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Two envelopes problem

• jdkee 4y ago
  Notably different from the Monty Haul problem where additional information is given to the player of the game prior to making the decision to switch. See https://en.wikipedia.org/wiki/Monty_Hall_problem https://en.wikipedia.org/wiki/Monty_Hall_problem
  • klodolph 4y ago
    The “Monty Haul” problem is where a dungeon master in D&D awards too much treasure to the party :-) https://dungeonsdragons.fandom.com/wiki/Monty_Haul https://dungeonsdragons.fandom.com/wiki/Monty_Haul https://tvtropes.org/pmwiki/pmwiki.php/Main/MontyHaul https://tvtropes.org/pmwiki/pmwiki.php/Main/MontyHaul
    • jdkee 4y ago
      Yikes! I see what you did there.
  • scrapheap 4y ago
    Personally I don't think you are given any new information in the Monty Hall problem that's worth anything. You always knew there was a 100% chance that one of the doors in that set has a goat behind it, which door that is is pretty much irrelevant. Or to put it another way how many people would stick with their chosen door if instead of opening a door and showing a goat they were simply asked "would you like to stick with your current choice of door, or switch and take the best result from the other doors?"
    • RHSeeger 4y ago
      > switch and take the best result from the other doors That is, effectively, what the choice boils down to. But that being said, since the problem is stated as only getting to pick one "other" door, the "extra information" received is knowing which of the other 2 doors has the best result.
    • arnarbi 4y ago
      Try thinking of it this way: The information is in which door the host does not open. 2/3rds of the time, your initial choice will be a goat. In those cases the host deliberately avoids opening the remaining door with a car - thereby telling you where it is. If you chose the car door initially, then indeed you get no new information. But that only happens 1/3rd of the time. I am curious if that changes your opinion.
      • tsimionescu 4y ago
        Read their second paragraph. That is exactly how they think of it, and they do think it makes it obvious you should switch.
        • arnarbi 4y ago
          Ah! Apologies, I did not read that properly.
    • dllthomas 4y ago
      When Monty reveals a goat, your odds increase from 1/3 to 1/2 if Monty is picking randomly and just happened not to reveal a goat (and if he revealed a car we start from the top, or you automatically lose, or you automatically win, or...), but to 2/3 if Monty will reliably reveal a goat. I think it's hard to say that you have no extra information in that latter case. It's true that we can add other options (pick both doors) of equal likelihood, but it's because of our increased information that the reduced option of picking just one door isn't actually worse off.
      • tsimionescu 4y ago
        The GP knows all this, and agrees it's better to switch (look at the second paragraph). They just dislike the common formulation (that is given in terms of the extra information from the boot being opened), and find another formulation more compelling (your original choice is one door, while the switch is actually to the best prize from the other two doors, not simply to another one door).
    • Schroedingersat 4y ago
      It becomes a lot clearer if you increase the number of goats. I have a deck of a billion cards, one entitles you to the car, the others have a picture of a goat. You pick one at random but do not get to see it. Of the remaining 999,999,999 cards, I discard 999,999,998 goats. Do you want your initial card, or the remaining card in my deck? Or to frame it another way, do you want the car iff you picked it first or iff you did not pick it first?
      • tsimionescu 4y ago
        The GP is not contesting this, they just think that the whole discarding/revealing of the goats is an unnecessarily complicated explanation for why this works. The way they put it, the strategy becomes obvious if you instead reformulate the problem like so: there are three doors with three different prizes. You get to pick one door, but are then given a choice: stick with your original choice, or choose to get the best prize from the other two doors. The solution (switch!) is more obvious in the second formulation. I have no idea if it's easier to convince someone else of the reasoning in the first formulation, or to convince them that the second formulation is equivalent to the original.
    • thekiptxt 4y ago
      I didn’t question the math, but what made me intuitively believe the Monty Hall problem was this: Me: I am thinking of a number between 1 and a million. If you guess correctly, you get a prize. you make your guess Me: Ok, I will reveal that the number is either $YOUR_GUESS or 452,871. You may keep your original guess or switch to 452,871. Of course, you would switch, because the chances you guessed correctly originally were 1 in a million.
      • junon 4y ago
        Wow. I don't think it's ever been explained to me this well before.
    • jordo 4y ago
      I always think of the Monty Hall problem with 100 doors. And with the game show host revealing 98 'goat' doors thus leaving two doors left (the original guess, and another one). You'd switch then right? :)
    • tsimionescu 4y ago
      Your first paragraph makes it seem like you don't accept the common solution to the Monty Hall problem, which is why people have responded explaining it to you, even though the second paragraph makes it clear that you do, and just dislike the way it is explained. Personally, I'm not sure that it would be easier to convince someone that the second formulation is equivalent to the original problem, but next time I find a dis-believer I will give it a shot!
    • CamperBob2 4y ago
      The Monty Hall problem is easily resolved by imagining there are a million doors instead of just three. The host opens 9,999,998 doors, revealing goats behind each. You'd obviously switch doors to the one the host left closed, and the reasoning you followed will be just as valid for any number of doors including 3. This problem is completely different, since the number of degrees of freedom remains fixed throughout the decision process. As someone else pointed out, it is trivially resolvable by realizing that no matter how many times you switch envelopes, you're really only making one choice, and doing so with 50-50 odds of picking the more profitable envelope. The offer to switch amounts to meaningless magician's patter, whether the offer is made once or a million times.
    • Elte 4y ago
      When I was initially exposed to Monty Hall, what confused me the most is that I didn't know that the host would always open a door with a goat, rather than just picking one of the remaining doors at random. In the latter case (i.e. the case where the host can pick a car), even provided that you are in the universe where he picked a goat, switching and staying both yield a 50% chance of success.
  • junon 4y ago
    A fascinating line that shocked me, tucked away at the end of a paragraph with no further additional context aside from its source: > Pigeons repeatedly exposed to the problem show that they rapidly learn to always switch, unlike humans.
• sjburt 4y ago
  I love this problem because it is so simple, and the false line of reasoning is so compelling that it would hardly raise an eyebrow if you saw it in an academic paper and yet the conclusion is so obviously wrong. Decision problems are tricky and in non intuitive ways.
  • Schroedingersat 4y ago
    Step 3 (and thus also 1 and 2 on reflection) stands out to me immediately as describing two different situations for the entire game. The case where switching gives you 2A and the case where switching gives you A/2 describe 2 completely different universes, not two different actions in one.
