15 ms·

A simple non-mathematical resolution is that no information has been gained by selecting the first envelope. Since no information about the contents was availab

by 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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- JoshCole 4y ago
  No, he articulated the problem correctly. He's just wrong that the math isn't with him. Game theory solves imperfect information games using information sets. The math is with him. In imperfect information games you are in an information set, not a subgame. When the value is 2A and you have to make a keep or switch decision, the value A is still part of how you calculate the solution. When the value A is the game you are in, it doesn't matter, the value 2A is still part of how you calculate the solution. It has to be this way, because you don't have a way of discriminating between which game state you are in. This is one of the notable and counterintuitive (apparently) results of game theory and it really matters, because when you don't notice it you run into a contradiction. Check out Noam's talk on solving imperfect information games at NIPS 2017 for a more detailed treatment of this subject. I'm going to handwave it to dive deeper into the exact logical contradiction. So 2A and A both have to exist still for the correct calculation. The subgames are still connected. But notice what happens? What the failed solution does though is it creates two different subgames! In one they set 2A. In another they have A/2. But its still two different subgames. Now why does this seem appropriate? Because this sort of reasoning is okay in perfect information games! You can reason about different subgames independently of each other. But we're not in a perfect information game. We're in a null information set. So 2A and A coexist. Since we can't state that we are in the 2A case or the A case, being in an information set that prevents us from discrimination between these two things, they error when they try to build their reasoning on the supposition they are able to do so. In particular, they create a logical contradiction. The dependence means you are required to have the subgame from the 2A as the other side of the equation. Required, but they don't do that. Instead they evaluate one subgame as if they are in a perfect information context. That is the missing equality symbol that people aren't seeing. That dependence. So what is really happening is this: They claim 2A=A; assuming A is not zero, they claim that 2=1. You can see this very clearly when you take the actual correct probabilities for expectation and place them beside the incorrect ones: Observe. They said: 5/4A = (1/2)2A + (1/2)A/2 Reality : 3/2A = (1/2)2A + (1/2)A Equality : (1/2)2A + (1/2)A/2 = (1/2)2A + (1/2)A (1/2)A/2 = (1/2)A A/2 = A A = 2A 1 = 2 It's a bit worse than that in the error, because they omit that expectation is a function of both policy and information set. Trying to make the decision based on expected value when expected value is not well defined without a policy choice is another very fundamental error. I mean seriously, ask yourself if you get the same result by slamming your head into a wall as opposed to doing something productive. You don't. Policy choice matters. In this case literally the only policy choice that has 0 EV is the one they claim is the best. Its silly. When you look at the game tree, when you actually evaluate the proper graph, its literally the only bad decision you can make. There are quite literally an infinite number of strategies that give the same EV solution. All except the one they chose. The reward for the game is only defined for R(KEEP) and it is defined as {0.5: A, 0.5: 2A}. R(SWITCH) ends up getting its valuation according to a bellman equation: R(SWITCH) = R(SWITCH) * P(SWITCH) + R(KEEP) * P(KEEP). This is obviously an infinite sequence if P(SWITCH) = 1. We actually tend to use a discount factor here like 0.9999999999*R(some_choice) because for this infinite sequence it lets us use limits to claim the case is defined as 0. But this is like, an extension of why this is wrong - because game theory with information sets already solves for probabilities over action space. It already acknowledges it. So when you point to the information set problem? You point to game theory. And you point to the root of why their calculations were wrong: in an imperfect information game, you play both the game you're in and the game you're not in at the same time.
camgunz 4y ago
Yeah I was looking for someone saying this to double down. I think the "paradox" comes down to information loss. The tricky bit is this framing: "when you're holding an envelope, the other envelope contains >= $MONEY, so there's no reason not to switch". But this omits (loses) the information of "both envelopes contain >= $MONEY". That is, as soon as you switch to the other envelope, the situation is still true, the previous envelope contains >= $MONEY. So if you forget that your current envelope also contains >= $MONEY you're stuck in an infinite loop, which you break out of only because you can only do a single iteration.
- not2b 4y ago
  The right resolution is: suppose the two amounts are $50 and $100. If the envelope you have contains $50 and you switch, your gain is $50. If it contains $100 and you switch, your gain is -$50. So your expected gain from switching is (0.5)(50) + (0.5)(-50) which is zero. The way that they fool you is by saying that you either double your money or you halve it, leading you to think that your expected ratio is 2.5/2. But that isn't correct. That ratio doesn't mean anything. That's the wrong step.
