, for any integer n, the integral of x^n from 0 to infinity, in a very specific sense, is zero - perhaps I'll do a layman's post on this phenomenon at some point, it's quite useful in some cases), so normalization doesn't really work out so well. Of course, probabilities can't be negative under the usual rule set, so all of that would be pointless, wouldn't it? :)
dennisgorelik 16y ago
So basically the fallacy is in counting gains only, without counting opportunity cost [of the switch].

sunir 16y ago

That is the right answer. The flaw is that the "A"s in the formula are different actual values depending on which envelope you have chosen, but the analysis has them as the same. In other words the paradoxical analysis is done relatively, which is wrong. You have to analyze probability from an absolute frame of reference.

InclinedPlane 16y ago
Exactly, A doesn't equal A for all parts of the formula, so it's silly to add it up.
miloshh 16y ago
Well, of course you're right, as is yanowitz. But there is something unsatisfying about this explanation. Intuitively it is obvious, but why is the relative reasoning incorrect? There is no formal distinction between absolute and relative quantities, and no theorem that says that expected values can only be taken from absolute quantities. There are just random variables, and these have some distributions, and they can be independent or not. Nothing prevents you from taking an expectation of a random variable that is a ratio of two other random variables. Another angle - I could say that our definition of expected value, based on weighted arithmetic average, is completely arbitrary, and instead define my own expected value G[X], based on the geometric average. Suddenly, the relative approach becomes correct: sqrt(2 * 0.5) = 1, so the expected relative improvement from switching is 1. What the hell is going on?
- yanowitz 16y ago
  You're combining two worlds and then adding up probabilities. But there's no world where all three possible values are in play at the same time. There's only two, and you either picked high or low. It's a 50/50 shot.
  - miloshh 16y ago
    Which three worlds? After you have picked one envelope, there are two possible relative outcomes - 2 and 0.5. The average is 1.25 - there is nothing wrong with the math, the only problem is that taking an average of two relative quantities is intuitively not useful for decision making. Putting the intuition on a formal grounding is not something you succeeded with.
- scotty79 16y ago
  > Intuitively it is obvious, but why is the relative reasoning incorrect? Because in the reasoning symbol A is used to denote "expected value of a random variable representing the amount in the envelope you picked" but later on "expected value of a random variable representing the amount in the envelope you picked provided that you picked envelope with more money" and "expected value of a random variable representing the amount in the envelope you picked provided that you picked envelope with less money". Error comes from splitting reasoning for two cases and failing to factor in the condition on which you split in your further calculation of the cases. If you solve some equation and you have to split your reasoning in two (or more) cases, the while reasoning those cases you have to remember the condition that you assumed for given case and factor it in (possibly toss away some solutions). I'd like to see some day less one dimensional way of writing down mathematic reasoning so one can see how information flows through through the course of a proof and errors like this would show up more easily.
  - miloshh 16y ago
    OK, now I see there are two slightly different possible formalizations of the problem: 1. You are told that the two envelopes contain amounts A and 2A, but you aren't told what A is. After you pick one envelope, you are allowed to open it, and then you're given the choice to switch. Here the optimal move depends on the distribution of A, and if you don't know it, you can't do much other than pick randomly. After some googling, this is the more common formalization, and it is analyzed in several math papers and blogs. 2. (The version I was assuming.) You are told that the envelopes have, say, $100 and $200. You pick one and you aren't allowed to open it yet. Now you're given the option to switch one last time. There is no problem with undefined priors and weird conditional probabilities in this version. However, the freaking paradox still holds! The expected value you get by switching is $150, no question about that. But the expected relative gain you get by switching is 1.25, there's also no question about that! This is the real paradox to me. Taking an expectation of a relative quantity is intuitively wrong, but why exactly?

techiferous 16y ago

That is the clearest, most succinct explanation I've heard. Thanks.

