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> I thought that this corresponds perfectly to my poker example. If it doesn't, and it means something completely different, ok, sure. I'm not a mathematician.

> I thought that this corresponds perfectly to my poker example. If it doesn't, and it means something completely different, ok, sure. I'm not a mathematician. I can't even read the notation that's used in subsequent part of the paper. I think you understood his point very well. I just think you're making a mistake in trying to recast his claim from "imperfect information games" have this property to "this narrow subset of imperfect information games" has this property. Both imperfect games with an opponent and imperfect games without an opponent have the property of needing to play as if you are in multiple subgames, because the definition of the problem is that you don't know which subgame you are in. His claim was that the policy in one subgame could influence the EV of another subgame - and in this problem, it does. I believe you're thinking of the EV of the envelope when you think you are proving this game doesn't have that property via calculation. To see his claim applies to this game consider the case where P(Switch)=1. Being able to calculate the EV of the envelope's contents is a bit different from solving the subgame. Here, we have two subgames E1 and E2. We can know a priori what the expected values of envelope's contents in E1 and E2 are. But if you change your policy in E1, it changes your expected value in E2. If you doubt this, remember the core of the paradox again - choose to always switch and your EV is no longer the EV of the envelopes. Ergo, the EV of the subgame E1 is dependent on the strategies and policies of another subgame, E2.

baobabKoodaa 4y ago
> I think you understood his point very well. I just think you're making a mistake in trying to recast his claim from "imperfect information games" have this property to "this narrow subset of imperfect information games" has this property. Both imperfect games with an opponent and imperfect games without an opponent have the property of needing to play as if you are in multiple subgames, because the definition of the problem is that you don't know which subgame you are in. If we take the poker example and we modify it by "locking" our opponent's strategy, then this property is removed. Suddenly the optimal strategy for us no longer includes any GTO-like thinking such as "balancing our range", we should simply maximize our EV for each hand "in a vacuum" without any thought to other subgames. We no longer care how our range looks to our opponent, because they are no longer able to change their strategy. > To see his claim applies to this game consider the case where P(Switch)=1. The policy you chose in one subgame just changed the EV of another subgame. Sorry, but I don't understand this. This sounds to me like we fix the probability of switching to 1, which means that we end up in infinite loop and the outcome can not be computed. I don't understand this premise, nor its implications for the EV of the other subgame. > This is a subtle distinction that I mentioned earlier - the EV of the envelope as in the wikipedia problem isn't the EV of the subgame. So you're not solving the subgame if you figure out the EV of the envelope. The expression you use "EV of the envelope" is ambiguous in this context. I'm not sure if you mean EV relative to the value of the total amount of money in the game, or if you mean EV relative to the value of the firstly-chosen envelope. When we're talking about "solving a game", we're talking about finding the optimal decisions within a game to maximize the expected value from the game as a whole. In the Two Envelope game we have just one decision: switch or not. So we need to find out if EV(switch) > EV(stay), where both EVs are relative to the total amount of money in the game (not relative to the value of firstly-chosen envelope). We have many ways to conclude that both of these actions have expected value zero. We don't have to be able to compute the "EV of the envelope". We only need to know the relative difference between the EV of these 2 actions. There's many ways of computing them, and all of those ways lead us to the conclusion that the EV of both actions is zero. I'm not aware of any "incorrect" way of computing those EVs such that we would get a nonzero result.
- JoshCole 4y ago
  > without any thought to other subgames I guarantee you that your solution is going to have to incorporate a set somewhere that includes both subgames. You might forget it does, because you simplify to a scalar, but it is going to be there. In perfect information, it isn't there. In imperfect information it is.
  - baobabKoodaa 4y ago
    > I guarantee you that your solution is going to have to incorporate a set somewhere that includes both subgames. You might forget it does, because you simplify to a scalar, but it is going to be there. In perfect information, it isn't there. In imperfect information it is. I'm having a lot of trouble identifying what exactly it is that we disagree about. If you have identified what it is that we disagree on, can you please formulate the disagreement as a wager that can be simulated with code? That way we can easily resolve the disagreement (or conclude that we actually don't have a disagreement).
    - JoshCole 4y ago
      Okay. Your opponents strategy in poker is fixed; they will always play the nash equilibrium strategy, they will never play another strategy. Their strategy is fixed. They will never change it from this setting. You've claimed that subgame perfect play can be calculated without respect to the subgame you aren't in, because you can make a choice on the basis of the EV of the subgame you are in without respect to the subgames you aren't in. I disagree. I think you still need to account for every subgame you are in as if you are in all of them. Let the subgame you are in be you having KK and your opponent having AA. However, obviously - you only know that you have KK. Therefore, you should be able to compute the strategy which is the best response to 37 suited and according to your logic it should be equal to the best response to AA. After all, you have no means of determining which subgame you are in. So you have to have the same response in both subgames. So compute the best response for KK to AA and prove that this is also the best response to 37 suited. However, you've claimed you don't need to calculate this with respect to other subgames. So your computation of 37 suited and your computation for AA must not be equal to each other - if they are, then you share terms. You calculated them with respect to each other. Let Br = Best response. Write a program which shows Br(p1, p2, I[KK]) != Br(p1, p2, I[KK]) and Br(p1, p2, I[KK]) = Br(p1, p2, I[KK]) simultaneously. (That is to say, both your policy and your opponents policy are fixed) My contention is that you can't do this. You claim you can. You are free to use a simpler variant of poker - Kuhn poker - so that the computation becomes more tractable.

