7 ms·

Two Envelopes Problem

112 points by atomon 16y ago

omaranto 16y ago
I like Raymond Smullyan's version even better, it removes all probability issues and is so simple you feel it must be easy to figure out. The setup is the same: two envelopes, once with twice as much money as the other, you pick one and before you open you are offered to switch. If you do switch you might either win money or lose money relative to what was in the envelope you chose first. Let's compare the possible gain to the possible loss (without worrying about the probability of having a gain or a loss): 1. The possible gain is equal to the possible loss. Proof: The two envelopes contain x and 2x, so you'd gain x if switching from x to 2x, and you'd lose x if switching from 2x to x. The possible gain and the possible loss are both equal to x. 2. The possible win is larger than the possible loss. Proof: Say the envelope you chose first has y. Then the other envelope has either y/2 or 2y. So if you gain money by switching you gain 2y - y = y, but if you lose in the switch, you lose y - y/2 = y/2. Clearly y > y/2. This version kept up at night for days when I first read it. I never figured it out --not that I had any chance of doing so, given that Smullyan couldn't figure it out either--, I just got used to not understanding it...
- jeromec 16y ago
  This doesn't eliminate probability, and it's really the same flaw, just with different wording. The flaw is still that the variable is undefined (or infinite probability) in one part of the word problem, but treated as defined in another. So, start with when you first chose the envelope which has y. At this point "y" is undefined, and can be the higher or lower amount. That's the key. Yet, you then continue to propose that by switching you gain 2y - y = y. That equation is not okay, because "y" could have been the higher amount, and the equation breaks down. Once you assert the equation holds true, you have thus defined the original value of "y" from that first envelope choice (i.e. "y" was lower). The same flaw surfaces by asserting your switch to lose equation is true, as well. Edit: BTW, the reason #1 works okay is because you define upfront the variable value ("x") as being either the higher or lower of the two possibilities, before constructing your equation(s).
  - omaranto 16y ago
    You mention this doesn't eliminate probability but don't explain why you think that might be the case. I just meant that the argument Smullyan presents does not involve probabilities, random variables, distributions, expected values, etc. (This can be verified by reading it again and checking those concepts do not appear. ;)). I don't find your argument convincing either, maybe it helps to remove the "undefined variable" you complain about: The first envelope chosen has some definite amount of money, let's say $10 (but the same argument can be adapted to any amount). Then the other envelope must have either $5 or $20. If it is $5, switching loses you $5. If it is $10, switching gains you $10. Imagine actually doing this experiment with the two envelopes, opening the first and seeing $10. Wouldn't you say the other contains either $5 or $20?
    - jeromec 16y ago
      It doesn't eliminate probability because it's the exact same envelope experiment. ;) In other words, there is still a 50% probability of gaining or losing. My post above isn't an argument, it's an explanation. The "undefined variable" is established by your own proposal, i.e., in #2 you first choose an envelope which has an amount we call "y". At that point "y" is undefined, because you specifically explain it could be either the higher or lower amount. I think your confusion lies in not considering carefully enough the meaning of your wording. In other words, you're performing a play on words to get a specific result, akin to the "heads I win, tails you lose" word play. Let me try to explain further below: Imagine actually doing this experiment with the two envelopes, opening the first and seeing $10. Wouldn't you say the other contains either $5 or $20? No. The other envelope contains the amount the person who prepared the envelopes placed in it. Try defining both values explicitly and this issue may become clearer. Imagine one envelope contains $10 and the other contains $5. Go through your above proposed equations using strictly those values. Remember, in a real experiment that would be completely valid.
    - Confusion 16y ago
      I just meant that the argument Smullyan presents does not involve probabilities, random variables, distributions, expected values, etc. (This can be verified by reading it again and checking those concepts do not appear. ;)). The fact that he doesn't uses those words does not mean he doesn't use those concepts. The Smullyan version is simply less precisely stated and anyone solving it would first have to go through the trouble to convert the phrasing to incorporate the usual, well defined mathematical words. Solving it any other way is just handwaving.
- Amnon 16y ago
  Statement 1 isn't true. If you have 10$ in the envelope, and it's the smaller envelope, you gain 10$. But if it's the larger envelope you don't lose 10$. The fallacy is that "x" refers to different things in each case. Statement 2 is true but it isn't relevant, since you don't know the posterior probability given y. If the envelope contains the smaller amount of money then there is no way you can lose -- it's not a 50-50 chance.
deleted 16y ago
[deleted]
klochner 16y ago
I got that in a tech interview and hadn't seen it before. Dirty.
- dotBen 16y ago
  In what context was the interview (ie type of position). I'm all for logical problem solving questions in interviews, even ones that can't be solved but are interesting to see how people tackle them. However this one doesn't seem to lend itself much to programming or true problem solving - just probability theorem. In other words, I'm not sure what useful info I'd get out of a candidate by asking them to work on this (other than their understanding of probability, which is not that useful for most programming jobs)
aliston 16y ago
Interesting... here's my "gut reaction" as a non-statistician. The problem is that 'A' has a probabilistic component as well. If you let X be the "lesser of the 2 dollar amounts" and let A = X/2 + 2X, then you come out with equal expected values for switching and non-switching.
imajes 16y ago
