Two envelopes problem

You don't know what A is even if he tells you the first envelope is 60. A and 2A exist together counterfactually, if you have A, you have 2A, if you have 2A, you have A. This is a fundamental property of imperfect information games. You don't have State=A, ever, and you are lying to yourself when you pretend that you do. It is /why/ the reasoning error happens. State based reasoning only works in perfect information games because in that framing you don't have counterfactual subgames forcing them to depend on each other - forcing you to play them both simultaneously, because you don't know which subgame you are in.

A and 2A are together, so you can't propose A/2 without violating the A and 2A codependency imposed by not knowing which subgame you are in.

There are multiple A when you get told 60. They are 30, 60, and 120. This leads to three potential solutions to the expected value: 3/2(30), 3/2(60), 3/2(120). You can't tell which of these solutions you are in, because you don't know. Regardless, since the terminating R(KEEP) condition is the only one that is defined P(SWITCH) = 1 still is either undetermined or 0 depending on how you solve the bellman equations. Your policy decides your expected value. So earlier I was a little loose when I claimed the 3/2A relationship.

Honestly, the wikipedia article is really terrible. It demands you stick in its formalism and declares you a no true scotssman if you don't, buts its approach is fundamentally wrong. Throw off its chains and consider the framing where you remove the requirement of thinking about imperfect information. People are bad at it and if this is confusing you then it is /because/ you are struggling with the imperfect information.

The way you convert between the two game types is this: Instead of states -> information sets. So you only have one move in this game. You always have the Null information set. You always have the same situation. So you're only allowed to have one choice and that is it every time forever. So you always have to make the same decision. Secondly instead of actions -> probability vectors over actions. To simplify and make it tractable assume [0.0, 0.1, 0.2, ... 1.0] so the action space is small. You are choosing 'an action' to take and its going to create multiple subgames. Focus on the recurrence relationship between those games. In particular look at the base case: R(KEEP) = {0.5: A, 0.5: 2A}.

Everything heads toward that base case except one thing: P(SWITCH)=1. It is a markov process and R(SWITCH) 'drains' so as to be R(KEEP) because any probability in it inevitably becomes R(KEEP) after enough iterations.

Notice that {0.5: A, 0.5: 2A} is always true! It is true for both you have A and you keep it and it is true for both you have 2A and you keep it, because notice - you never ever have A. You have the null information set. You can't ever see a difference between these two things.

The moment you force in the idea that you can actually define A to be a particular thing, you smash all over that recurrence relationship. You destroy the relationship. You claim there is no relationship between the policy function and the expected value even though /there is/. There is so much wrong with replacing R(SWITCH) with a fake reality where you can you know you're actually in 1/2A when you can't. And that is what the equations are doing. They're saying you can replace the dependence on each other with a subgame that exists in a different reality. Lets say 120 was when you got 60 - there is no 30 in existence, but your equation calls to replace the recurrence relationship with the idea there is one. It doesn't make sense.

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) 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.

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.

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.

Doesn't matter. Define A = 'the lowest' and rewrite the equations you will see the same expectation.

The problem is that they interchange A being the lowest and the highest in the same equation.