baobabKoodaa 4y ago

> I think you understood his point very well. I just think you're making a mistake in trying to recast his claim from "imperfect information games" have this property to "this narrow subset of imperfect information games" has this property. Both imperfect games with an opponent and imperfect games without an opponent have the property of needing to play as if you are in multiple subgames, because the definition of the problem is that you don't know which subgame you are in. If we take the poker example and we modify it by "locking" our opponent's strategy, then this property is removed. Suddenly the optimal strategy for us no longer includes any GTO-like thinking such as "balancing our range", we should simply maximize our EV for each hand "in a vacuum" without any thought to other subgames. We no longer care how our range looks to our opponent, because they are no longer able to change their strategy. > To see his claim applies to this game consider the case where P(Switch)=1. The policy you chose in one subgame just changed the EV of another subgame. Sorry, but I don't understand this. This sounds to me like we fix the probability of switching to 1, which means that we end up in infinite loop and the outcome can not be computed. I don't understand this premise, nor its implications for the EV of the other subgame. > This is a subtle distinction that I mentioned earlier - the EV of the envelope as in the wikipedia problem isn't the EV of the subgame. So you're not solving the subgame if you figure out the EV of the envelope. The expression you use "EV of the envelope" is ambiguous in this context. I'm not sure if you mean EV relative to the value of the total amount of money in the game, or if you mean EV relative to the value of the firstly-chosen envelope. When we're talking about "solving a game", we're talking about finding the optimal decisions within a game to maximize the expected value from the game as a whole. In the Two Envelope game we have just one decision: switch or not. So we need to find out if EV(switch) > EV(stay), where both EVs are relative to the total amount of money in the game (not relative to the value of firstly-chosen envelope). We have many ways to conclude that both of these actions have expected value zero. We don't have to be able to compute the "EV of the envelope". We only need to know the relative difference between the EV of these 2 actions. There's many ways of computing them, and all of those ways lead us to the conclusion that the EV of both actions is zero. I'm not aware of any "incorrect" way of computing those EVs such that we would get a nonzero result.

JoshCole 4y ago
> without any thought to other subgames I guarantee you that your solution is going to have to incorporate a set somewhere that includes both subgames. You might forget it does, because you simplify to a scalar, but it is going to be there. In perfect information, it isn't there. In imperfect information it is.
- baobabKoodaa 4y ago
  > I guarantee you that your solution is going to have to incorporate a set somewhere that includes both subgames. You might forget it does, because you simplify to a scalar, but it is going to be there. In perfect information, it isn't there. In imperfect information it is. I'm having a lot of trouble identifying what exactly it is that we disagree about. If you have identified what it is that we disagree on, can you please formulate the disagreement as a wager that can be simulated with code? That way we can easily resolve the disagreement (or conclude that we actually don't have a disagreement).
  - JoshCole 4y ago
    Okay. Your opponents strategy in poker is fixed; they will always play the nash equilibrium strategy, they will never play another strategy. Their strategy is fixed. They will never change it from this setting. You've claimed that subgame perfect play can be calculated without respect to the subgame you aren't in, because you can make a choice on the basis of the EV of the subgame you are in without respect to the subgames you aren't in. I disagree. I think you still need to account for every subgame you are in as if you are in all of them. Let the subgame you are in be you having KK and your opponent having AA. However, obviously - you only know that you have KK. Therefore, you should be able to compute the strategy which is the best response to 37 suited and according to your logic it should be equal to the best response to AA. After all, you have no means of determining which subgame you are in. So you have to have the same response in both subgames. So compute the best response for KK to AA and prove that this is also the best response to 37 suited. However, you've claimed you don't need to calculate this with respect to other subgames. So your computation of 37 suited and your computation for AA must not be equal to each other - if they are, then you share terms. You calculated them with respect to each other. Let Br = Best response. Write a program which shows Br(p1, p2, I[KK]) != Br(p1, p2, I[KK]) and Br(p1, p2, I[KK]) = Br(p1, p2, I[KK]) simultaneously. (That is to say, both your policy and your opponents policy are fixed) My contention is that you can't do this. You claim you can. You are free to use a simpler variant of poker - Kuhn poker - so that the computation becomes more tractable.