  - twinkletwinkle_ 4y ago
    I think you nailed it. If the question were posed as "Two envelopes with $A and $(A+1)" then there would be no talk of ratios and no "paradox" would ever emerge.
  - camgunz 4y ago
    Exactly. The other envelope might be the good one or the bad one. The "paradox" is basically just sophistry.
  - lovecg 4y ago
    That’s a different problem though. In the original problem you don’t know the two amounts ahead of time. All you see is $50 in your envelope - you don’t know if the other one is $100 or $25 and you’re back to square one.
ncmncm 4y ago
The goalposts here are different from a normal "paradox". The actual solution is explained right there on the page, the expectation for both envelopes (one has x, one has 2x) is x*3/2, so switching changes nothing. The "solution" demanded is supposed to be pointing out the specific logical error in the erroneous calculation that says switching should win. Citing the correct logic is not considered sufficient.
- amalcon 4y ago
  The paradox is supposed to be in the contradiction between this (incorrect) reasoning and the correct reasoning, not in the counterintuitive conclusion of the incorrect reasoning.
- myhf 4y ago
  It seems like this "solution" is saying that the error is in step 1: > Denote by A the amount in the player's selected envelope. which is information that we have not actually gained.
  - JoshuaDavid 4y ago
    Variant: I tell you "These two envelopes contain money. One of the envelopes contains twice as much money as the other one. Pick one.". You pick one. I tell you the one you picked contains $60. You've now determined the amount in your selected envelope -- should you switch?
    - ncmncm 4y ago
      Now you know x is either 30 or 60. The expectation is either 45 or 90. Switching has equal likelihood of increasing or decreasing your take. The correct calculation will show the expected value of the other envelope as 60.
      - mrow84 4y ago
        Inserting A=60 into the original line of reasoning (just to show that it doesn't immediately resolve the paradox): 1. Denote by A=60 the amount in the player's selected envelope. 2. The probability that A=60 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=120 or A/2=30. 4. If A=60 is the smaller amount, then the other envelope contains 2A=120. 5. If A=60 is the larger amount, then the other envelope contains A/2=30. 6. Thus the other envelope contains 2A=120 with probability 1/2 and A/2=30 with probability 1/2. 7. So the expected value of the money in the other envelope is: 1/2(120) + 1/2(30) = 75 8. This is greater than A=60 so, on average, the person reasons that they stand to gain by swapping. 9. ... edit: If anything, to me this obscures things further, because it makes it non-obvious that we are actually talking about ratios (i.e. 2A, 1/2A) rather than absolute numbers, and so the expectation calculated in step 7 is inappropriate - a geometric mean recovers the correct answer.
pontus 4y ago
There actually is a paradox: Select a power of 2 according to the probability P(2^k) = 0.2 * 0.8^k i.e. p(1) = 20%, p(2) = 16%, p(4) = 12.8%, etc. Then place this amount in one envelope and twice that I'm the other envelope. Then, randomly shuffle the envelopes and select one and open it. Say you see the number 16. It could have landed there in two ways: Either 16 was generated and the other envelope has 32. This happens with probability 0.2*0.8^4 ~ 8.192%. Alternatively 8 was generated and the envelope you have is already the largest one. This happens with probability 0.2*0.8^3 ~ 10.24%. Given that you see 16 already, the relative probability becomes 4/9 vs 5/9. In other words the chance that you have the largest envelope already is 5/9. The expected value of the other envelope is then 5/9*8 + 4/9*32 = 18.67 > 16 so you should switch. In fact, try this for different initial envelopes and you'll see that you should always switch. But, if you should always switch, why even look inside the envelope? But, then you're back at the symmetry / no information argument.
- kmod 4y ago
  Note that the problem requires allowing arbitrarily small amounts of money in the envelopes, otherwise you would know at some point that you have the smallest amount of money possible. So in your formulation, you have to let k be negative as well.
  - pontus 4y ago
    No, k >= 0 in my example. In one case you know for sure that you have the smallest envelope since you see that it has $1 in it. I'm this case you should absolutely switch. In all other cases, you should still switch although you can't be completely sure. That being said, even if you let k go negative I think it's still a paradox. If you're curious the resolution to this is that before you look in the envelope the expected value of both envelopes is infinite. Only after you look inside one of them does that collapse to two finite values that you can subtract.
    - kmod 4y ago
      If you disallow negative values of k then you break the symmetry argument, and you do indeed get quite a bit of information by opening an envelope. I assume you meant to use a exponential base <0.5 so that the mean is finite
posterboy 4y ago
> This is in contrast to the three-door problem where information is gained by opening the first door. This is in contrast to the earlier problem statement of Gardiner's wealthy men playing against each other as per the article, actually.