I think this is better posed as two random-drawn sets of numbers for the first drawing of an as-of-yet un-played lottery. Then, there's no value in considering extraneous data such as the mass of the envelopes, looking through them or other potential evidence. Instead, you're left with two identical chances. Since there is no data to suggest that either has an advantage over the other, you have a probability of exactly 0.5 that you have the better chance of winning. On the question of the switch, it's important to consider that you have no new data about the one you picked, so the odds are _exactly the same_ as they were at the beginning: there is no conditionality, nothing has changed since you picking it. The probability then of the winning ticket is still 0.5. This is where the infinite probability problem kicks in. I don't think this is an actual problem needing to be solved, unless we can find a way to mathematically express the probability modifiers of our gut feeling, which is impossible, as it comes into our own personal experiences. Still, Gladwell might have something to say about it :)
jatenate 16y ago
The swapping indefinitely does not work. For the first iteration, if I have A in my envelope, then yes, the other envelope will either have 1/2A or 2A, meaning the expected value is indeed 5/4A. The switching argument wrongly assumes this is again the case after I switch. After I have switched, the value of the original envelope is unchanged (meaning it is A with probability 1), so I should not switch back.
- bmm6o 16y ago
  Are you seriously arguing that you should switch exactly once?
btilly 16y ago
First a random observation. People use "expected value" to evaluate risky decisions far too often. In many investment situations a more appropriate measure is "expected value of the log of my worth". (See http://elem.com/~btilly/kelly-criterion/ http://elem.com/~btilly/kelly-criterion/ for an explanation of why.) That measure, in this problem, says that you should be indifferent about switching. Admittedly that is coincidence - the investment reasoning for that rule has little to do with why that rule works for this problem. But it turns out that there is an even crazier twist to this problem. Change the problem to say that you're allowed to look at the envelope you receive before deciding whether to switch. Would you believe that there is an algorithm for deciding whether to switch that results in your getting the larger envelope more than 50% of the time? The catch is that those odds depend on the unknown amounts in the envelopes. But, no matter what is in the envelopes, you are guaranteed to get the larger one more than 50% of the time. See http://www.perlmonks.org/?node_id=39366 http://www.perlmonks.org/?node_id=39366 for an explanation. Read it very carefully, because the problem is very subtle. Even a slight change in what is meant tends to render the problem ill defined and indeterminate. (And our intuition is very bad.)
- harryh 16y ago
  That blog post is wrong. If N is the difference between the values in the two envelopes, then the chance that you will guess a value in this range is N/∞ which is 0.
  - sesqu 16y ago
    Not so. The only stipulation was that you must be able to guess a number in any range (which is not possible in practice). It doesn't matter that the probability of picking any particular number will be 0, as long as the probability of picking any one between the two in the letters isn't 0. Any time your guess hits the correct range, you win. Any other time, it depends on whether you got the big envelope or the small one. So knowing just one endpoint of a range is, oddly, evidence as to which way the other endpoint lies - as long as it might be either one. You just take a guess that depends on the number you know in such a way that the smaller the number you see, the more likely you are to guess it's the smaller of the two, but are never certain. Presto, guaranteed you'll get more than 50% right, as long as you get fed both the small and the large envelopes in equal proportion. If you can't be sure the other guy is playing fair and might favor giving you the low envelope, you need to look at something like the Monty Hall problem.
geuis 16y ago
Ok, I'm not a huge math wiz and after reading this article 2x, I'm still at a loss as to what the answer is. So after we have the nice breakdown in the form of a mathematical proof, we should keep switching envelopes forever? I'm the kind of guy that isn't as interested as to the joy of solving the problem. I just want the money! If I were to break this down into layman's terms, would it be appropriate to say the following? "There's equal odds whether you keep the envelope or exchange it for the other one. There's no difference either way, so flip a coin and decide."
samd 16y ago
Key takeaway: In the words of [David] Chalmers this is "just another example of a familiar phenomenon, the strange behaviour of infinity".
jwegan 16y ago
I think the problem is they effectively change a problem from having 2 possibilities (either you get $X or you get 2 * $X) to a problem that has 3 possibilities: A/2 (1 possible amount in the other envelope), A (the amount in your envelope), and 2A (the other possible amount in the other envelope) by fixing A relative to the hypothetical amount you could have in your envelope.
- Uhhrrr 16y ago
  Yeah - it's not really (0.5A + 2A)/2, it's (0.5A1 + 2A2)/2, where A1=2*X (as you have above) and A2=X. Substituting, your expected value upon changing is 3X/2, which is, surprise surprise, the same as the expected value for the envelope you already have.
risotto 16y ago
The 'paradox' that is presented leads back to the right conclusion: there is no way to inform which envelope is better so just pick one at random. These are non-causal random events, and the probability is always 50/50. There is a twist to this problem, which is more interesting, where you open the first envelope, then decide whether or not to switch to the second envelope. Believe it or not, there is a trick to beating 50% in this variant. I'm too tired to try to fully explain it, but this conclusion comes from the fact that once you look at an envelope that presents information that can be used for deciding to switch. So it becomes a causal system that does not follow a random probability density function. But think: if the envelope you opened had $7 million dollars in it, you probably wouldn't and shouldn't switch, and you'd probably be able to skew the odds in your favor on scenarios like this.
yanowitz 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? 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?
CJefferson 16y ago
The problem with this is very subtle. The problem is saying that it is equally likely that the other envelope contains half, or double, the money of the envelope we have chosen. It is impossible to choose a value from an infinite series (the list of possible monies), with every value having an equal opportunity of being chosen. Therefore the probability of money distribution must be non-even in some way, and the argument falls to pieces. It is not reasonable to say "Well, I don't know the probability, so I'll treat it as 50/50".
- philh 16y ago