    • NamTaf 4y ago
      Agreed. Step 6 is where it breaks down, because it's redefined A as being one of the two envelopes to being the halfway point between those two envelopes. Until step 6, it's describing two parallel views of the universe (depending on which envelope you first get - hence steps 4 and 5), and then mashes them together in a way that doesn't work. Take the practical example of $100 and $200: A is either $100 or $200 depending on which envelope you receive first, so the equation is either 0.5*A + 0.5*2A, or it is 0.5*A + 0.5*(A/2). It can never be 0.5*2A + 0.5*(A/2)
      • RHSeeger 4y ago
        Even after looking at those numbers, it still felt wrong to me. It eventually occurred to me that this was because my intuition was telling me the total should add up to A. And it doesn't; it adds up to 9/8 A. It took me a while to realize that my intuition was wrong, and that there's no reason the total should be A. The thing that helped me reconcile this was realizing that the A in the two equations are different values; then replacing them. (0.5⋅(0.5⋅A + 0.5⋅2.A)) + (0.5⋅(0.5⋅A + 0.5⋅0.5⋅A)) = 9/8 A ^ but the A in the left grouping (where it's 100) is different than the A in the right grouping (where it's 200). Replacing the As with their actual values 0.5⋅(0.5⋅100+0.5⋅2⋅100)+0.5⋅(0.5⋅200+0.5⋅0.5⋅200) = 150 And, since one envelope has 100 and the other has 200, an expected outcome across both envelopes is, as calculated, 150.
• mountainwalker 4y ago
  change your mind as many times as you like, there will only ever be one actual choice
  • djmips 4y ago
    It's a simple as that. If you know you pick A and then you will switch to B then it's the same as if you had just decided to pick B in the first place.
• pkdpic 4y ago
  > you are given two identical envelopes [...] containing money. One contains twice as much as the other > having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch? > because the person stands to gain twice as much money if they switch, while the only risk is halving what they currently have, a case can be made for switching the envelope I dig this, but it seems like an unavoidable nonsense pattern that congeals in pools of otherwise practically useful logic / language systems. Like Neo in the Matrix. Remember the Architect? That guy was cool.
• antognini 4y ago
  There is a related puzzle which I've also heard called the Two Envelopes Problem. The premise is that you are handed two envelopes. Inside each envelope is a piece of paper with a number written on it. I have written down these numbers by sampling from some probability distribution, but you don't know what it is. All you know is that the two numbers are different. You get to pick one envelope and then look at the paper inside. You then need to guess whether or not the number in the other envelope is larger or smaller than the one you are looking at. Can you guess with better than 50% chance?
  • Nihmie 4y ago
    Is there actually some way to guess with better than 50% chance? I'm pretty sure this could be reduced to the secretary problem where there's a pool of 2 candidates, which yields an optimal hire with probability 50%. Maybe you play word games and say, "I guess the other number is not higher," or "I guess the other number is not lower." Since the number you observed was produced once, there is some non-zero probability that it was produced twice, in which case the other number is neither nigher nor lower.
    • CaptainNegative 4y ago
      The reduction is leaky in that it requires you to assume that there is nothing to learn from looking at the first number. Here's a hint that is not a full solution. If you knew that the numbers in both envelopes were independently sampled from the same Gaussian (but not necessarily which Gaussian), then there is a simple strategy that wins more than 50% of the time: pick an envelope to peek at uniformly at random and guess that it is the larger one if and only if it shows a positive number. Why does this work? If both numbers are positive, your strategy is equivalent to "pick a random envelope" and you're left with a uniformly random choice of envelope. If both numbers are negative, you're also left with a uniformly random choice of envelope. But if one is positive and the other is negative, you win 100% of the time. What are the odds of the third case? Well every Gaussian has at least some mass on positive numbers and some mass on negative numbers. So there is some nonzero chance, even if you don't know what it is. So while the strategy might not always do better than a coin toss, some positive percent of the time you win every time. And so you get a distinct advantage over random guessing. The problem is now how would one transfer such an approach to the original problem without the Gaussian assumption. Somehow the two problems are less different than they initially seem.
• kazinator 4y ago
  It is very simple. Let's represent the poorer envelope by the value 0.5 and the more bountiful one by 1.0. The expected value from the envelope draw is 0.75: 0.5 drawn with a 50% probability contributing 0.25 to the expected value, and 1.0 drawn with a 50% probability contributing another 0.5 to the expected value for a total of 0.75. If you were to do a large number N of these draws, the value you will obtain will be from a distribution that is centered on 0.75 N. So far so good? So the way to look at a draw is this: you're going to be getting the expected/average value of 0.75, together with either a 0.25 bonus if you pick the better envelope, or a else -0.25 penalty if you pick the poorer envelope. So you see, those two situations are equal and opposite. By switching, you either turn a 0.25 bonus into a -0.25 penalty (- 0.5) or vice versa (+ 0.5) with equal probability. If you need any more complex analysis than this, you need to be thwacked on the head. The rhetoric about doubling versus halving the money is a distracting red herring. If you switch from the 1.0 envelope to the 0.5 envelope, you're down 0.5, and if you switch the other way, you're up 0.5. That's it. Accounting is based on debits credits not halving and doubling. You would never update a ledger by crediting double some amount to one account, and debiting half the amount from another account. It's all purely additive. You have to look at what you're potentially gaining or losing on its own not as a fraction of some guaranteed fixed portion that you're getting from either envelope.
  • thematrixturtle 4y ago
    You are missing the point. The problem of whether you should actually switch is obvious and trivial (you should not). The challenge here is explaining why the plausible 'proof' that you should switch is wrong.
    • kazinator 4y ago
      I read that in the Wikipedia page. There is some obvious argument involving repeatedly switching sides and whatnot, and it has some subtle logical flaw. Why bother with that at all. OK, but let's look at it: 1. Denote by A the amount in the player's 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. 3. The other envelope may contain either 2A or A/2. OK, here is where this argument got seduced by the multiplicative thinking. The correct reasoning is: let's call the smaller base amount A and the larger one 2A. One envelope contains A, the other one 2A. That's it. 3. If A is the smaller amount, then the other envelope contains 2A. 4. If A is the larger amount, then the other envelope contains A/2. Here, the argument is splitting into cases because of the IF, and in these two cases, A has a different meaning. 5. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2. 6. So the expected value of the money in the other envelope is: This is wonky. The overall two-envelope situation has an expected value, not the individual envelope. The correct view is that if this is the richer envelope, it contains X more money. If it is the less endowed one, then it contains X less. 1 1 5 - 2A + - (A/2) = - A 2 2 4 7. This is greater than A so, on average, the person reasons that they stand to gain by swapping. That calculation is wrong though, by itself. That issue should be attacked directly. Separate calculations have to be performed for the two cases: A is the smaller amount and A is the bigger amount. For each of these cases, the expected value from both envelopes has to be calculated. The way the above formula is doing the expected value calculation is off because it's equivocating on A. It ends up working with two unrelated quantities: 2A and A/2. 2A is four times bigger than A/2. A is simultaneously denoting the smaller amount and the larger one. 8. After the switch, denote that content by B and reason in exactly the same manner as above. Yes, and if you do, you will end up with 5/4B. So that can't be right and the whole remaining argument about infinite switching being irrational compared to just opening an envelope is superfluous.