_delirium 16y ago

Under a frequentist interpretation of probability that's correct (and the paradox doesn't arise), but Bayesian decision theory does allow you to reason in the way the paradox proposes, starting in a situation where you've already picked an envelope, and estimating the utility of switching to the other one by multiplying the probability that you're in the high->low configuration with the value of a high->low switch, and likewise with a low->high switch. So you have 50% chance of halving your money, 50% of doubling, and thus estimated 1.25x return. Frequentist probability doesn't let you do that, because it doesn't let you say "there's a 50% likelihood I'm in this state, and 50% that I'm in the other". Instead, it says you must be in one or the other state (non-probabilistically), and the probabilities are only attached to a previous process that generated that outcome. But Bayesian probability does let you interpret the probabilities as belief in each current state, so you have a 50% belief you're in one state, and a 50% belief you're in the other, and the value of any decision is thus 0.5 * value_if_I_was_in_situation_A + 0.5 * value_if_I_was_in_situation_B. (If you don't allow that kind of computation, Bayesian decision theory has to be revised in some other cases as well.)

deleted 16y ago
[deleted]
john7 16y ago
No, Bayesian decision theory does not allow you to reason in that way. Suppose the amounts in the envelopes are A and B. A Bayesian would say: Prob(I have A) = 1/2 and Prob(I have B) = 1/2 The expected gain from switching is: Prob(I have A)x(what I'd gain from switching = B-A) + Prob(I have B)x(what I'd gain from switching = A-B) which is: 1/2 x (B-A) + 1/2 x (A-B) = 0
- _delirium 16y ago
  Ah yes, this one can be solved that way; oops. My understanding of the literature is that variants are much more problematic, though, and require more complex restrictions on typical Bayesian probability frameworks to exorcise them. This is one good review, with a proposed solution (from a Bayesian perspective): http://books.google.com/books?id=14_ykEOAZ6AC&pg=PA49 http://books.google.com/books?id=14_ykEOAZ6AC&pg=PA49

afusiak 16y ago

Heard on the street?

robinhouston 16y ago

“You can't take an unknown and pretend it‘s a known.” is reasonable. However, the paradox still stands if you are allowed to open your chosen envelope and count the money inside before deciding whether to switch. The problem genuinely does have to do with the use of an impossible probability distribution. It’s not just a straightforward mistake (as in the wrong answer to the Monty Hall problem, say).

yanowitz 16y ago
Let's say you open the envelope. And there's $20. That means the set of two envelopes is either $20 and $10 or $20 and $40. In one case, the amount of money in play is $30, in the other, $60. Now, you have no idea which it is. And the fact that you're holding a $20 doesn't tell you anything. The $20 is fake information. It hasn't revealed any real info, as you knew you'd open something. The key is not to combine a world where the total envelope value is $30 and where the total envelope value is $60 which is what happens if you add up probabilities. Instead, there's a 50% chance you're holding 1/3 of the total value, 50% chance you are holding 2/3. So if you switch, half the time, you add 1/3, half the time, you lose 1/3. net expected gain from switching: 0.
miloshh 16y ago
"use of an impossible probability distribution" - this sounds interesting. Which exact distribution do you mean? To me it seems (though I might be wrong) that the distributions are perfectly valid. You have a pair of random variables (X, Y) that take values (100, 200) or (200, 100) with equal probability. Then E[X] = E[Y] = 150, there is no question about that. Also, E[X/Y] = 1.25, there is also no question. The only question is why E[X] is useful and E[X/Y] is not useful for our decision making - and I honestly don't know why.
- brazzy 16y ago
  If it were (100, 200) or (200, 100) with equal probability, you'd switch iff the first envelope contains 100. Nothing interesting about that scenario. The problem states that (and is only interesting because) one of the envelopes contains twice as much as the other, but not how much, so that the amount in the first envelope tells you nothing. This is actually impossible because there is no information about how the amount (let's say the bigger of the two) is distributed, which usually implies uniform distribution. But a uniform distribution is only possible if you assume an upper limit (otherwise, what's the expected value?). If there is an upper limit, the question again becomes quite easy (you switch if the first envelope contains less than half the upper limit).

btilly 16y ago

Isn't the expected value of both envelopes, prior to picking either, 0.5 X, where X is the sum of the envelopes' value?* Only if the value of X is determinate. As matters stand for the person picking, it is not. It is simply undefined. The solution to the "paradox" is simply that expected value doesn't always mean what we'd like it to mean.