  No, that's not the problem. You don't know how much money is in either envelope, but it needn't have been selected randomly. The only randomness is whether you the envelope you choose has the greater or lesser value, which is just a uniform distribution on two points. That being said, >It is impossible to choose a value from an infinite series (the list of possible monies), with every value having an equal opportunity of being chosen. this is also incorrect. You're right to say that you can't have a uniform distribution over the possible values of money, but that's not just because it's infinite. You can have a uniform distribution over [0,1] for example, which is larger than the set of possible money values.
  - cdavidcash 16y ago
    >You don't know how much money is in either envelope, but it needn't have been selected randomly. This is exactly wrong, at least in the mathematical formalization of this problem that everyone is assuming. The definition of "don't know how much money" must be taken to be "has a uniform distribution". If you want to use some reasoning that treats the amount of money as an "unknown in an equation" then you are instead giving a solution that differs for each possible value of the money and does not capture the random choices we are trying to model. By the way, no one said that uniform distributions on infinite sets don't exist. He said "infinite series (the list of possible monies)", a slightly mis-used term that clearly excludes [0,1], your chosen example.
- robinhouston 16y ago
  This is precisely the correct answer — and it's disappointing to see a less good answer getting so many votes when this one hasn't had any.
Dove 16y ago
The wiki page doesn't explain the flaw in the logic. I think it is this: Let X denote the value of the larger envelope, and A the value of the envelope we have in hand. If we have the larger envelope, the other envelope is worth A/2 (where A=X). If we have the smaller envelope, the other envelope is worth 2A'. (where A'=X/2) So far so good, and this is all legal. The error comes when we try to add: (0.5 A) + (0.5 A'/2) /= 5/4 A The A's are not the same. You can't add them as if they were.
- sp332 16y ago
  The wiki page used to have the solution. Apparently there is some arguing going on about sources, but this explanation is correct: http://en.wikipedia.org/wiki/Talk:Two_envelopes_problem#Solution_to_original_paradox_is_much_simpler http://en.wikipedia.org/wiki/Talk:Two_envelopes_problem#Solu... Also, yanowitz in this HN thread has a good (equivalent) explanation.
  - btilly 16y ago
    And yet, that explanation isn't correct.
- crpatino 16y ago
  Yes, I think you are correct. I would word it this way: Let: {X, 2X} := The amount of $$$ in each envelop A := 1st envelop has $2X B := 2nd envelop has $2X; The flaw in the paradox is to assume that events A and B are independent, when in fact they are mutually exclusive. Now, we can rewrite the whole "(0.5 A) + (0.5 A'/2) /= 5/4 A" nonsense as: P(A|B)2X + P(B|A)X = 02X + 0X = 0 Which is zero because it only covers half of the probabilities space, the other half is where all the density of the probability function lies. P(A|B')2X + P(B|A')X = 0.52X + 0.5X = 3X/2; which is what we would expect by intuition, halfway between the 2 prizes.
- lbrandy 16y ago
  i've realized my math was wrong and removed my comment :( i hate this problem
richtofen 16y ago
Trouble is, we are making it a mathematical problem. It need not be. The same logic applies to human [as opposed to mathematic] problems like: 1. Don't you wish your girlfriend was hot like me? 2. How much money is in the first envelope, to begin with? Whatever there is, I can get it to work to better use in my business, than some class 8 formula. 3. How many birds are there in the bush? 4. What if... All right. Comments notwithstanding. Seriously - how much money?
- richtofen 16y ago
  This was a seriously crass comment. What was I thinking? My apologies fir trivializing the original inquiry of the thread.
DannoHung 16y ago
Because you're negating a random choice and it's not like Monty Hall where a bad choice is removed for you.
- lionhearted 16y ago
  I didn't get that Monty Hall problem for a while after seeing it, but then it clicked and I got it. It's really beautiful to think about: http://en.wikipedia.org/wiki/Monty_Hall_problem http://en.wikipedia.org/wiki/Monty_Hall_problem
mirkules 16y ago
There's a similar problem with three closed doors, one of which contains a prize behind it. You are asked to choose a door, and then shown one of the two remaining doors which does NOT contain the prize. The question is should you switch doors, given the opportunity? What is the probability that of getting the prize if you switch, and if you don't switch? To illustrate, let A, B, C be doors, and door C be the door with the prize. You choose door A; the host tells you the prize is not in door B. What is the probability you will get the prize if you switch your door?
- necubi 16y ago
  No, this problem is different, and has a different solution. In your problem it makes sense to switch when given the choice, whereas in this one it does not. With the three doors you're given additional information once you choose a door, which allows you to choose a door with 1/2 probability of being right, rather than the 1/3 chance you started out with. In this problem you are given no additional information, so the probability of choosing the right envelope is the same when given a choice to switch as it is at the beginning (i.e., 1/2).
  - mirkules 16y ago
    Actually, your probability of choosing the right door is 2/3 if you switch. I agree though that this problem is not the same, it just reminded me of it -- I guess you could say it's a similar class of problems.
- InclinedPlane 16y ago
  It's similar in that it's easy to fool people into thinking about the problem incorrectly and coming up with the wrong conclusion. People are very touchy about the monty hall problem and will refuse to believe that it's better to switch, even when confronted with numerical simulations of the problem. As for this problem, it's more of a riddle than a paradox, the reasoning is erroneous but you may not realize that if you're not careful. It's more similar to the 1 = 0 riddle or the 3 guys paying $10 dollars for rent riddle, tricking you into making a mathematical error that isn't immediately obvious.
techiferous 16y ago
I found this paradox so compelling that I blogged about it a while ago. I tried to tease out the "intuitive" explanation without resorting to much statistics. http://techiferous.com/2010/06/the-two-envelope-paradox/ http://techiferous.com/2010/06/the-two-envelope-paradox/
Ramfjord 16y ago