• thematrixturtle 4y ago
  Clearly the solution here is three envelopes. http://wikibon.org/wiki/v/Prepare_three_envelopes http://wikibon.org/wiki/v/Prepare_three_envelopes
• jansan 4y ago
  A related problem, but actually interesting, because there is a beneficial strategy, is the Monty Hall Problem. https://en.wikipedia.org/wiki/Monty_Hall_problem https://en.wikipedia.org/wiki/Monty_Hall_problem
  • xcambar 4y ago
    Also, it is a good opportunity to remind that the Monty Hall Problem was mentioned in a Brooklyn 99 episode, in case someone missed on Brooklyn 99. https://m.imdb.com/title/tt6026160/movieconnections https://m.imdb.com/title/tt6026160/movieconnections
• chii 4y ago
  https://www.youtube.com/watch?v=_NGPncypY68 https://www.youtube.com/watch?v=_NGPncypY68 this video explains the paradox properly, with a very reasonable explanation.
  • IshKebab 4y ago
    Yeah the Wikipedia explanation is terrible! Someone rewrite that.
  • ossopite 4y ago
    that's an excellent treatment. I'm curious: this video picks a particular example of a distribution of the amounts and shows that the expected profit from switching is not defined. Can one somehow prove that for all possible distributions, either the expected profit from switching is 0 or it is undefined?
    • chii 4y ago
      the point of the video is that given a distribution, the total expected profit cannot be defined, because the infinite sum of the probabilities of each case don't have an order, and also adds up to positive infinity and negative infinity. it doesn't really matter what the distribution of the amounts are, as long as there's an infinite number of possibilities in the distribution (ie., it's not a finite amount of possible envelopes).
      • ossopite 4y ago
        > as long as there's an infinite number of possibilities in the distribution how would one prove that claim? it's not obvious to me. it seems like the specific conditions under which the series of terms in the total expectation has a well defined sum is important, but I didn't fully grasp those conditions
      • zeroonetwothree 4y ago
        That’s not true for all distributions. For example if instead of halving the probability of each value as he does in the video we take 1/20 then we get the expected value as the sum of 9/20^2 - 9/20^2 + 90/20^3 - 9/20^3 + … Here the series is absolutely convergent since the positive terms sum to 9/10 and the negative terms sum to 9/10.
• beanlog 4y ago
  What an interesting trick. A starting point to intuit a resolution: yes you have a 50% chance of doubling and a 50% of halving. But the doubling only happens if you have x dollars, and the halving only happens if you have 2x. So you can see you either gain x or lose x, so your expected return is 0. When the amount you multiply your money by depends on the amount of money you currently have, you have to factor in your initial money to each case to compute expected return.
  • cyborgx7 4y ago
    Oh, that actually really helped me getting a grasp on the problem. I feel like other comments and sections of the article were trying to get at that idea, but your phrasing of it made it finally click for me. Thanks.
  • IshKebab 4y ago
    This is a far better explanation than the confused hand waving in the Wikipedia article.
• accountLost 4y ago
  My take : let M be the maximum amount of money that a envelope can contain (at worst it can be the total amount of money in the world). You don't know the value of this upper bound M. Most of the time, you get 2*x or x/2 with 50% chance if you choose the other envelope (which on average is a win) But when x=M you get M/2 instead of M and therefore you loose M/2 with 100% chance. And when it happens you have lost much more than in the other cases (because exponentiel...) If you do the math, you can see that what you win for 1 2 4 8 16 ... M/2 is equal to what you loose for M. And so on average what win from switching is 0.
• m463 4y ago
  This reminds me of the Monty Hall Problem: https://en.wikipedia.org/wiki/Monty_Hall_problem https://en.wikipedia.org/wiki/Monty_Hall_problem (except that problem has a more definite solution)
  • YashdharaC 4y ago
    Exactly what I thought too. I find human behavior in such situations fascination. Have penned some thoughts on it - have a read if you like https://ycmusings.com/why-wont-we-switch/ https://ycmusings.com/why-wont-we-switch/
  • mike_hock 4y ago
    In fact, the fake/paradoxical solution probably plays on the reader's familiarity with the Monty Hall problem. ("Ah yeah I remember this. Switching is better!")
  • tgv 4y ago
    The difference is of course that you do get information in the Monty Hall setup, and here you get none, if I read that correctly.
• davidguetta 4y ago
  The bug is obvious when they say "one envelope contains 2*A and the other A/2". Well no, one doesn't contain one fourth of the other.
  • xmprt 4y ago
    That's not what it's saying. It's saying one envelope contains A and the other contains either 2*A or A/2.
    • qwertox 4y ago
      Yes, but doesn't the formula treat it a such?
    • davidguetta 4y ago
      yes but the formula at bullet point 7 says it. That's the point. The formula is wrong and doesn't correspond to reality
  • mike_hock 4y ago
    Basically the sleight of hand trick is using a random variable A as though it was a constant. Yes, B = A/2 or B = 2A with probability .5 each, but not for the same value of A.
    • rawling 4y ago
      Why not?
      • Tepix 4y ago
        A is the starting value, in the one case it's half as much as in the other case.
• JustFinishedBSG 4y ago
  And this is why we try to hammer into students' head that you can only manipulate series after you've established convergence / that the quantity actually exists in general.
• paganel 4y ago
  And what is the solution? I would just take the money at point A, i.e. at the very beginning, you never know when the experiment supervisor decides that he has enough of potentially giving even more free money (in case I'm smarter compared to him/her). Plus, free money at moment t is better than potentially double free money at moment t+t1
  • quickthrower2 4y ago
    At some before point memorise enough random generated entropy so that when you enter this situation you combine that with the Schelling point (the left hand envelope for example) to pick one at random. Then whatever trick is being played you still have a 50-50 chance.
• billpg 4y ago
  Back when I made videos for YouTube, I did a little video on this paradox. I used a pair of cash boxes instead of envelopes because I thought opening boxes looked better on screen than opening an envelope. I know I didn't explain the issue very well because a lot of commenters thought I was discussing the TV game show "Deal Or No Deal". https://www.youtube.com/watch?v=0yjVv_iB1h0 https://www.youtube.com/watch?v=0yjVv_iB1h0 (Interestingly, a lot of those DoND comments have gone. Did they delete them?)
• robocat 4y ago
  > Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch? Normally we would assume there is someone who owns the envelopes of money. Now their incentive could be to offer to switch only if the envelope you chose contains more money (so they have a chance of keeping more). More like Monty Hall. Even without that, after switching, should you switch back? The equation given argues you should switch back, so obviously the equation must be wrong. It is easier to think: if I switch I have a 50% chance of making $X and a 50% chance of losing $X. Switching doesn’t change expected $. I actually think the later problem in the Wikipedia article is more clever: Two people, equally rich, meet to compare the contents of their wallets. Each is ignorant of the contents of the two wallets. The game is as follows: whoever has the least money receives the contents of the wallet of the other (in the case where the amounts are equal, nothing happens). One of the two men can reason: "I have the amount A in my wallet. That's the maximum that I could lose. If I win (probability 0.5), the amount that I'll have in my possession at the end of the game will be more than 2A. Therefore the game is favourable to me." The other man can reason in exactly the same way. In fact, by symmetry, the game is fair. Where is the mistake in the reasoning of each man? — "Martin Gardner: Aha! Gotcha"
  • feintruled 4y ago
    > Even without that, after switching, should you switch back? The equation given argues you should switch back, so obviously the equation must be wrong. Ah, that observation is very clever! Even if you let yourself be bamboozled by the logic for switching, there is no reason then not to switch again by using the same logic. And again.. and again!
