# 2 envelopes problem revisited **URL:** \ **Category:** Factual Questions\ **Created:** [June 20, 2012, 3:23pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755 "2012-06-20T15:23:33Z")\ **Posts on this page:** 19\ **Page:** 6
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [June 26, 2012, 7:29pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/101 "2012-06-26T19:29:58Z")
> [@Indistinguishable](#): > > The paradox is this: Conditioned on any particular value for your envelope, the expected profit from switching is positive, which leads us to say that unconditionally, the expected profit from switching is positive. Conditioned on any particular value for the other envelope, the expected profit from switching is negative, which leads us to say that unconditionally, the expected profit from switching is negative. And conditioned on any particular unordered pair of monetary values for the two envelopes, the expected profit from switching is zero, which leads us to say that unconditionally, the expected profit from switching is zero. > > And the explanation is this: Yup. That’s right. The sign of the expected profit (equivalently, the sign of the total profit over all cases) changes based on how you calculate it; it is given by a conditionally convergent summation. If a summation has infinitely many positive and negative terms, it is indeed possible for its value to change based on how you group the calculation, as you have just shown with this example. > > It’s the same phenomenon as that 1 - 1 + 1 - 1 + 1 - 1 + … = (1 - 1) + (1 - 1) + (1 - 1) + … = 0 + 0 + 0 + … = 0, while also 1 - 1 + 1 - 1 + 1 - 1 + … = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + … = -1 + 0 + 0 + 0 + … = -1, and also 1 - 1 + 1 - 1 + 1 - 1 + … = 1 - (1 - 1 + 1 - 1 + …), and thus must equal 1/2. > > Any finite truncation “resolves” the paradox because a finite summation cannot give two different values based on how you group it. But this is essentially to ignore the actual underlying phenomenon (after all, there is more than one way to truncate; we can specify a lowest and highest value altogether (giving an expected profit of zero by switching), a lowest and highest value for your envelope alone (giving a positive expected profit by switching), or a lowest and highest value for the other envelope alone (giving a negative expected profit by switching)). Okay. But what should you _do_? ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 26, 2012, 7:38pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/102 "2012-06-26T19:38:01Z")
Not make all your decisions based on average profit. I would happily keep my envelope or swap it; I am indifferent between the two. But not because of an average profit calculation. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [June 26, 2012, 7:38pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/103 "2012-06-26T19:38:38Z")
What about the modified version where you actually have a value for the envelope you initially selected? There are two envelopes, one has twice as much as the other, you pick up an envelope, open it, and see that it contains $100. So, it would seem, there’s a .5 probability that the other envelope contains $200, and a .5 probability that it contains $50. By switching, you stand equal chances to either gain a hundred or lose fifty, so it seems that if you want to undertake an action which, if repeated in the long run under relevantly similar circumstances, would tend to cause you to gain money rather than lose it, then you should switch. Suppose I make that argument, and say, “because of this reasoning, I hereby switch.” Have I made a mistake? If not in the reasoning, then in the application of the reasoning to action? Or have I not made a mistake–should I in fact switch? ETA: This was posted before **Indistinguishable** ’s post, just to be clear. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [June 26, 2012, 7:41pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/104 "2012-06-26T19:41:58Z")
> [@Indistinguishable](#): > > Not make all your decisions based on average profit. And if I have a reason to make this particular decision based on average profit, is there anything in particular that I should do? I guess you’re saying no–that average profit isn’t well defined here, right? But isn’t average profit just “how much I would tend to gain if I undertook this action under relevantly similar circumstances?” Well, for more along that line see my other post… ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 26, 2012, 7:46pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/105 "2012-06-26T19:46:45Z")