I believe there's some problem with assuming that, given an envelope containing X dollars, that the other envelope has a 50% chance of containing 2x, and a 50% chance of containing X/2. This step is clearly what leads us into the paradox. Suppose you open the first envelope, and find $100, and are given the option to switch. If there is a 50/50 chance of the other envelope containing 200/50, we could easily model this problem a million times and find that it averages to 125. The problem here is that the numbers have not been pre-determined in the start. If a naive player repeatedly played a game where the numbers are guaranteed to be 100 and 200, then the always switching strategy would be exactly the same as the always staying strategy. In this case, let the difference between the two envelopes be x. When you switch, you have a 50% chance of losing x and a 50% chance of gaining x. There is really no problem if you define the two envelopes as differing by a fixed number, rather than one being a multiple of the other. When the two numbers have fixed, predetermined values, the two problems are the same.
jpwagner 16y ago
for all $X > p, (price to play) myLifeQuality(p,X) > myLifeQuality(X,2X) > myLifeQuality(X,X/2) easy
robinhouston 16y ago
That Wikipedia article is really bad. Here is a good article about the Two Envelopes Paradox, by Keith Devlin: http://www.maa.org/devlin/devlin_0708_04.html http://www.maa.org/devlin/devlin_0708_04.html
trominos 16y ago
The flaw in the argument for switching comes in its second line: "1. I denote by A the amount in my selected envelope. 2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2." If we're gonna be taking expected values we need to assume that the monetary amounts in the two envelopes are generated from a probability distribution on the nonnegative reals. Then the probability distribution for the amount of money in the smaller envelope is going to be some kind of curve (call it S), and obviously the probability distribution for the amount of money in the bigger envelope is going to be the same curve only "stretched out" and "squashed" by a factor of two (we'll call it B). Now if you say "my envelope contains exactly A dollars," you can tell what the probability is that your envelope is the smaller envelope by comparing the relative heights of the two curves at the value A; let's denote these heights by S(A) and B(A). It is certainly possible that the two heights are the same, in which case it's fifty-fifty that you have the bigger or smaller envelope, the rest of the logic holds, and you should indeed switch, getting an expected return of 5/4 * A. But it is obviously impossible that in general the two heights are the same for any given A, because then the two probability distributions are the same -- and that can't be true, because B is a squashed, stretched-out version of S, and for a variety of fairly obvious reasons you can't squash and stretch out a finite curve on the positive reals and get the same curve (unless that curve is 0 everywhere). And so we can't conclude that in general you should switch, which is good because if you're not allowed to look at the money before you switch it obviously doesn't matter whether you do or not (unless the people running the game are messing with you).
te_platt 16y ago
While not exactly the same problem this is what cleared it up for me: First pick a positive integer at random. Now pick a second integer at random. What is the probability that the second number is larger? The answer seems to be 1 because after the first number is chosen there are finitely many numbers smaller but infinitely many larger. But what if two people (A and B) choose two positive integers at random but don't tell anyone just yet. What is the probability that A>B? By symmetry we can see it should be 0.5 . So A tells his number. Now the probability that A>B seems to jump to 0 because there are finitely many numbers smaller than A but infinitely many larger. But why should the probability change when we don't even know what A is, just that it known? The problem lies with the phrase "choose a positive integer at random". However you choose the number you will be hopelessly biased toward 0 because every number you choose will be finite. Now think about the envelope problem but add the condition that the largest amount that any envelope will have is X. Then if your envelope has N in it there is a chance that 2N is larger than X. That cut's down on the chance that switching gets you more. Magically (because I'm too lazy to actually write it all out) it turns out it cuts it down just so the probability of getting more money by switching works out to .5
- jeromec 16y ago
  So A tells his number. Now the probability that A>B seems to jump to 0 because there are finitely many numbers smaller than A but infinitely many larger. This is not correct. You said B chose a positive integer. Therefore, there are not infinitely many larger numbers B's choice can be.
  - InclinedPlane 16y ago
    Sure there are. No matter what finite positive integer you choose there are infinitely more finite positive integers greater than that number.
    - jeromec 16y ago
      That's true, but what te_platt said was "two people (A and B) choose two positive integers at random" If two people, A and B, have chosen -- that's choose in the past tense -- two positive integers, there are not infinitely more positive integers B's number can be greater than A's number. Why? Again, because B has already made a choice.
pvg 16y ago
And that's Numberwang!
dmor 16y ago
can't eat your cake and have it too
uio 16y ago
You denote with A two different things, and this is not correct. Suppose you have a bag with two animals a dog and a fox. You choice an animal A ten times, and then another animal A ten times, so you have 20 A. The question is: Do you have twenty dogs or twenty foxs?
praptak 16y ago
I believe the root of the paradox is that there exists no distribution of two variables A,B for which: a) (A,B) always fulfills A=2B or A=0.5B and b) For any value of A, the conditional probability of B=2A is equal to the conditional probability of B=0.5A.
drhodes 16y ago
The example seems to miss something critical. If the dollar amount of the first envelope is odd, then it can't be the doubled sum...
mitko 16y ago
A friend of mine told me a similar problem that he got at D.E. Shaw interview. Person A writes random real numbers in two envelopes. Then person B randomly 50/50 picks one envelope and sees the number X written there. Then, B has to guess whether X is the bigger of the two numbers or not. Show that B has a strategy, that even if A knows it, B can guess correctly in more than 50% of the cases. The mathematical beauty of this problem is that you can derive what are the non-trivial steps towards the solution.
mitko 16y ago
For me, what is interesting in that problem (besides the obvious flaw that people here discussed) is that the answer changes is the value A is observed. I.e. : If I have two envelopes, one of which has twice the money of the other, you pick an envelope AND YOU SEE the money inside, then it becomes better to switch (in expected money).