  • t_mann 4y ago
    hmm, I don't really see where the probability of 50% comes from? even if we assume that we don't have any contextual knowledge about what amounts would typically be carried in wallets, the probability of winning would surely have to depend in some way on the amount currently in my wallet? in the extreme case, if I had zero cash, my probability of winning (not just drawing) would be zero, since we can't carry negative amounts of cash.
• shannifin 4y ago
  I must admit I don't find the "compelling line of reasoning" all that compelling; to me it seems the "paradox" has more to do with the meaning of probability than with its calculation. That is, the probability is 1/2. That's all that matters to the decision making. Calculating an "expected value" at all is completely useless, whether or not you do it "correctly". Am I missing something?
  • ghusbands 4y ago
    Calculating an expected value is meaningful and regular and does affect decisions. If you were told that you could pay £1 to flip a coin and heads would pay out 50p and tails would pay out £2, then playing it you would expect to make £1.25 (£0.25 profit) per play, on average. Many people interpret the envelope problem as being the same situation.
    • shannifin 4y ago
      > Many people interpret the envelope problem as being the same situation. It just seems obvious to me that it's not the same situation. That is, I think the "paradox" is ultimately about the temptation to treat these as similar problems in the first place.
      • wccrawford 4y ago
        Right, the difference here is that the calculation that determines the expected payout is the same as the calculation that determines how much it costs to switch. With the coin, the cost to play is set and is lower than the expected outcome.
        • shannifin 4y ago
          I was thinking more along the lines of there being no need to actually calculate anything at all because the only relevant information is the 50-50 probability of choosing the higher envelope. And since you gain no new information after choosing, there's no need to calculate anything afterward either. The "paradox" is in making one think there's something to be calculated beyond the 50-50 chance.
• quickthrower2 4y ago
  The paradox is resolved by realising you are doubling a smaller amount than you are halving.
• oli5679 4y ago
  If you treat this as a Bayesian problem, your context and experiences could lead you to specify a prior which has a finite mean, in which case for some values you will have a posterior probability of > 2/3 of it being the larger value. E.g. if the maximum volume of currency that could fit in an envelope is $10k, it would be possible to have a uniform prior between 0-10k, at which point I would stick if my envelope had $5.01k. A more extreme prior could be one capped at the largest ever academic research grant (I bet it's not more than $100m). It could also be that you have a prior that has infinite mean, in which case it is not surprising that you will always want to switch if you draw a specific value. If you have risk aversion, this further increases the range that you would like to stick, since switching could decrease your expected utility, even if it increases your expected return. There is a separate paradox in decision theory that we should be approximately risk-neural for 'small' amounts.
• ghusbands 4y ago
  One mental model I apply to this and other similar problems is tracking what 'universe' you're in. In the flawed solution, the problem is that there's a universe in which the amounts are A and 2A and a universe in which the amounts are A and A/2, but they aren't the same universe, and the two possibilities will have different values for A and so you can't easily average A/2 from one universe with 2A from the other. (The usual term is sample space; it's just more fun to split and eliminate universes than states.)
• one-more-minute 4y ago
  This line of reasoning isn't possible if you have specific amounts (eg £50 and £100) for the envelopes, which suggests that the variable A is being misused somehow. `B = (2A if A=50, A/2 if A=100)`. Simplifying this to `B = 2A or A/2` loses important information: namely that when B is smaller, you expect A to be larger. Or alternatively, treating A as fixed (say A=100) conflates two different situations: one where the amounts are `100, 200` and one where they are `100, 50`. So you end up thinking B must have (200+50)/2 in it, which is incoherent. If you compute the expected B correctly then `A` is different in each branch. It is subtle though. If this appeared in a paper, arguing something not obviously wrong, it'd be hard to convince everyone that the reasoning is bad.
  • lisper 4y ago
    The correct answer is described here: https://www.youtube.com/watch?v=_NGPncypY68 https://www.youtube.com/watch?v=_NGPncypY68 TL;DR (Spoiler alert): the expected value of the amount of money you end up with is an infinite series whose sum changes depending on the order in which you add up the terms, and so you can choose an order that makes this value come out to be positive, negative, or zero.
    • deleted 4y ago
      [deleted]
    • hammock 4y ago
      Right away I was taken aback by the wikipedia article’s quick dive into mathematical solutions to the problem, when it seemed to me that what would be missing is consideration/framing of the initial setup. That initial setup being, the act of starting the game with someone is(?) confers value in and of itself. So I’m glad the real solution is wonky.
      • ghaff 4y ago
        Welcome to Wikipedia. Where people who love using equation editors outnumber those predisposed to writing cogent explanations.
• pontus 4y ago
  I wrote up a pretty detailed analysis of this problem a while back here: https://mindbowling.wordpress.com/2020/09/14/two-envelope-paradox/ https://mindbowling.wordpress.com/2020/09/14/two-envelope-pa... At the end I constructed a super weird example where regardless of what you see in your envelope, you should switch (this seems to go against the symmetry argument, but is sort of saved by a divergence in the expectation value).
• andrewshawcare 4y ago
  Once you pick an envelope, you no longer stand to only gain money. You can lose money and that has to be reflected in the potential value of each envelope. After the first selection, you must express the envelope value as the potential of what each envelope holds (the probabilities from the initial selection) which makes selecting again a wash. Let A = 50 Envelope 1 is 100 Envelope 2 is 25 First selection 1/2(100) + 1/2(25) = 62.5 Great! Do it! 62.5 is bigger than 0, which is the expected value of not playing at all. After that, it makes no difference to switch (the envelope value is recursive): 1/2(62.5) + 1/2(62.5) = 62.5 or 1/2(1/2(2A) + 1/2(A/2)) + 1/2(1/2(2A) + 1/2(A/2)) = 1/2(2A) + 1/2(A/2) = 5/4A and the trick is that’s now the potential expected value you have (you now have 5/4A in your envelope), so switching is a wash. You never really “had” A as a value to compare against (by evaluating that 5/4A is bigger than A), we just get tripped up with the doubling and halving at the outset. Said another way, the fact that 5/4A is bigger than A is irrelevant, no envelope contains A, they both contain the expected value of 5/4A.
  • jph 4y ago
    Excellent answer. Worth adding to the wiki page.
  • lisper 4y ago
    > Let A = 50 Envelope 1 is 100 Envelope 2 is 25 No. The problem specifies that E1 is twice E2, but here you have E1 = 4 x E2. A is the amount in one of the envelopes, so if A=50 then either E1 is 50 or E2 is 50, and the other E is 25 or 100. But under no circumstances can E1=100 and E2=25 at the same time.
    • andrewshawcare 4y ago
      Impossible. Schrodinger’s envelope, then. You can’t reflect the probability to be 2A AND 1/2A if A represents the value of an envelope. The value of the other envelope must include the probability that it is the original A (not some sleight of hand new A’ that, itself, is based on an expected value).