A few random responses: > [@Chronos](#): > > And the fact that folks refer to it as a “paradox” is itself a sign that the mathematics is being related to the real world. All of the mathematical “paradoces” ultimately come from some bit of mathematics that isn’t actually modeling something in our world, but which looks almost like it is, so that the differences between the mathematical results and what actually happens are called into contrast. I wanted to come back to this and say I do agree with everything in it, except the phrase “the real world”. In its place could go anything people have built up strong intuitions about, which is sometimes the real world and sometimes not (except in the sense that everything is part of the real world). All the paradoxes come from supposing system A to act like system B, which it is similar to but not exactly the same as, and being surprised when they act differently. > [@Chronos](#): > > Put another way, whenever you open the first envelope, you’re always going to be disappointed. Before you opened either envelope, the expected value for both was infinite. Then, you opened one, and found some finite value. No matter what finite value it was, it was certainly less than infinite, and so you always do worse than your expectation. Hah, this is a really interesting observation, that there are situations where _everything_ is below average. The Lake Wobegon effect made real! I don’t think it actually explains what happens in the two-envelope paradox (it’s not the case that, conditioned on the value of your envelope, the other envelope’s average value is still infinite), but it does suggest a host of other nice paradoxes. [E.g., most similarly to this one, we could have two envelopes filled at independent random from “Wobegon” distributions (such as uniform random natural numbers, if such things are to be tolerated, or flipping coins and doubling the prize till a tail comes up). Conditioned on any possible value for either envelope, the average value of the other envelope is infinitely higher.] ---
**Author:** ![aux203](https://avatars.discourse-cdn.com/v4/letter/a/77aa72/32.png) [@aux203](https://boards.straightdope.com/u/aux203)\ **Post date:** [June 26, 2012, 7:58pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/106 "2012-06-26T19:58:23Z")
> [@Frylock](#): > > What about the modified version where you actually have a value for the envelope you initially selected? > > There are two envelopes, one has twice as much as the other, you pick up an envelope, open it, and see that it contains $100. So, it would seem, there’s a .5 probability that the other envelope contains $200, and a .5 probability that it contains $50. By switching, you stand equal chances to either gain a hundred or lose fifty, so it seems that if you want to undertake an action which, if repeated in the long run under relevantly similar circumstances, would tend to cause you to gain money rather than lose it, then you should switch. > > Suppose I make that argument, and say, “because of this reasoning, I hereby switch.” Have I made a mistake? If not in the reasoning, then in the application of the reasoning to action? Or have I not made a mistake–should I in fact switch? > > ETA: This was posted before **Indistinguishable** ’s post, just to be clear. Whether you made a mistake or not depends now what you think the chances are that the person stuffing the envelopes put more than 199$ in one of them. If you think that’s the max that could possibly be in there, then you shouldn’t switch. If you think there’s more than that, then you should. Now you could say that’s not part of the original question but that’s the point: The scenario is incomplete. That’s why it get so confusing. In general, without a defined limit? Switch every time. Now, Indistinguishable also pointed out that utility is a factor. We’re specifically ignoring it for this problem. If you don’t, things make a bit more sense in the real world. But that brings us to the St Petersburg paradox. One paradox at a time though… ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 26, 2012, 8:28pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/107 "2012-06-26T20:28:55Z")
> [@Frylock](#): > > By switching, you stand equal chances to either gain a hundred or lose fifty, so it seems that if you want to undertake an action which, if repeated in the long run under relevantly similar circumstances, would tend to cause you to gain money rather than lose it, then you should switch. Do you intend to repeat your action in the long run? I don’t think there’s anything wrong with someone saying “Action A has a 99.99% chance of losing me twenty bucks, and a 0.01% chance of winning me a million dollars. Action B has a 99.99% chance of me winning twenty bucks, and an 0.01% chance of me losing a hundred bucks. Forced to choose between these two as one-off actions, I would rather take Action B, where I almost certainly end up better off, than Action A, where I almost certainly end up worse off.” Of course, the game “Play A a million times and add the results” and the game “Play B a million times and add the results” have