Confusion 16y ago
Yanowitz defines X as the sum of the value of both envelopes. That amount is not undefined and not indeterminate. The amount inside an individual envelope isn't undefined either: it has a probability distribution.
- btilly 16y ago
  Sorry, but no. No probability distribution has been defined for the amount inside an individual envelope, and you can't just pull one out of thin air. That's not how mathematics works. If you had a probability distribution for that, you could make all sorts of arguments from it. But we're not given one, and we're given no information from which we can imply the existence of one. Therefore there is no such distribution. Period. (Interestingly, assuming such a distribution gives useful information even when it is the wrong distribution, but that is a complicated variant on the problem.) Now you can argue until the cows come home about what probability "really" means, or what is "really" true in a made up problem. Plenty of philosophers are willing to argue that with you. But the way that mathematicians look at this is quite simple. You can only work with the information you are given, and you can't make up information that you weren't given. Arguing about the fine points of reality within a piece of fiction is, not to put too fine a point on it, just plain silly. After all it is a made up example. So don't do that. We are given certain information. We look in the one envelope if we wish, and we get a dollar value. Based on that we can draw inferences about what might be in the other envelope. We can create an expected value for said envelope. All that is fine. But our intuition about what the expected value has to say about what we should do is wrong. Nothing is wrong in the math - it is our intuition that is wrong here. Our intuition is based on what happens if you encounter large numbers of similar, but independent, events. However in this case we have a singular, and extremely unlikely to be repeated, event. Therefore the basis for our intuition is lacking here. And we are mislead.

ewjordan 16y ago

Your math is right, but there's more to the story here, specifically to understand why the switching argument is wrong - this is commonly misunderstood, and has more to do with the implied "fairness" of the putting-numbers-in-the-envelope process than the calculation of the odds afterwards. The super quick version: the number drawing process can't be fair, and that makes those .5/.5 hi/low probabilities wrong in general. The real impossibility is the idea that we can "pick a random number" such that all numbers are equally probable, with no constraints on the size of that number. If this isn't intuitively obvious, then just try to normalize a uniform probability distribution over the positive real line, you'll see the problem. In order to select the two amounts, the original problem statement implies that we need to randomly pick A (which we can take as the bigger envelope) in exactly such a way, so basically, the "paradox" is dead-on-arrival, because the true prior distribution that A was picked from must be different - for example, there might be an upper bound, say $100, in which case if we see that our envelope has any number over $50 on it, we are 100% sure that we hold the larger envelope. Otherwise, without some bounds or tapering of the distribution, we can't normalize it, and we can't use it to draw numbers from, so we have no numbers in envelopes to even discuss. tl;dr version: If you can write me an algorithm that "randomly selects a number" (let's even make it an integer to make it even simpler - of course I mean an Actual Integer, not those short bit sequences that we pretend are integers when we program), such that no integer is more likely to be chosen than any other, then I've got two envelopes that I'd like to show you. :) On another note, the truly insane might notice that in the same way that we can formally sum series like 1+2+3+... = -1/12 (http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%C2%B7%C2%B7%C2%B7 http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%C2%B7%...), and 1+1+1+... = -1/2 (http://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%C2%B7_%C2%B7_%C2%B7 http://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%C2%B7_...), we might consider the probabilistic analogues of these formulas, and "formally normalize" a PDF that is uniform over the positive integers, with a constant probability of -2 for every integer (whatever that means). Similar things could be tried over the real numbers, though it gets messier because the integral analogues of those formal sums tend to come out to zero (for instance, for any integer n, the integral of x^n from 0 to infinity, in a very specific sense, is zero - perhaps I'll do a layman's post on this phenomenon at some point, it's quite useful in some cases), so normalization doesn't really work out so well. Of course, probabilities can't be negative under the usual rule set, so all of that would be pointless, wouldn't it? :)

dennisgorelik 16y ago

So basically the fallacy is in counting gains only, without counting opportunity cost [of the switch].