      • 4y ago
• twic 4y ago
  Just keep swapping the envelopes. Infinite money hack!
• ogogmad 4y ago
  Also see the Surprise Execution Paradox. It's a favourite of mine. [edit] Now submitted: https://news.ycombinator.com/item?id=31568880 https://news.ycombinator.com/item?id=31568880
  • deleted 4y ago
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• GeorgeRichard 4y ago
  > Imagine you are given two identical envelopes, each containing money. One contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch? Unless the imaginary “giver” shows you what was in the other imaginary envelope you’ll never know. We don’t even know if this imaginary person is telling the truth. How do you know that “One contains twice as much as the other?” I have never been offered a choice of cash containing envelopes in my life – maybe because I have never worked in a job where bribes could be usefully employed. The whole “problem” is about as worthwhile a subject of contemplation as, say, “Do flying pigs taste more like partridge than earth-bound pigs?” Does anyone know of a paradox that works/exists in the real world? I mean something a bit more substantial than whether some random dude claims to be a liar and you supposedly bother to wonder whether or not he’s telling the truth.
  • cedilla 4y ago
    It's a thought experiment. It's not about envelopes of cash, it's about logic and game theory. Their usefullness can hardly be overstated. But for what it's worth, there are a ton of thought experiments inspired by or lifted from the real world. The Monty hall problem is taken from a game show - it even has a contestant who must choose one of three doors...
    • GeorgeRichard 4y ago
      Why is it a thought experiment? After all envelopes, cash, experimenters etcetera exist. It could be a real experiment. On the other hand what theory or hypothesis would it be testing? What’s the problem with Monty Hall? The solution (switch to the other door) is demonstrably true. I believe it’s called a veridical paradox.
• the_sleaze9 4y ago
  I have to admit I remain unconvinced. Has anyone ever run a real world experiment to verify the math? For whatever reason I can't get past my intuitive feeling that this problem is just a simple 50/50 and these proofs are just fancy window dressing.
  • andrewclunn 4y ago
    It's much like the Monte Hall problem, in that people often forget to include the portion about the revealed door ALWAYS being a loser. If it were a randomly opened door then the option to switch would actually just be 50/50. This can be shown by just running the simulation over thousands of iterations. The same is true here. It's not actually going to yield better outcomes to switch. Sometimes logicians just need to own up to the fact that empiricism and applied mathematics is where the rubber meets the road.
    • selimthegrim 4y ago
      I’m not sure how literally applying Bayes’ theorem and reducing the sample space accordingly is owning logicians but you do you.
    • rtkwe 4y ago
      The difference between this and the Monte Hall problem is there's no new information revealed or pool reduction here.
  • pontus 4y ago
    I wrote a pretty detailed analysis about this problem here in case you're interested: https://mindbowling.wordpress.com/2020/09/14/two-envelope-paradox/ https://mindbowling.wordpress.com/2020/09/14/two-envelope-pa... It all depends on how the envelopes are prepared.
  • olalonde 4y ago
    You remain unconvinced that switching is pointless? Unlike most paradoxes, this is one where the intuitive answer is the correct one. It's a paradox because of the flawed, but convincing, reasoning that is given in favor of switching.
• motohagiography 4y ago
  Intuitively, and ignoring the constraint of the values being (A|2A) if I just assign a random number value to each envelope, does the order in which I have assigned the random numbers change the amount of information I have? I suspect not. However, I now have two random integers with no bits of information about either, where I have now removed information from the system (the relationship between the number of envelopes and the contents of one being twice the other), while leaving the states of the system intact. The envelope I in effect already have yields no bits of information because I didn't choose it, it is just the random (no information) one I start with, whereas the fact of the existence of the second envelope contains more information than my current envelope selection, and I have a decision to make about whether to switch. The null state of the system is me holding an envelope with no information, and addition of a second envelope is a new state of possibilities, where the information in this two envelope system is now non-zero. The decision with the most information is to switch. This additional information is independent of the implied desire for (A|2A), whereas the original problem is defined as the best possible decision is the one with the most information. Further, if I perform some operation with those two random integers (Za and Zb), the result has less "entropy" than the two random inputs in isolation, as by correlating them via a function I have produced some non-zero negentropy between them, say using xor, which is how I added information back into the system after removing it using the random numbers. I get this new number N composed of my two random numbers ( N = xor(Za, Zb) ), and I need to apply it to my choice of, not which envelope to select first (A|B) - but whether I should switch. The only way to get this number with information in it is to have assigned a random number to both envelopes and used the two bits of negative information to cancel each other - and therefore the only choice with non-zero information is the one which includes both envelopes - hence the decision to switch is the only one that yields the non-zero information outcome. Again, intuitively it seems related to how multiplying two negative integers yields a positive one, multiplying two entropy (negative information) sources yields negentropy (positive information). That could further imply that the information in the decision to switch is also based on whether the number of envelopes is even or odd, or maybe prime or composite? Sounding this out, it may also be related to graphs where the number of nodes/edges (where edges represent switching decisions and nodes envelopes) either forms a closed path or open one, and in the case of an even number of envelopes you get the max information by swtiching n-times, and in the case of odd ones, you switch n+1 times. Short version is, the choice with the highest number of bits of information in the two envelope case is to switch because it is predicated on there being more envelopes than the one you have started with.
• civilized 4y ago
  I'm going to present a critique similar to this one ( https://news.ycombinator.com/item?id=31568047 https://news.ycombinator.com/item?id=31568047) but I'll try to slightly rephrase. Suppose the amount in the envelope you choose is $100. If your amount is the smaller one, the other one has $200. If your amount is the larger one, the other one has $50. So the expectation supporting the switching argument is computed over two different scenarios, where the amounts in the envelopes are {50, 100} and {100, 200} respectively. This calculation seems to violate a hidden logical constraint: the set of amounts in the two envelopes must be the same no matter what conditional branches we consider for the expectation. Another related issue - if you switch, the argument for switching back assumes the amount in your original envelope is 4x what you originally said it was.
  • one-more-minute 4y ago
    To expand on this, the switching argument is actually correct if you happen to know that the envelopes contain either {x, 2x} or {2x, 4x} dollars (where x = 50 or whatever), and you observe that A=100. These are effectively hidden assumptions of the argument. If you wanted to, you could explicitly model all possible values of the envelopes, and all possible observations of A – and you'd get the right answer again, including in the case above. You just can't define A in a path-dependent way, and then treat it as having the same value in all paths.
  • phkahler 4y ago
    So the issue is assigning the value A to the selected envelope and then doing conditional calculations for only the other envelop.