different better off vs. worse off probabilities, and I might make a different choice regarding those. Yes, there’s a soritical paradox lurking here regarding “Well, if you prefer B to A once, you should prefer it twice, and three times, all the way through…”. And there are paradoxes regarding the framing of a game: if I’m in a cube lottery, trying to get the largest cube possible, should I focus on the average edge-length, average side-area, or average volume (which do not track each other)? Well, I don’t really care about money or cubes, I care about utils, say some. If I’m VNM-rational, we can construct utils, and by definition, I seek to maximize their average value. Then what happens with situations with indeterminate average value? If I were in one you might observe the fact, which I knew from the start, that I am not VNM-rational. If God offered me a blank check and said “Hey, write yourself as many utils as you like”, would I write a value, and then kick myself for not writing a value one util higher? I don’t know; I’ve never been in that situation. I suppose I would have to. Or perhaps there’s a maximum on utils? I don’t know; I’ve never been in that situation either. I’m being a bit glib, but I am not generally of the position that there is some Platonic fact of the matter as to a labelling of each decision in each situation as either “correct” or “incorrect”. I do what I want, and my wants (and thus my decisions) react to argument, but they don’t necessarily constrain themselves into any kind of theoretical consistency. It is a true fact about me that I can be coaxed into different responses to the same choice by different arguments. If I become aware of this possibility for contradiction, I may throw up my hands and follow my gut, heart, or other trustworthy internal organ. If I remain unaware of it… I remain unaware of it. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [June 26, 2012, 8:29pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/108 "2012-06-26T20:29:17Z")
> [@aux203](#): > > Whether you made a mistake or not depends now what you think the chances are that the person stuffing the envelopes put more than 199$ in one of them. If you think that’s the max that could possibly be in there, then you shouldn’t switch. If you think there’s more than that, then you should. So if it’s a generous rich man who I trust, then I should switch. But isn’t that exactly what’s supposed to be counterintuitive? ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [June 26, 2012, 8:30pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/109 "2012-06-26T20:30:22Z")
> [@Indistinguishable](#): > > Understanding the whole problem helps one cut through the fog of particular questions. When OP asks about a fallacy which applies even to the trivial case where the envelopes are known to contain exactly $1 and $2, going beyond that is obfuscation. The more intriguing problems may be worth discussing, after OP’s question is answered. (I’m still waiting for the clearest one-sentence debunking of OP’s fallacy; obviously mine was unsatisfactory.) > [@Indistinguishable](#): > > Come now, this is a cheap shot. Using Google, I see that expected values, arithmetic, and cash are all used in quantum physics. Perhaps no one should mention those, either. 😕 😕 😕 I’m sorry if I gave offence. My question was sincere. I find both quantum and non-Kolmogorovian probabilities confusing and might benefit from a brief explication. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 26, 2012, 9:02pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/110 "2012-06-26T21:02:30Z")
> [@septimus](#): > > When OP asks about a fallacy which applies even to the trivial case where the envelopes are known to contain exactly $1 and $2, going beyond that is obfuscation. The more intriguing problems may be worth discussing, after OP’s question is answered. (I’m still waiting for the clearest one-sentence debunking of OP’s fallacy; obviously mine was unsatisfactory.) The problem is that the OP erroneously assumed the average profit from switching, conditioned on the value of their envelope, to be zero. And, fair enough, your example shows that the average profit from switching needn’t be zero if, for example, your envelope contains the lowest or highest values possible. But the OP might still say “That’s not what I’m interested in. What I’m interested in has no lowest or highest value; every value is on equal footing.” The answer is the same “You assumed the average profit from switching to be zero, without grounds for this assumption”, but the further task is to figure out why they supposed this assumption, and dissolve their urge to do so. > [@](#): > > 😕 😕 😕 > I’m sorry if I gave offence. My question was sincere. I find both quantum and non-Kolmogorovian probabilities confusing and might benefit from a brief explication. Sorry, I misinterpreted