It is trivial! It is trivial! The game defines a graph and your path through that graph is going to lead to an outcome. You have a strategy, a choice of how to

It is trivial! It is trivial! The game defines a graph and your path through that graph is going to lead to an outcome. You have a strategy, a choice of how to move through the graph. However, you have no information about where you are in the graph. So when we think about how we move through the graph, we must always think the same way. We can never think differently, because no action changes the information we have about where we are. However, the terminal case is defined thusly. R(STAY) = {50%: 2A, 50% A}. There is never a reward for R(SWITCH). Anyone telling you to always choose switch when that is the only solution that can never have a reward is obviously wrong. But why? Why were they wrong? Simple. In imperfect information games things work differently than in perfect information games. In chess when you make a move it is that move. If you make a move it is a different move. The subgames are independent of each other. In poker, it isn't so! If you make a bet in Poker, you don't know what subgame you are in. You have to play as if you're in multiple at the same time. Now look at the fast con that seems to be tripping people up. They are pretending we have perfect information. 2A, A/2? Phaw. We can't have that, because that supposes we know which subgame we are in, but we can't know which subgame we are in, because we were never given that information. Even if we were - it doesn't matter. We're in an imperfect information game! The places we didn't explore are relevant to our calculation! We can't ignore them! But we did, by insisting on an inequality which didn't hold. This is probably counterintuitive. Its trivial, but people argue about it. So lets talk about this more concretely. This is what they pulled on you. They claimed that these two different game trees could be treated as the same game tree: -> Keep -> 2A 2A -> Switch -> A 0.5 -> A -> Keep -> A -> Switch -> 2A -> Keep -> A A -> Switch -> A/2 0.5 -> A/2 -> Keep -> A/2 -> Switch -> A Two different game trees! Two different game trees and they acted like they were the same. But they aren't! They can't be! You can't tell which subgame you are in you have to treat them as equal in imperfect information games! So what is our actual situation with their reasoning? Our actual situation is a contradiction. Since subgames depends on each other when they do this combination they are secretly speaking nonsense. Look closely: They are claiming that 2A=A; that A=A/2. They have to be, because in imperfect information games we have to be in every subgame at once, but they made two different games, but then they combined them. Since the correct formulation has to have every solution treated as the same, but they took something from a different subgame, they declared that two subgames were equal, but they weren't. It is so trivial. That's why the correct calculation looks like only an expectation on one game tree. That's why it seems to make sense - your brain knows this sort of reasoning can be appropriate with perfect information. I realize a lot of people have argued about this, but frankly they should stop. And if you're interested in exploring why you have to treat subgames as different in imperfect information, why you have to operate on information sets rather than state spaces in imperfect information, here is a great talk on the NIPS 2017 Best Paper award which happens to touch on the subject: https://www.youtube.com/watch?v=tRiaGahlyy4 https://www.youtube.com/watch?v=tRiaGahlyy4 It is about solving poker, which is a much more complicated imperfect information game. But even the simple imperfect information games have this property.