• kloch 4y ago
  A simple non-mathematical resolution is that no information has been gained by selecting the first envelope. Since no information about the contents was available before selecting the first envelope switching envelopes is a neutral act. While this sounds like just a lazy intuitive explanation, it calls out the key component - a lack of information about the outcome - that can help avoid the paradox in more complex situations. This is in contrast to the three-door problem where information is gained by opening the first door. https://en.wikipedia.org/wiki/Monty_Hall_problem https://en.wikipedia.org/wiki/Monty_Hall_problem
  • t_mann 4y ago
    There's an amusing amount of probability puzzles with seemingly paradoxical solutions out there, to the point that it's almost more surprising to see one where the intuitive answer is correct: - https://en.wikipedia.org/wiki/Boy_or_Girl_paradox https://en.wikipedia.org/wiki/Boy_or_Girl_paradox - https://en.wikipedia.org/wiki/Penney%27s_game https://en.wikipedia.org/wiki/Penney%27s_game - Waiting time paradox - Honorary mention: https://en.wikipedia.org/wiki/St._Petersburg_paradox https://en.wikipedia.org/wiki/St._Petersburg_paradox (feel free to add more)
    • deelly 4y ago
      https://en.wikipedia.org/wiki/Monty_Hall_problem https://en.wikipedia.org/wiki/Monty_Hall_problem
  • sorokod 4y ago
    This is not a resolution, a resolution is where you point out the faulty step in the reasoning that leads to the paradox.
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• jonath_laurent 4y ago
  Some excellent critiques have been provided in this thread already but I would like to offer a different perspective. In programming terms, the switching argument (see original link) is incorrect because it does not typecheck. And it does not typecheck because variable A is used out of its scope when writing down the (5/4)*A expected value. Indeed, variable A is tied to a specific random outcome and so it simply does not make sense to refer to it in an expected value ranging over this same outcome. Trying to formalize the argument in a proof assistant makes the mistake clear and obvious. However, the ambiguity of the English language makes it possible for such typing errors to sneak in undetected.
  • lisper 4y ago
    > it does not typecheck ... because variable A is used out of its scope when writing down the (5/4)*A expected value But that's simply not true. > variable A is tied to a specific random outcome and so it simply does not make sense to refer to it in an expected value ranging over this same outcome No, it isn't. It is the value of the chosen envelope -- which is known. You can open the envelope and look inside. > Trying to formalize the argument in a proof assistant makes the mistake clear and obvious. Have you actually tried this? Consider the following alternative scenario: I give you $20 and offer you the following wager: we will flip a fair coin. If it lands heads you lose $10, tails you win (an additional) $20. Would you take that wager? If so, how is that different from the envelope problem? (Note that the $20 grant is a no-op, all it does is serve to fix the amount of the payouts after the coin flip.)
• cosmojg 4y ago
  Comparing this to the Monty Hall problem: Probability of choosing the correct door at time of choice: Two Envelopes: 1/2 -> 1/2 Monty Hall: 1/3 -> 1/2 ...because Monty opens one of the doors, providing information.
  • devman0 4y ago
    Monty opening the door condenses the probabilities of the two unpicked doors in to the one unpicked door. This occurs because Monty only opens a door with a goat and never the car. In the original three door version you always had a 2/3 chance of winning if you switched because you'd only lose if you correctly picked the car on the original 1/3 pick.
• m3kw9 4y ago
  Whatever formulation that compels the player to swap is flawed.
• ChrisLomont 4y ago
  This reminds me of a fascinating problem (not sure if there's a name for it): You play the following game (say for many rounds). Someone puts different values in two envelopes (they can pick the values arbitrarily, even maliciously, each round). You want to pick the higher valued envelope. You pick an envelope, they show you the value of the other. You can hold your value or you can switch. It also works if the game is they show you your own value and you can choose to hold or switch. Actually, now that I think about it, you can even make the game such that you pick an envelope and tell them which envelope to show you, then you can hold or switch. More interesting :) The questions is: is there a strategy where you get the higher value strictly more than 50% of the time? The surprising answer is, yes, there is such a strategy. It's really neat, and I've tried to find other places to apply the reasoning, but have not found another interesting place. And yes, this problem and strategy are mathematically correct, not cheating, not "out of the box" stuff. Good luck :)
  • metacritic12 4y ago
    Generate a random number T from the arctan distribution. If the number in the other envelope is greater than T, switch. For envelopes with two different numbers n, m, the above is guaranteed to give you the higher one strictly more than 50% of the time.
    • zeroonetwothree 4y ago
      To expand on this, you can use any distribution that has nonzero density over every interval.
    • ChrisLomont 4y ago
      Yep - any distribution that is nonzero over the set of choices works, so over the reals, anything that is everywhere nonzero works. For example, exp^{-x^2} with normalization factor is a simple choice.
      • emerongi 4y ago
        Is there a complete explanation for this somewhere? Asking for a friend, I obviously fully comprehend what you're talking about here.
• johnrob 4y ago
  Sounds like a misuse of relative return (growth rate). If you gain 100% today, then lose 50% tomorrow, you are back where you started. Absolute numbers give you the correct answer: x -> 2x is a gain of x, 2x -> x is a loss of x, so expected value is zero (x is a hard number not a ratio).
  • metacritic12 4y ago
    Exactly! The logic of "the average of +100% and -50% is positive" only applies if the denominator is the same. In this case it isn't. The relative percentage "rule of thumb" doesn't even feel like a primitive in math. It seems wrong to reason from that.
    • platz 4y ago
      why isn't this listed in wikipedia as an answer.
      • vehementi 4y ago
        It is, it's the one where you treat both as having the same expected value
        • platz 4y ago
          that the expected value is the same that is an outcome of all solutions, so just stating that is reductive. What is at question is the reason for getting to the expected value is the same.
      • whatshisface 4y ago
        Because if the larger envelope is changed to contain four times as many dollars, then /2 vs. *4 points back towards swapping indefinitely.
        • MrBlueIncognito 4y ago
          If one envelope contains 4 times the money of the other envelope, then by swapping you either gain 3x or lose 3x, where x is the amount of money in the smaller envelope. Even in this case, the average return is +3x -3x = 0.
• vzaliva 4y ago
  Just write a python script simulating this :)
• praptak 4y ago
  My solution is this: In point 2. the solution starts talking about probabilities without defining the sample space. And this is why the solution is nonsensical. "One envelope contains twice as much as the other" does not define a sample space, so it is too early to start talking about probabilities. If you say "the sample space is all pairs of natural numbers where one is twice as big as the other", then what are the probabilities assigned to each pair? They cannot be pairwise equal because the space is infinite. So even if probabilities for (X, 2*X) and (X, X/2) happen to be equal for a particular X, they definitely cannot be equal for every X, so the "infinite swap" doesn't hold anymore.
• Gehinnn 4y ago
  The mistake is in step 7. You cannot calculate the expected value (of the envelope you get, when you pick one randomly and then swap) like that, as A is not a constant and actually depends on if you took the less or more valuable envelope initially. If you use A1=X in case you took the less valuable envelope initially and A2=2X in the other case, the expected value is (1/2)*(2*A1) + (1/2)*(A2/2) = 1.5X and if you don't swap, you get: (1/2)*A1 + (1/2)*A2 = 1.5X ... which makes sense. If you change the experiment so that the value of the second envelope does not depend on the first one (if you first chose and only then the value of the other envelope is randomly either double or half), A is constant and you actually should swap.