112 points by atomon 16y ago

I like Raymond Smullyan's version even better, it removes all probability issues and is so simple you feel it must be easy to figure out. The setup is the same: two envelopes, once with twice as much money as the other, you pick one and before you open you are offered to switch. If you do switch you might either win money or lose money relative to what was in the envelope you chose first. Let's compare the possible gain to the possible loss (without worrying about the probability of having a gain or a loss): 1. The possible gain is equal to the possible loss. Proof: The two envelopes contain x and 2x, so you'd gain x if switching from x to 2x, and you'd lose x if switching from 2x to x. The possible gain and the possible loss are both equal to x. 2. The possible win is larger than the possible loss. Proof: Say the envelope you chose first has y. Then the other envelope has either y/2 or 2y. So if you gain money by switching you gain 2y - y = y, but if you lose in the switch, you lose y - y/2 = y/2. Clearly y > y/2. This version kept up at night for days when I first read it. I never figured it out --not that I had any chance of doing so, given that Smullyan couldn't figure it out either--, I just got used to not understanding it...

jeromec 16y ago
This doesn't eliminate probability, and it's really the same flaw, just with different wording. The flaw is still that the variable is undefined (or infinite probability) in one part of the word problem, but treated as defined in another. So, start with when you first chose the envelope which has y. At this point "y" is undefined, and can be the higher or lower amount. That's the key. Yet, you then continue to propose that by switching you gain 2y - y = y. That equation is not okay, because "y" could have been the higher amount, and the equation breaks down. Once you assert the equation holds true, you have thus defined the original value of "y" from that first envelope choice (i.e. "y" was lower). The same flaw surfaces by asserting your switch to lose equation is true, as well. Edit: BTW, the reason #1 works okay is because you define upfront the variable value ("x") as being either the higher or lower of the two possibilities, before constructing your equation(s).
- omaranto 16y ago
  You mention this doesn't eliminate probability but don't explain why you think that might be the case. I just meant that the argument Smullyan presents does not involve probabilities, random variables, distributions, expected values, etc. (This can be verified by reading it again and checking those concepts do not appear. ;)). I don't find your argument convincing either, maybe it helps to remove the "undefined variable" you complain about: The first envelope chosen has some definite amount of money, let's say $10 (but the same argument can be adapted to any amount). Then the other envelope must have either $5 or $20. If it is $5, switching loses you $5. If it is $10, switching gains you $10. Imagine actually doing this experiment with the two envelopes, opening the first and seeing $10. Wouldn't you say the other contains either $5 or $20?
  - jeromec 16y ago
    It doesn't eliminate probability because it's the exact same envelope experiment. ;) In other words, there is still a 50% probability of gaining or losing. My post above isn't an argument, it's an explanation. The "undefined variable" is established by your own proposal, i.e., in #2 you first choose an envelope which has an amount we call "y". At that point "y" is undefined, because you specifically explain it could be either the higher or lower amount. I think your confusion lies in not considering carefully enough the meaning of your wording. In other words, you're performing a play on words to get a specific result, akin to the "heads I win, tails you lose" word play. Let me try to explain further below: Imagine actually doing this experiment with the two envelopes, opening the first and seeing $10. Wouldn't you say the other contains either $5 or $20? No. The other envelope contains the amount the person who prepared the envelopes placed in it. Try defining both values explicitly and this issue may become clearer. Imagine one envelope contains $10 and the other contains $5. Go through your above proposed equations using strictly those values. Remember, in a real experiment that would be completely valid.
  - Confusion 16y ago
    I just meant that the argument Smullyan presents does not involve probabilities, random variables, distributions, expected values, etc. (This can be verified by reading it again and checking those concepts do not appear. ;)). The fact that he doesn't uses those words does not mean he doesn't use those concepts. The Smullyan version is simply less precisely stated and anyone solving it would first have to go through the trouble to convert the phrasing to incorporate the usual, well defined mathematical words. Solving it any other way is just handwaving.
Amnon 16y ago
Statement 1 isn't true. If you have 10$ in the envelope, and it's the smaller envelope, you gain 10$. But if it's the larger envelope you don't lose 10$. The fallacy is that "x" refers to different things in each case. Statement 2 is true but it isn't relevant, since you don't know the posterior probability given y. If the envelope contains the smaller amount of money then there is no way you can lose -- it's not a 50-50 chance.

deleted 16y ago

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klochner 16y ago

I got that in a tech interview and hadn't seen it before. Dirty.

dotBen 16y ago
In what context was the interview (ie type of position). I'm all for logical problem solving questions in interviews, even ones that can't be solved but are interesting to see how people tackle them. However this one doesn't seem to lend itself much to programming or true problem solving - just probability theorem. In other words, I'm not sure what useful info I'd get out of a candidate by asking them to work on this (other than their understanding of probability, which is not that useful for most programming jobs)

aliston 16y ago

Interesting... here's my "gut reaction" as a non-statistician. The problem is that 'A' has a probabilistic component as well. If you let X be the "lesser of the 2 dollar amounts" and let A = X/2 + 2X, then you come out with equal expected values for switching and non-switching.