your intent. “Kolmogorovian probability” is just the name for a particular technical formalization of probability; the salient points for us are that it assumes probabilities are real numbers (in the sense that a probability is entirely determined by which ratios of integers are below and above it), and that any way of carving outcomes up into a countable infinity of exclusive and exhaustive possibilities is such that the probabilities of these possibilities form an infinite series summing to 1. One problem this creates is that it implies you cannot have a countably infinite set of equiprobable exhaustive and exclusive possibilities. So you can’t have a uniformly random integer or such things. “Non-Kolmogorovian probability” is just any account of probability which doesn’t demand this particular technical formalization. So, for example, one might have a non-Kolmogorovian account of probability on which the probability of a natural number having property p conditioned on property q is the limiting ratio, as n goes to infinity, of the probability of a uniformly random natural number less than n having property p conditioned on property q. This gives some account of what a “uniformly random natural” behaves like, but that account doesn’t hew strictly to the Kolmogorov axioms. There’s no one particular formalization of what non-Kolmogorovian probability means; it just means any discussion of probability that isn’t strictly in adherence to the Kolmogorov axioms. Quantum mechanics goes even further askew… Probability theory is in some sense the study of weighted averages where the weights are semipositive real numbers summing to 1 (the theory of convex spaces). One could generalize this in various fashions. For example, one could allow negative weights (making probability theory essentially the same as the theory of affine spaces). Quantum mechanics uses complex number weights, and demands that their squared magnitudes sum to 1, rather than the weights themselves. But that sense of “non-Kolmogorovian probability” is not what I mean; I mean something that still tries to model the same thing Kolmogorovian probability is supposed to model, but just does so with slightly different technicalities. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [June 26, 2012, 9:18pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/111 "2012-06-26T21:18:06Z")
> [@Indistinguishable](#): > > The problem is that the OP erroneously assumed the average profit from switching, conditioned on the value of their envelope, to be zero. And, fair enough, your example shows that the average profit from switching needn’t be zero if, for example, your envelope contains the lowest or highest values possible. But the OP might still say “That’s not what I’m interested in. What I’m interested in has no lowest or highest value; every value is on equal footing.” The answer is the same “You assumed the average profit from switching to be zero, without grounds for this assumption”, but the further task is to figure out why they supposed this assumption, and dissolve their urge to do so. Sorry to be so basic, but the argument that average profit from switching is zero, in the $1/$2 case, is: When you have the $1 and switch, you gain a dollar, and when you have the $2 and switch, you lose a dollar. So half the time you gain a dollar, and half the time you lose a dollar. Hence, average profit from switching is zero. What’s wrong with that reasoning? ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 26, 2012, 9:22pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/112 "2012-06-26T21:22:22Z")
The OP’s argument was about the average profit from switching conditioned on the value of their envelope. You have given the average profit from switching conditioned on the minimum value of the two envelopes. Those are different things. The OP is looking at the average profit of, say, the cases ($1, $2) and ($1, $0.50) while you are looking at the average profit from cases ($1, $2) and ($2, $1). ---
**Author:** ![septimus](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/septimus/32/410_2.png) [@septimus](https://boards.straightdope.com/u/septimus)\ **Post date:** [June 26, 2012, 9:38pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/113 "2012-06-26T21:38:58Z")
> [@Indistinguishable](#): > > The OP is looking at the average profit of, say, the cases ($1, $2) and ($1, $0.50) while you are looking at the average profit from cases ($1, $2) and ($2, $1). Perhaps I’m belaboring the obvious, but my point seems still overlooked. **OP’s “paradox” exists even when the only two cases are ($1, $2) and ($2, $1)** (assuming of course, as he did, that both envelopes are still sealed. That’s the point I keep trying to make. E = E[Y] = 1.5 in this trivial case, yet OP still derives E[Y] = 1.25 E. We all know that’s incorrect and why it’s incorrect but, AFAICT, no one has presented the five-words-or-less name for the fallacy. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 26, 2012, 9:44pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/114 "2012-06-26T21:44:30Z")