• dang 4y ago
  Related: The Two Envelopes Problem: the most boring paradox in the world - https://news.ycombinator.com/item?id=6388645 https://news.ycombinator.com/item?id=6388645 - Sept 2013 (12 comments) Two envelopes problem - https://news.ycombinator.com/item?id=6387044 https://news.ycombinator.com/item?id=6387044 - Sept 2013 (88 comments) Two Envelopes Problem - https://news.ycombinator.com/item?id=1582219 https://news.ycombinator.com/item?id=1582219 - Aug 2010 (88 comments)
• jonath_laurent 4y ago
  I would argue that this Wikipedia article is misleading and that it confuses more than it clarifies when it comes to resolving the paradox. In particular, I am disputing the fact that "no proposed solution is widely accepted as definitive" (exact quote). Indeed, the switching argument (or at least the argument as it is presented in the Wikipedia article) makes a clear and precise mistake that I claim any trained mathematician will immediately agree on once pointed to it. This mistake is explained in the following comments: https://news.ycombinator.com/item?id=31566226 https://news.ycombinator.com/item?id=31566226, https://news.ycombinator.com/item?id=31567251 https://news.ycombinator.com/item?id=31567251 and https://news.ycombinator.com/item?id=31569991 https://news.ycombinator.com/item?id=31569991. Once you have identified it, this mistake is all but subtle. It is not about infinite series or Bayesian reasoning. It does not require a 30 minutes rebuttal talk. It is simply about misusing the very definition of an expected value in a way that could be qualified as a typing error (see linked comments). This error can only go undetected because of the ambiguity of the English language. Most of the fancy mathematical discussions I've been reading are distractions. This paradox is about language and not about any deep mathematical fact of probability theory.
  • Chinjut 4y ago
    This is not the issue. When expected value is well-defined, it is perfectly fair to compute E[X] = E[E[X | A]], where E[X | A] is a function of random variable A ("conditional expectation"). If E[X | A] > A for every particular value of A, then E[X] > E[A], whenever the expected values E[X], E[A], and E[X - A] are all well-defined. The issue in this case is that E[X - A] is not well-defined. The expected profit from switching is given by an infinite series whose positive and negative components are each of infinite total magnitude, thus yielding different sums when bracketed differently ("conditional convergence"; just a coincidence that the word "conditional" comes up here again).
    • jonath_laurent 4y ago
      This is a good comment and I believe your justification is equally valid. The reason for the apparent disagreement is that in order to point out a logical flaw in an argument, one must make the argument formal and explicit enough first. There are several plausible ways that the Wikipedia argument can be translated into a machine-checkable formal proof (and this ambiguity is at the core of the paradox). When I make such an attempt in my head, I am not interpreting the expectation as a conditional one and the problematic step is about conflating a random variable with a fixed constant. In your attempt to formalize the Wikipedia argument, you try and use the iterated expectation theorem and the problematic step becomes different. Seeing from the comments, several people agree with my interpretation but clearly there is an ambiguity here that is part of the paradox. Thanks for prompting me to add nuance to my comment.
• GeorgeKangas 4y ago
  Why is no one looking at the chance-giver, and considering his motive? One would reasonably assume he'd like to minimize the payout -- so he'd offer you the chance to switch only if you had chosen the richer envelope. The description explicitly allows this possibility: "Having chosen... you are given the chance to switch" Nowhere does it say, you'll be offered the switch no matter which envelope you choose. Same idea applies, to the very similar Monty Hall problem.
• mathogre 4y ago
  Take 10 minutes to model it and you'll find there is no benefit in changing. As a couple others have pointed out, it isn't the Monte Hall problem. Below I did two runs with a million cases. One envelope has 1 unit; the other has 2 units. I shuffle the envelopes, and the "player" chooses an envelope. "noChange" means this is the payoff if the player doesn't switch envelopes. "yaChange" means this is the payoff if the player switches envelopes. $ python3 ./twoEnvelopes_00.py 1000000 noChange = 1500023 yaChange = 1499977 $ python3 ./twoEnvelopes_00.py 1000000 noChange = 1499984 yaChange = 1500016 When modeling Monte Hall, the payoff for changing is obvious.
• WhyOhWhyQ 4y ago
  The Monty Hall problem has a simple resolution. At the beginning you had a 1/3 chance of picking the good door, and 2/3 chance of picking a bad door, meaning you probably picked a bad door at the start, so you should switch.
  • entropicgravity 4y ago
    You can say 'simple' but a whole pantheon of academics famously got it wrong. If you really want to understand the problem use 100 doors instead of three and then it becomes very clear whats going on.
    • orthoxerox 4y ago
      Or a million scratch-off tickets.
    • WhyOhWhyQ 4y ago
      I'm also an academic, if that matters for some reason, and I think it doesn't matter how many people didn't understand something. They must have been confusing themselves by seeking answers in the wrong direction. What matters is that the resolution I've given truly is simple. You have a 1/3 chance of being in a situation where changing doors is unfavorable, you have a 2/3 chance of being in a situation where changing doors is favorable. If you picked the door with the car, which happens with 1/3 probability, at the next stage switching is a bad option. If you picked a door with the goat, which happens with 2/3 probability, at the next stage switching is a good option. This is objectively simple, regardless of what a large number of people might think. I also don't think changing the problem to an analogous one is as helpful as this direct and simple solution. Using 100 doors could mean multiple things. Maybe it means at the next stage you have 99 doors left. Maybe it means at the next stage you have 2 doors left. Why should I think either is the correct analog? You can just see the simple answer I've stated and be done with it.
• bergenty 4y ago
• kmod 4y ago
  If it is truly the case that the second envelope has equal probability of having half or double the amount of money as the first, then you can prove that the probability distribution of money in each envelope is not well-formed (has total mass either zero or infinity). This means the premise is already contradictory, and further false conclusions (such as "P and not-P" for P="it is better to switch") are not surprising.
  • baobabKoodaa 4y ago
    > If it is truly the case that the second envelope has equal probability of having half or double the amount of money as the first, then you can prove that the probability distribution of money in each envelope is not well-formed (has total mass either zero or infinity). This means the premise is already contradictory... What? I can literally take 2 envelopes right now - physical envelopes - and I can literally put $10 in one envelope and $20 in the other envelope. Then I can shuffle the envelopes and hand one of them to you. Now the other envelope has equal probability of having half or double the amount of money as the first. How is this premise contradictory in your opinion? I can physically arrange objects in this fashion, so where's the contradiction?
    • kmod 4y ago
      If I open an envelope with $10 then I know that there is $20 in the other envelope. As soon as you assign a well-formed probability distribution to the money in the envelopes you will find that opening the first envelope is informative.
      • baobabKoodaa 4y ago
        Opening the envelope is something that you added to the problem just now. It wasn't present in the original problem, and it wasn't present in your earlier post. Here's a quote from Wikipedia: "Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?" Note the part that says "before inspecting it". That means you can't open it.
• vivegi 4y ago
  Reminds me of the "splitting the dinner bill" problem we had fun with in middle school. It goes like this. Three friends go out for dinner and decide they will split the bill threeways. The bill comes to $25. Each friend gives a $10 note with $30 pooled together to pay the bill. The waitress returns with five $1 bills in change. They leave a $3 tip for the waitress with $2 in change remaining. The question is: Of the original $30 paid if we deduct $3 in tips paid, we get $27. Adding the $2 in change remaining we get $29. Where is the other $1?