imajes 16y ago

I think this is better posed as two random-drawn sets of numbers for the first drawing of an as-of-yet un-played lottery. Then, there's no value in considering extraneous data such as the mass of the envelopes, looking through them or other potential evidence. Instead, you're left with two identical chances. Since there is no data to suggest that either has an advantage over the other, you have a probability of exactly 0.5 that you have the better chance of winning. On the question of the switch, it's important to consider that you have no new data about the one you picked, so the odds are _exactly the same_ as they were at the beginning: there is no conditionality, nothing has changed since you picking it. The probability then of the winning ticket is still 0.5. This is where the infinite probability problem kicks in. I don't think this is an actual problem needing to be solved, unless we can find a way to mathematically express the probability modifiers of our gut feeling, which is impossible, as it comes into our own personal experiences. Still, Gladwell might have something to say about it :)

jatenate 16y ago

The swapping indefinitely does not work. For the first iteration, if I have A in my envelope, then yes, the other envelope will either have 1/2A or 2A, meaning the expected value is indeed 5/4A. The switching argument wrongly assumes this is again the case after I switch. After I have switched, the value of the original envelope is unchanged (meaning it is A with probability 1), so I should not switch back.

bmm6o 16y ago
Are you seriously arguing that you should switch exactly once?

btilly 16y ago

First a random observation. People use "expected value" to evaluate risky decisions far too often. In many investment situations a more appropriate measure is "expected value of the log of my worth". (See http://elem.com/~btilly/kelly-criterion/ http://elem.com/~btilly/kelly-criterion/ for an explanation of why.) That measure, in this problem, says that you should be indifferent about switching. Admittedly that is coincidence - the investment reasoning for that rule has little to do with why that rule works for this problem. But it turns out that there is an even crazier twist to this problem. Change the problem to say that you're allowed to look at the envelope you receive before deciding whether to switch. Would you believe that there is an algorithm for deciding whether to switch that results in your getting the larger envelope more than 50% of the time? The catch is that those odds depend on the unknown amounts in the envelopes. But, no matter what is in the envelopes, you are guaranteed to get the larger one more than 50% of the time. See http://www.perlmonks.org/?node_id=39366 http://www.perlmonks.org/?node_id=39366 for an explanation. Read it very carefully, because the problem is very subtle. Even a slight change in what is meant tends to render the problem ill defined and indeterminate. (And our intuition is very bad.)

harryh 16y ago
That blog post is wrong. If N is the difference between the values in the two envelopes, then the chance that you will guess a value in this range is N/∞ which is 0.
- sesqu 16y ago
  Not so. The only stipulation was that you must be able to guess a number in any range (which is not possible in practice). It doesn't matter that the probability of picking any particular number will be 0, as long as the probability of picking any one between the two in the letters isn't 0. Any time your guess hits the correct range, you win. Any other time, it depends on whether you got the big envelope or the small one. So knowing just one endpoint of a range is, oddly, evidence as to which way the other endpoint lies - as long as it might be either one. You just take a guess that depends on the number you know in such a way that the smaller the number you see, the more likely you are to guess it's the smaller of the two, but are never certain. Presto, guaranteed you'll get more than 50% right, as long as you get fed both the small and the large envelopes in equal proportion. If you can't be sure the other guy is playing fair and might favor giving you the low envelope, you need to look at something like the Monty Hall problem.

geuis 16y ago

Ok, I'm not a huge math wiz and after reading this article 2x, I'm still at a loss as to what the answer is. So after we have the nice breakdown in the form of a mathematical proof, we should keep switching envelopes forever? I'm the kind of guy that isn't as interested as to the joy of solving the problem. I just want the money! If I were to break this down into layman's terms, would it be appropriate to say the following? "There's equal odds whether you keep the envelope or exchange it for the other one. There's no difference either way, so flip a coin and decide."

samd 16y ago

Key takeaway: In the words of [David] Chalmers this is "just another example of a familiar phenomenon, the strange behaviour of infinity".

jwegan 16y ago

I think the problem is they effectively change a problem from having 2 possibilities (either you get $X or you get 2 * $X) to a problem that has 3 possibilities: A/2 (1 possible amount in the other envelope), A (the amount in your envelope), and 2A (the other possible amount in the other envelope) by fixing A relative to the hypothetical amount you could have in your envelope.

Uhhrrr 16y ago
Yeah - it's not really (0.5A + 2A)/2, it's (0.5A1 + 2A2)/2, where A1=2*X (as you have above) and A2=X. Substituting, your expected value upon changing is 3X/2, which is, surprise surprise, the same as the expected value for the envelope you already have.

risotto 16y ago

The 'paradox' that is presented leads back to the right conclusion: there is no way to inform which envelope is better so just pick one at random. These are non-causal random events, and the probability is always 50/50. There is a twist to this problem, which is more interesting, where you open the first envelope, then decide whether or not to switch to the second envelope. Believe it or not, there is a trick to beating 50% in this variant. I'm too tired to try to fully explain it, but this conclusion comes from the fact that once you look at an envelope that presents information that can be used for deciding to switch. So it becomes a causal system that does not follow a random probability density function. But think: if the envelope you opened had $7 million dollars in it, you probably wouldn't and shouldn't switch, and you'd probably be able to skew the odds in your favor on scenarios like this.

yanowitz 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? 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?