Sure, if those are the only two cases, the starting point “E[Y | X] = 1.25X” is false, because “P(Y = 2X | X) = P(Y = X/2 | X)” is false. We needn’t name the fallacy; we just point to the fact that “P(Y = 2X | X) = P(Y = X/2 | X)” is false, if ($1, $2) and ($2, $1) are the only cases. [Note: All equalities here are intended as equalities between functions of X; i.e., they denote equality for every value of X] But if we let ourselves have all infinitely many cases such that “P(Y = 2X | X) = P(Y = X/2 | X)” is true, then “E[Y | X] = 1.25X” is true, and “E[Y] = E[E[Y | X]] = E[1.25X] = 1.25E” correctly follows. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 26, 2012, 9:53pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/115 "2012-06-26T21:53:44Z")
> [@septimus](#): > > yet OP still derives E[Y] = 1.25 E The OP, incidentally, does not derive E[Y] = 1.25 E, except in presenting the classical two-envelope paradox (where it is given that Y is equally likely to be twice or half of X, independently of X, and this result then correctly follows). The OP’s own question is about starting with the assumption that E[Y | X] = X, and deriving that Y must be twice as likely to be half X as to be twice X, independently of X (the flaw here being in the starting assumption). ---
**Author:** ![md2000](https://avatars.discourse-cdn.com/v4/letter/m/73ab20/32.png) [@md2000](https://boards.straightdope.com/u/md2000)\ **Post date:** [June 26, 2012, 10:01pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/116 "2012-06-26T22:01:57Z")
The heck with math. If someone gives you an envelope with $100 and says “you could get $200 by asking for the other one, or it could cost you $50 to find out that you were wrong…” Odds are 50-50 either way. The value of getting more is worth the risk of losing a lesser amount. ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [June 26, 2012, 10:04pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/117 "2012-06-26T22:04:47Z")
You are doing math. [Indeed, you are doing the same math everyone does, to (correctly) determine that the average profit of switching, conditioned on any particular value in your envelope, is positive (given that the other envelope is equally likely to be half or double yours). This is given as part of the presentation of the paradox. It’s familiar to everyone. The whole paradox is about reconciling this with other calculations or symmetries.] ---
**Author:** ![doorhinge](https://avatars.discourse-cdn.com/v4/letter/d/a5b964/32.png) [@doorhinge](https://boards.straightdope.com/u/doorhinge)\ **Post date:** [June 26, 2012, 10:34pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/118 "2012-06-26T22:34:45Z")
> [@Saffer](#): > > …So it turns out that the probability of the other envelope containing twice as much cash as the one in your hand is always 1/3 :dubious: > > What is wrong with this reasoning? From a practical viewpoint, there is no paradox. You have a choice of envelope “A” or envelope “B”. Envelope “A” has $1 (or add as many zeros as you wish) and envelope “B” has $2 (or $20, $200, $200,000). You are unaware of which envelope is which. You pick either envelope and you’re guarenteed to have X more money than you had. If you picked correctly, you could have 2X more money. It’s a “can’t lose” situation. Unless you’re greedy. Then the “other” envelope will ALWAYS look like the better choice. The “Chicago” version of this puzzle would have a different resolution. Assuming there was enough money to make it worth our while, let’s say $300K total, split $200K/$100K. You suggest to the envelope holder that you will give them $50K “IF”, and only if, you pick the envelope with the larger amount. Take the envelope he suggests you take. ---
**Author:** ![Frylock](https://avatars.discourse-cdn.com/v4/letter/f/ce7236/32.png) [@Frylock](https://boards.straightdope.com/u/Frylock)\ **Post date:** [June 26, 2012, 11:39pm UTC](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755/119 "2012-06-26T23:39:03Z")
> [@Indistinguishable](#): > > The OP’s argument was about the average profit from switching conditioned on the value of their envelope. You have given the average profit from switching conditioned on the minimum value of the two envelopes. Those are different things. The OP is looking at the average profit of, say, the cases ($1, $2) and ($1, $0.50) while you are looking at the average profit from cases ($1, $2) and ($2, $1). No yeah, sorry, I meant to be explicit that I was talking about a different kind of case. And yet, even so, I was confused, and I see that now. [Previous page](https://boards.straightdope.com/t/2-envelopes-problem-revisited/625755.md?page=5)