by kloch 4y ago

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.

deleted 4y ago
[deleted]
deleted 4y ago
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deleted 4y ago
[deleted]
JoshCole 4y ago
No, he articulated the problem correctly. He's just wrong that the math isn't with him. Game theory solves imperfect information games using information sets. The math is with him. In imperfect information games you are in an information set, not a subgame. When the value is 2A and you have to make a keep or switch decision, the value A is still part of how you calculate the solution. When the value A is the game you are in, it doesn't matter, the value 2A is still part of how you calculate the solution. It has to be this way, because you don't have a way of discriminating between which game state you are in. This is one of the notable and counterintuitive (apparently) results of game theory and it really matters, because when you don't notice it you run into a contradiction. Check out Noam's talk on solving imperfect information games at NIPS 2017 for a more detailed treatment of this subject. I'm going to handwave it to dive deeper into the exact logical contradiction. So 2A and A both have to exist still for the correct calculation. The subgames are still connected. But notice what happens? What the failed solution does though is it creates two different subgames! In one they set 2A. In another they have A/2. But its still two different subgames. Now why does this seem appropriate? Because this sort of reasoning is okay in perfect information games! You can reason about different subgames independently of each other. But we're not in a perfect information game. We're in a null information set. So 2A and A coexist. Since we can't state that we are in the 2A case or the A case, being in an information set that prevents us from discrimination between these two things, they error when they try to build their reasoning on the supposition they are able to do so. In particular, they create a logical contradiction. The dependence means you are required to have the subgame from the 2A as the other side of the equation. Required, but they don't do that. Instead they evaluate one subgame as if they are in a perfect information context. That is the missing equality symbol that people aren't seeing. That dependence. So what is really happening is this: They claim 2A=A; assuming A is not zero, they claim that 2=1. You can see this very clearly when you take the actual correct probabilities for expectation and place them beside the incorrect ones: Observe. They said: 5/4A = (1/2)2A + (1/2)A/2 Reality : 3/2A = (1/2)2A + (1/2)A Equality : (1/2)2A + (1/2)A/2 = (1/2)2A + (1/2)A (1/2)A/2 = (1/2)A A/2 = A A = 2A 1 = 2 It's a bit worse than that in the error, because they omit that expectation is a function of both policy and information set. Trying to make the decision based on expected value when expected value is not well defined without a policy choice is another very fundamental error. I mean seriously, ask yourself if you get the same result by slamming your head into a wall as opposed to doing something productive. You don't. Policy choice matters. In this case literally the only policy choice that has 0 EV is the one they claim is the best. Its silly. When you look at the game tree, when you actually evaluate the proper graph, its literally the only bad decision you can make. There are quite literally an infinite number of strategies that give the same EV solution. All except the one they chose. The reward for the game is only defined for R(KEEP) and it is defined as {0.5: A, 0.5: 2A}. R(SWITCH) ends up getting its valuation according to a bellman equation: R(SWITCH) = R(SWITCH) * P(SWITCH) + R(KEEP) * P(KEEP). This is obviously an infinite sequence if P(SWITCH) = 1. We actually tend to use a discount factor here like 0.9999999999*R(some_choice) because for this infinite sequence it lets us use limits to claim the case is defined as 0. But this is like, an extension of why this is wrong - because game theory with information sets already solves for probabilities over action space. It already acknowledges it. So when you point to the information set problem? You point to game theory. And you point to the root of why their calculations were wrong: in an imperfect information game, you play both the game you're in and the game you're not in at the same time.

camgunz 4y ago

Yeah I was looking for someone saying this to double down. I think the "paradox" comes down to information loss. The tricky bit is this framing: "when you're holding an envelope, the other envelope contains >= $MONEY, so there's no reason not to switch". But this omits (loses) the information of "both envelopes contain >= $MONEY". That is, as soon as you switch to the other envelope, the situation is still true, the previous envelope contains >= $MONEY. So if you forget that your current envelope also contains >= $MONEY you're stuck in an infinite loop, which you break out of only because you can only do a single iteration.

not2b 4y ago
The right resolution is: suppose the two amounts are $50 and $100. If the envelope you have contains $50 and you switch, your gain is $50. If it contains $100 and you switch, your gain is -$50. So your expected gain from switching is (0.5)(50) + (0.5)(-50) which is zero. The way that they fool you is by saying that you either double your money or you halve it, leading you to think that your expected ratio is 2.5/2. But that isn't correct. That ratio doesn't mean anything. That's the wrong step.
- twinkletwinkle_ 4y ago
  I think you nailed it. If the question were posed as "Two envelopes with $A and $(A+1)" then there would be no talk of ratios and no "paradox" would ever emerge.
- camgunz 4y ago
  Exactly. The other envelope might be the good one or the bad one. The "paradox" is basically just sophistry.
- lovecg 4y ago
  That’s a different problem though. In the original problem you don’t know the two amounts ahead of time. All you see is $50 in your envelope - you don’t know if the other one is $100 or $25 and you’re back to square one.