CJefferson 16y ago

The problem with this is very subtle. The problem is saying that it is equally likely that the other envelope contains half, or double, the money of the envelope we have chosen. It is impossible to choose a value from an infinite series (the list of possible monies), with every value having an equal opportunity of being chosen. Therefore the probability of money distribution must be non-even in some way, and the argument falls to pieces. It is not reasonable to say "Well, I don't know the probability, so I'll treat it as 50/50".

philh 16y ago
No, that's not the problem. You don't know how much money is in either envelope, but it needn't have been selected randomly. The only randomness is whether you the envelope you choose has the greater or lesser value, which is just a uniform distribution on two points. That being said, >It is impossible to choose a value from an infinite series (the list of possible monies), with every value having an equal opportunity of being chosen. this is also incorrect. You're right to say that you can't have a uniform distribution over the possible values of money, but that's not just because it's infinite. You can have a uniform distribution over [0,1] for example, which is larger than the set of possible money values.
- cdavidcash 16y ago
  >You don't know how much money is in either envelope, but it needn't have been selected randomly. This is exactly wrong, at least in the mathematical formalization of this problem that everyone is assuming. The definition of "don't know how much money" must be taken to be "has a uniform distribution". If you want to use some reasoning that treats the amount of money as an "unknown in an equation" then you are instead giving a solution that differs for each possible value of the money and does not capture the random choices we are trying to model. By the way, no one said that uniform distributions on infinite sets don't exist. He said "infinite series (the list of possible monies)", a slightly mis-used term that clearly excludes [0,1], your chosen example.
robinhouston 16y ago
This is precisely the correct answer — and it's disappointing to see a less good answer getting so many votes when this one hasn't had any.

Dove 16y ago

The wiki page doesn't explain the flaw in the logic. I think it is this: Let X denote the value of the larger envelope, and A the value of the envelope we have in hand. If we have the larger envelope, the other envelope is worth A/2 (where A=X). If we have the smaller envelope, the other envelope is worth 2A'. (where A'=X/2) So far so good, and this is all legal. The error comes when we try to add: (0.5 A) + (0.5 A'/2) /= 5/4 A The A's are not the same. You can't add them as if they were.

sp332 16y ago
The wiki page used to have the solution. Apparently there is some arguing going on about sources, but this explanation is correct: http://en.wikipedia.org/wiki/Talk:Two_envelopes_problem#Solution_to_original_paradox_is_much_simpler http://en.wikipedia.org/wiki/Talk:Two_envelopes_problem#Solu... Also, yanowitz in this HN thread has a good (equivalent) explanation.
- btilly 16y ago
  And yet, that explanation isn't correct.
crpatino 16y ago
Yes, I think you are correct. I would word it this way: Let: {X, 2X} := The amount of $$$ in each envelop A := 1st envelop has $2X B := 2nd envelop has $2X; The flaw in the paradox is to assume that events A and B are independent, when in fact they are mutually exclusive. Now, we can rewrite the whole "(0.5 A) + (0.5 A'/2) /= 5/4 A" nonsense as: P(A|B)2X + P(B|A)X = 02X + 0X = 0 Which is zero because it only covers half of the probabilities space, the other half is where all the density of the probability function lies. P(A|B')2X + P(B|A')X = 0.52X + 0.5X = 3X/2; which is what we would expect by intuition, halfway between the 2 prizes.
lbrandy 16y ago
i've realized my math was wrong and removed my comment :( i hate this problem

richtofen 16y ago

Trouble is, we are making it a mathematical problem. It need not be. The same logic applies to human [as opposed to mathematic] problems like: 1. Don't you wish your girlfriend was hot like me? 2. How much money is in the first envelope, to begin with? Whatever there is, I can get it to work to better use in my business, than some class 8 formula. 3. How many birds are there in the bush? 4. What if... All right. Comments notwithstanding. Seriously - how much money?

richtofen 16y ago
This was a seriously crass comment. What was I thinking? My apologies fir trivializing the original inquiry of the thread.

DannoHung 16y ago

Because you're negating a random choice and it's not like Monty Hall where a bad choice is removed for you.

lionhearted 16y ago
I didn't get that Monty Hall problem for a while after seeing it, but then it clicked and I got it. It's really beautiful to think about: http://en.wikipedia.org/wiki/Monty_Hall_problem http://en.wikipedia.org/wiki/Monty_Hall_problem

mirkules 16y ago

There's a similar problem with three closed doors, one of which contains a prize behind it. You are asked to choose a door, and then shown one of the two remaining doors which does NOT contain the prize. The question is should you switch doors, given the opportunity? What is the probability that of getting the prize if you switch, and if you don't switch? To illustrate, let A, B, C be doors, and door C be the door with the prize. You choose door A; the host tells you the prize is not in door B. What is the probability you will get the prize if you switch your door?

necubi 16y ago
No, this problem is different, and has a different solution. In your problem it makes sense to switch when given the choice, whereas in this one it does not. With the three doors you're given additional information once you choose a door, which allows you to choose a door with 1/2 probability of being right, rather than the 1/3 chance you started out with. In this problem you are given no additional information, so the probability of choosing the right envelope is the same when given a choice to switch as it is at the beginning (i.e., 1/2).
- mirkules 16y ago
  Actually, your probability of choosing the right door is 2/3 if you switch. I agree though that this problem is not the same, it just reminded me of it -- I guess you could say it's a similar class of problems.
InclinedPlane 16y ago
It's similar in that it's easy to fool people into thinking about the problem incorrectly and coming up with the wrong conclusion. People are very touchy about the monty hall problem and will refuse to believe that it's better to switch, even when confronted with numerical simulations of the problem. As for this problem, it's more of a riddle than a paradox, the reasoning is erroneous but you may not realize that if you're not careful. It's more similar to the 1 = 0 riddle or the 3 guys paying $10 dollars for rent riddle, tricking you into making a mathematical error that isn't immediately obvious.

techiferous 16y ago

I found this paradox so compelling that I blogged about it a while ago. I tried to tease out the "intuitive" explanation without resorting to much statistics. http://techiferous.com/2010/06/the-two-envelope-paradox/ http://techiferous.com/2010/06/the-two-envelope-paradox/

