5 ms·

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 ho

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.
  - newaccount74 4y ago
    But the two choices aren't equally likely, because someone has chosen the amounts in the envelopes through a non-random process. You could make inferences based on what you expect the person to spend on prices, but it's no longer a simple mathematical problem at that point. Which is why the problem as stated doesn't let you look in the envelope before deciding whether to switch or not. Then it's obvious switching makes no difference.
- JoshCole 4y ago
  Yep! You're in both subgames at once. You're playing both the A state and the 2A state at the same time, because the information set I you are in contains both of them. This is the key difference between perfect information and imperfect information. In perfect information you can do the assumption of just one subgame and it works because you know what game you are in - its the one you are in. So you are playing over states and actions. In imperfect information you have to assume you are in both at the same time - so you can't assume one subgame, because you don't know which one you are in. You are playing over information sets - every state contained within that information set, not just one state. I have a feeling that saying it like this, in the general way, is a lot more confusing to people than your phrasing. So thanks for sharing your phrasing.
- not2b 4y ago
  Oh, I just realized what I missed. You can use the ratio, but if you do you need to take the geometric mean, since it is a multiplying factor. The geometric mean of 2 and 0.5 is sqrt(2 * 0.5) or 1. So the expected ratio is 1.
JoshCole 4y ago
I think you're confusing yourself. Watch something like Naom Brown's NIPS 2017 lecture on subgame solving in imperfect information games and you'll realize you should be very wary of someone who is trying to play around with solutions to imperfect information games which use state based reasoning - not information set based reasoning. If you have a tree: 0.5 / \ A 2A In a perfect information game you can do this thing where your like "assume a" and think ahead and then "assume 2a" and think ahead. But in the the imperfect information game world you don't know whether you have a or 2a. So you have to "assume a 2a" and you have to deal with your solution space using counterfactual reasoning. Picture a chain coming out from the A to the 2A. They are married. They cannot be parted. It is really counterintuitive, but the way I think about it is "you are playing every game at once" and "you are playing a game over information sets and strategy choices" not "state space and actions" so since you are playing every game at once already when they come in with that other game - well, you were already playing another game. They're overwriting that part of your solutions memory. Which in math terms declares an equality. In that case that equality is 2=1 because they say A = A/2 or that 2A=A depending on which framing you use to build the real game tree. The actual recurrence relationship is obviously R(KEEP) = {0.5: A, 0.5: 2A} = 3/2A. R(SWITCH) = P(KEEP)*R(KEEP) + P(SWITCH)*R(SWITCH) Solving the game from there is really trivial - you're only allowed one strategy selection per information set and you only one set the null set. You literally get to make only one decision as to a scalar between 0 and 1 to define P(KEEP) and P(SWITCH) which is going to be 1-P(KEEP). `R(SWITCH)` becomes `R(KEEP)` because of the way the markov chain drains into it. Literally the only thing you can't pick is `P(SWITCH)=1`. Everything else is an equivalent EV solution. P(SWITCH) <= P(KEEP) because there exists one P(SWITCH=1) is undefined or else 0 (in practice we take the limit after a horizon by multiplying by some term like 0.9999999 to avoid the infinite regress). And since the problem didn't make this clear, your policy __influences your expected value__ and so their reasoning process is just so broken at a fundamental level. They literally declared 2=1 and therefore undefined solution is the best answer despite infinite solutions that aren't undefined. Its... really bad reasoning. Don't listen to the idea that switching is better. It can't be. Look at the actual relationship. Look at the actual graph. They skip some really important steps, like using the same gametree for the entire calculation.
- 4y ago

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.
- newaccount74 4y ago
  But the two choices aren't equally likely, because someone has chosen the amounts in the envelopes through a non-random process. You could make inferences based on what you expect the person to spend on prices, but it's no longer a simple mathematical problem at that point. Which is why the problem as stated doesn't let you look in the envelope before deciding whether to switch or not. Then it's obvious switching makes no difference.
JoshCole 4y ago
Yep! You're in both subgames at once. You're playing both the A state and the 2A state at the same time, because the information set I you are in contains both of them. This is the key difference between perfect information and imperfect information. In perfect information you can do the assumption of just one subgame and it works because you know what game you are in - its the one you are in. So you are playing over states and actions. In imperfect information you have to assume you are in both at the same time - so you can't assume one subgame, because you don't know which one you are in. You are playing over information sets - every state contained within that information set, not just one state. I have a feeling that saying it like this, in the general way, is a lot more confusing to people than your phrasing. So thanks for sharing your phrasing.
not2b 4y ago
Oh, I just realized what I missed. You can use the ratio, but if you do you need to take the geometric mean, since it is a multiplying factor. The geometric mean of 2 and 0.5 is sqrt(2 * 0.5) or 1. So the expected ratio is 1.

JoshCole 4y ago

I think you're confusing yourself. Watch something like Naom Brown's NIPS 2017 lecture on subgame solving in imperfect information games and you'll realize you should be very wary of someone who is trying to play around with solutions to imperfect information games which use state based reasoning - not information set based reasoning. If you have a tree: 0.5 / \ A 2A In a perfect information game you can do this thing where your like "assume a" and think ahead and then "assume 2a" and think ahead. But in the the imperfect information game world you don't know whether you have a or 2a. So you have to "assume a 2a" and you have to deal with your solution space using counterfactual reasoning. Picture a chain coming out from the A to the 2A. They are married. They cannot be parted. It is really counterintuitive, but the way I think about it is "you are playing every game at once" and "you are playing a game over information sets and strategy choices" not "state space and actions" so since you are playing every game at once already when they come in with that other game - well, you were already playing another game. They're overwriting that part of your solutions memory. Which in math terms declares an equality. In that case that equality is 2=1 because they say A = A/2 or that 2A=A depending on which framing you use to build the real game tree. The actual recurrence relationship is obviously R(KEEP) = {0.5: A, 0.5: 2A} = 3/2A. R(SWITCH) = P(KEEP)*R(KEEP) + P(SWITCH)*R(SWITCH) Solving the game from there is really trivial - you're only allowed one strategy selection per information set and you only one set the null set. You literally get to make only one decision as to a scalar between 0 and 1 to define P(KEEP) and P(SWITCH) which is going to be 1-P(KEEP). `R(SWITCH)` becomes `R(KEEP)` because of the way the markov chain drains into it. Literally the only thing you can't pick is `P(SWITCH)=1`. Everything else is an equivalent EV solution. P(SWITCH) <= P(KEEP) because there exists one P(SWITCH=1) is undefined or else 0 (in practice we take the limit after a horizon by multiplying by some term like 0.9999999 to avoid the infinite regress). And since the problem didn't make this clear, your policy __influences your expected value__ and so their reasoning process is just so broken at a fundamental level. They literally declared 2=1 and therefore undefined solution is the best answer despite infinite solutions that aren't undefined. Its... really bad reasoning. Don't listen to the idea that switching is better. It can't be. Look at the actual relationship. Look at the actual graph. They skip some really important steps, like using the same gametree for the entire calculation.

4y ago