ncmncm 4y ago

The goalposts here are different from a normal "paradox". The actual solution is explained right there on the page, the expectation for both envelopes (one has x, one has 2x) is x*3/2, so switching changes nothing. The "solution" demanded is supposed to be pointing out the specific logical error in the erroneous calculation that says switching should win. Citing the correct logic is not considered sufficient.

amalcon 4y ago
The paradox is supposed to be in the contradiction between this (incorrect) reasoning and the correct reasoning, not in the counterintuitive conclusion of the incorrect reasoning.
myhf 4y ago
It seems like this "solution" is saying that the error is in step 1: > Denote by A the amount in the player's selected envelope. which is information that we have not actually gained.
- JoshuaDavid 4y ago
  Variant: I tell you "These two envelopes contain money. One of the envelopes contains twice as much money as the other one. Pick one.". You pick one. I tell you the one you picked contains $60. You've now determined the amount in your selected envelope -- should you switch?
  - ncmncm 4y ago
    Now you know x is either 30 or 60. The expectation is either 45 or 90. Switching has equal likelihood of increasing or decreasing your take. The correct calculation will show the expected value of the other envelope as 60.
    - mrow84 4y ago
      Inserting A=60 into the original line of reasoning (just to show that it doesn't immediately resolve the paradox): 1. Denote by A=60 the amount in the player's selected envelope. 2. The probability that A=60 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=120 or A/2=30. 4. If A=60 is the smaller amount, then the other envelope contains 2A=120. 5. If A=60 is the larger amount, then the other envelope contains A/2=30. 6. Thus the other envelope contains 2A=120 with probability 1/2 and A/2=30 with probability 1/2. 7. So the expected value of the money in the other envelope is: 1/2(120) + 1/2(30) = 75 8. This is greater than A=60 so, on average, the person reasons that they stand to gain by swapping. 9. ... edit: If anything, to me this obscures things further, because it makes it non-obvious that we are actually talking about ratios (i.e. 2A, 1/2A) rather than absolute numbers, and so the expectation calculated in step 7 is inappropriate - a geometric mean recovers the correct answer.

pontus 4y ago

There actually is a paradox: Select a power of 2 according to the probability P(2^k) = 0.2 * 0.8^k i.e. p(1) = 20%, p(2) = 16%, p(4) = 12.8%, etc. Then place this amount in one envelope and twice that I'm the other envelope. Then, randomly shuffle the envelopes and select one and open it. Say you see the number 16. It could have landed there in two ways: Either 16 was generated and the other envelope has 32. This happens with probability 0.2*0.8^4 ~ 8.192%. Alternatively 8 was generated and the envelope you have is already the largest one. This happens with probability 0.2*0.8^3 ~ 10.24%. Given that you see 16 already, the relative probability becomes 4/9 vs 5/9. In other words the chance that you have the largest envelope already is 5/9. The expected value of the other envelope is then 5/9*8 + 4/9*32 = 18.67 > 16 so you should switch. In fact, try this for different initial envelopes and you'll see that you should always switch. But, if you should always switch, why even look inside the envelope? But, then you're back at the symmetry / no information argument.

kmod 4y ago
Note that the problem requires allowing arbitrarily small amounts of money in the envelopes, otherwise you would know at some point that you have the smallest amount of money possible. So in your formulation, you have to let k be negative as well.
- pontus 4y ago
  No, k >= 0 in my example. In one case you know for sure that you have the smallest envelope since you see that it has $1 in it. I'm this case you should absolutely switch. In all other cases, you should still switch although you can't be completely sure. That being said, even if you let k go negative I think it's still a paradox. If you're curious the resolution to this is that before you look in the envelope the expected value of both envelopes is infinite. Only after you look inside one of them does that collapse to two finite values that you can subtract.
  - kmod 4y ago
    If you disallow negative values of k then you break the symmetry argument, and you do indeed get quite a bit of information by opening an envelope. I assume you meant to use a exponential base <0.5 so that the mean is finite

posterboy 4y ago

> This is in contrast to the three-door problem where information is gained by opening the first door. This is in contrast to the earlier problem statement of Gardiner's wealthy men playing against each other as per the article, actually.