Ramfjord 16y ago

I believe there's some problem with assuming that, given an envelope containing X dollars, that the other envelope has a 50% chance of containing 2x, and a 50% chance of containing X/2. This step is clearly what leads us into the paradox. Suppose you open the first envelope, and find $100, and are given the option to switch. If there is a 50/50 chance of the other envelope containing 200/50, we could easily model this problem a million times and find that it averages to 125. The problem here is that the numbers have not been pre-determined in the start. If a naive player repeatedly played a game where the numbers are guaranteed to be 100 and 200, then the always switching strategy would be exactly the same as the always staying strategy. In this case, let the difference between the two envelopes be x. When you switch, you have a 50% chance of losing x and a 50% chance of gaining x. There is really no problem if you define the two envelopes as differing by a fixed number, rather than one being a multiple of the other. When the two numbers have fixed, predetermined values, the two problems are the same.

jpwagner 16y ago

for all $X > p, (price to play) myLifeQuality(p,X) > myLifeQuality(X,2X) > myLifeQuality(X,X/2) easy

robinhouston 16y ago

That Wikipedia article is really bad. Here is a good article about the Two Envelopes Paradox, by Keith Devlin: http://www.maa.org/devlin/devlin_0708_04.html http://www.maa.org/devlin/devlin_0708_04.html

trominos 16y ago

The flaw in the argument for switching comes in its second line: "1. I denote by A the amount in my selected envelope. 2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2." If we're gonna be taking expected values we need to assume that the monetary amounts in the two envelopes are generated from a probability distribution on the nonnegative reals. Then the probability distribution for the amount of money in the smaller envelope is going to be some kind of curve (call it S), and obviously the probability distribution for the amount of money in the bigger envelope is going to be the same curve only "stretched out" and "squashed" by a factor of two (we'll call it B). Now if you say "my envelope contains exactly A dollars," you can tell what the probability is that your envelope is the smaller envelope by comparing the relative heights of the two curves at the value A; let's denote these heights by S(A) and B(A). It is certainly possible that the two heights are the same, in which case it's fifty-fifty that you have the bigger or smaller envelope, the rest of the logic holds, and you should indeed switch, getting an expected return of 5/4 * A. But it is obviously impossible that in general the two heights are the same for any given A, because then the two probability distributions are the same -- and that can't be true, because B is a squashed, stretched-out version of S, and for a variety of fairly obvious reasons you can't squash and stretch out a finite curve on the positive reals and get the same curve (unless that curve is 0 everywhere). And so we can't conclude that in general you should switch, which is good because if you're not allowed to look at the money before you switch it obviously doesn't matter whether you do or not (unless the people running the game are messing with you).

te_platt 16y ago

While not exactly the same problem this is what cleared it up for me: First pick a positive integer at random. Now pick a second integer at random. What is the probability that the second number is larger? The answer seems to be 1 because after the first number is chosen there are finitely many numbers smaller but infinitely many larger. But what if two people (A and B) choose two positive integers at random but don't tell anyone just yet. What is the probability that A>B? By symmetry we can see it should be 0.5 . So A tells his number. Now the probability that A>B seems to jump to 0 because there are finitely many numbers smaller than A but infinitely many larger. But why should the probability change when we don't even know what A is, just that it known? The problem lies with the phrase "choose a positive integer at random". However you choose the number you will be hopelessly biased toward 0 because every number you choose will be finite. Now think about the envelope problem but add the condition that the largest amount that any envelope will have is X. Then if your envelope has N in it there is a chance that 2N is larger than X. That cut's down on the chance that switching gets you more. Magically (because I'm too lazy to actually write it all out) it turns out it cuts it down just so the probability of getting more money by switching works out to .5

jeromec 16y ago
So A tells his number. Now the probability that A>B seems to jump to 0 because there are finitely many numbers smaller than A but infinitely many larger. This is not correct. You said B chose a positive integer. Therefore, there are not infinitely many larger numbers B's choice can be.
- InclinedPlane 16y ago
  Sure there are. No matter what finite positive integer you choose there are infinitely more finite positive integers greater than that number.
  - jeromec 16y ago
    That's true, but what te_platt said was "two people (A and B) choose two positive integers at random" If two people, A and B, have chosen -- that's choose in the past tense -- two positive integers, there are not infinitely more positive integers B's number can be greater than A's number. Why? Again, because B has already made a choice.

pvg 16y ago

And that's Numberwang!

dmor 16y ago

can't eat your cake and have it too

uio 16y ago

You denote with A two different things, and this is not correct. Suppose you have a bag with two animals a dog and a fox. You choice an animal A ten times, and then another animal A ten times, so you have 20 A. The question is: Do you have twenty dogs or twenty foxs?

praptak 16y ago

I believe the root of the paradox is that there exists no distribution of two variables A,B for which: a) (A,B) always fulfills A=2B or A=0.5B and b) For any value of A, the conditional probability of B=2A is equal to the conditional probability of B=0.5A.

drhodes 16y ago

The example seems to miss something critical. If the dollar amount of the first envelope is odd, then it can't be the doubled sum...

mitko 16y ago

A friend of mine told me a similar problem that he got at D.E. Shaw interview. Person A writes random real numbers in two envelopes. Then person B randomly 50/50 picks one envelope and sees the number X written there. Then, B has to guess whether X is the bigger of the two numbers or not. Show that B has a strategy, that even if A knows it, B can guess correctly in more than 50% of the cases. The mathematical beauty of this problem is that you can derive what are the non-trivial steps towards the solution.

mitko 16y ago

For me, what is interesting in that problem (besides the obvious flaw that people here discussed) is that the answer changes is the value A is observed. I.e. : If I have two envelopes, one of which has twice the money of the other, you pick an envelope AND YOU SEE the money inside, then it becomes better to switch (in expected money).