• nonameking2026 4y ago
  Sounds like the "free will" discussion - go have fun while other people solve real problems.
• JoshCole 4y ago
  The game looks like this in tree form: -> Keep -> 2A 2A -> Switch -> A 0.5 -> A -> Keep -> A -> Switch -> 2A Switch and keep both have an expectation of 3A/2 because (1/2)2A + (1/2)A. In the equation in the article they use (1/2)2A + (1/2)*A/2. They count A/2 as half of what the actual outcome should be. The mistake was abusing the A to have different meanings. They created two trees one with 2A. -> Keep -> 2A 2A -> Switch -> A 0.5 -> A -> Keep -> A -> Switch -> 2A And another with 1/2A -> Keep -> A A -> Switch -> 1/2A 0.5 -> 1/2A -> Keep -> 1/2A -> Switch -> A Than they treated the chance node as selecting from these two different trees at the same time with 1/2 probability. Which is nonsensical. You can't be in an entirely different game with 1/2 probability - it isn't the same game so for analysis purposes it has 0 probability of being reached.
  • JoshCole 4y ago
    So why is this happening? Some people are pointing at the error I mentioned above, but I think the problem is actually much deeper than that. Identifying this error is basically identifying the superficial reason for why they went wrong. It doesn't dig deep enough to systematically eliminate the error. I want to point out something. Notice their logic when they call out why it is a paradox. You can switch multiple times and it would never terminate. Mathematical formalism like markov decision processes and the bellman equations actually have the ability to capture this type of reasoning. They can even deal with infinite loops as happens here by including multiplication by a horizon term which makes the limit of the recursion equal to zero. So the root problem is that, actually, the entire problem wasn't even using the right theoretical grounding. To get really explicit on how it was not using the right grounding I want to point out the key error again, but stated a bit differently. The author neglects information sets. He creates two game trees, discriminating on the basis of knowledge of which subgame he is in. This is a huge red flag. You can do that in perfect information games, but you can't in imperfect information games! You aren't playing over states in imperfect information games. You're playing over information sets. You end up getting more insights into how this is wrong when you think about the problem in this way. For example, there might not be an error in this particular problem, but the choice to not express the action space as a probability vector is going to bite you in more complicated imperfect information decision problems.
• JoshCole 4y ago
  It is trivial! It is trivial! The game defines a graph and your path through that graph is going to lead to an outcome. You have a strategy, a choice of how to move through the graph. However, you have no information about where you are in the graph. So when we think about how we move through the graph, we must always think the same way. We can never think differently, because no action changes the information we have about where we are. However, the terminal case is defined thusly. R(STAY) = {50%: 2A, 50% A}. There is never a reward for R(SWITCH). Anyone telling you to always choose switch when that is the only solution that can never have a reward is obviously wrong. But why? Why were they wrong? Simple. In imperfect information games things work differently than in perfect information games. In chess when you make a move it is that move. If you make a move it is a different move. The subgames are independent of each other. In poker, it isn't so! If you make a bet in Poker, you don't know what subgame you are in. You have to play as if you're in multiple at the same time. Now look at the fast con that seems to be tripping people up. They are pretending we have perfect information. 2A, A/2? Phaw. We can't have that, because that supposes we know which subgame we are in, but we can't know which subgame we are in, because we were never given that information. Even if we were - it doesn't matter. We're in an imperfect information game! The places we didn't explore are relevant to our calculation! We can't ignore them! But we did, by insisting on an inequality which didn't hold. This is probably counterintuitive. Its trivial, but people argue about it. So lets talk about this more concretely. This is what they pulled on you. They claimed that these two different game trees could be treated as the same game tree: -> Keep -> 2A 2A -> Switch -> A 0.5 -> A -> Keep -> A -> Switch -> 2A -> Keep -> A A -> Switch -> A/2 0.5 -> A/2 -> Keep -> A/2 -> Switch -> A Two different game trees! Two different game trees and they acted like they were the same. But they aren't! They can't be! You can't tell which subgame you are in you have to treat them as equal in imperfect information games! So what is our actual situation with their reasoning? Our actual situation is a contradiction. Since subgames depends on each other when they do this combination they are secretly speaking nonsense. Look closely: They are claiming that 2A=A; that A=A/2. They have to be, because in imperfect information games we have to be in every subgame at once, but they made two different games, but then they combined them. Since the correct formulation has to have every solution treated as the same, but they took something from a different subgame, they declared that two subgames were equal, but they weren't. It is so trivial. That's why the correct calculation looks like only an expectation on one game tree. That's why it seems to make sense - your brain knows this sort of reasoning can be appropriate with perfect information. I realize a lot of people have argued about this, but frankly they should stop. And if you're interested in exploring why you have to treat subgames as different in imperfect information, why you have to operate on information sets rather than state spaces in imperfect information, here is a great talk on the NIPS 2017 Best Paper award which happens to touch on the subject: https://www.youtube.com/watch?v=tRiaGahlyy4 https://www.youtube.com/watch?v=tRiaGahlyy4 It is about solving poker, which is a much more complicated imperfect information game. But even the simple imperfect information games have this property.
• JoshCole 4y ago
  > The puzzle is to find the flaw in the very compelling line of reasoning above. This includes determining exactly why and under what conditions that step is not correct, in order to be sure not to make this mistake in a more complicated situation where the misstep may not be so obvious. In short, the problem is to solve the paradox. Thus, in particular, the puzzle is not solved by the very simple task of finding another way to calculate the probabilities that does not lead to a contradiction. There are two flaws. One is the omission of counterfactual reasoning. This produces the logical contradiction 2=1 at step 7, but you get there a bit earlier because the steps before it are where you do case based reasoning. The scenario where this happens is when you are in imperfect information contexts. You can reason about subgames in perfect information games without the subgames influencing each other. You can't reason about them in imperfect information games without the subgames influencing each other. The lessons is this - you have to consider counterfactuals. To give a concrete example of someone doing this, think Jeff Bezos when he talks about how he decided to found Amazon. He says he considered the situation in which he founded it and failed and the situation where he founded it and succeeded and the situation where he didn't found it. He didn't get his expectation conditioned on one state, but over an information set that contained many counterfactual outcomes. The second flaw is much deeper and is what leads to the paradox. The failure is that the entire framing fails to recognize that EV is a function of policy. You can see this by setting the policy to always switch, as they do, and noting that the actual EV for switch is now undefined. Therefore in claiming knowledge of EV without having policy as a higher order function of EV you get a contradiction. 3/2A=0 and/or 3/2A=undefined. The lesson in this is that policy function influences EV. This should hopefully be obvious to most people in more complicated situations. For example, slamming your head into your table has a lower EV than enjoying a drink of water. You have to account for this whenever your policy choice isn't defined.
• salawat 4y ago
  Sidestep the evil statistician/probability punter's conundrum, and walk off with your found money. Have lunch with the Free Real Estate guy.