Let’s say that TDT agents can be divided into two categories, TDT-A and TDT-B, based on a single random bit added to their source code in advance. Then TDT-A can take the strategy of always picking the first box in Problem 2, and TDT-B can always pick the second box.
Now, if you’re a TDT agent being offered the problem; with the aforementioned strategy, there’s a 50% chance that the simulated agent is different than you, netting you $1 million. This also narrows down the advantage of the CDT agent—now they only have a 50% chance of winning the money, which is equal to yours.
The argument is that the simulation is either TDT-A in this case, or TDT-B. Either way, the simulated agent will pick a single favourite box (1 or 2) with certainty, so the money is in either Box 2 or Box 1,
Though I can see an interpretation which leads to Box 3. Omega simulates a “new-born” TDT (which is neither -A nor -B) and watches as it differentiates itself to one variant or the other, each with equal probability. So the new-born picks boxes 1 and 2 with equal frequency over multiple simulations, and Box 3 contains the money. Is that what you were thinking?
Yes. I was thinking that Omega would have access to the agent’s source code, and be running the “play against yourself, if you pick a different number than yourself you win” game. Omega is a jerk :D
That’s not too bad, actually. One of my ideas while thrashing about here was that an agent should have a “favourite” number in the set {1, 2} and pick that number with certainty. That way, Omega will definitely put the $1 million in Box 1 or Box 2 and each agent will have 50% chance that their favourite number disagrees with the simulated agent’s.
This won’t work if Omega describes the source-code of the simulation (or otherwise reveals the simulation’s favourite number) - since then any agent with that exact code knows it can’t choose deterministically, and its best chance is to pick each box with equal chance, as described in the original analysis.
Let’s say that TDT agents can be divided into two categories, TDT-A and TDT-B, based on a single random bit added to their source code in advance. Then TDT-A can take the strategy of always picking the first box in Problem 2, and TDT-B can always pick the second box.
Now, if you’re a TDT agent being offered the problem; with the aforementioned strategy, there’s a 50% chance that the simulated agent is different than you, netting you $1 million. This also narrows down the advantage of the CDT agent—now they only have a 50% chance of winning the money, which is equal to yours.
Actually, the way the problem is specified, Omega puts the money in box 3.
The argument is that the simulation is either TDT-A in this case, or TDT-B. Either way, the simulated agent will pick a single favourite box (1 or 2) with certainty, so the money is in either Box 2 or Box 1,
Though I can see an interpretation which leads to Box 3. Omega simulates a “new-born” TDT (which is neither -A nor -B) and watches as it differentiates itself to one variant or the other, each with equal probability. So the new-born picks boxes 1 and 2 with equal frequency over multiple simulations, and Box 3 contains the money. Is that what you were thinking?
Yes. I was thinking that Omega would have access to the agent’s source code, and be running the “play against yourself, if you pick a different number than yourself you win” game. Omega is a jerk :D
If it’s your own exact source being simulated, then it’s probably impossible to do better than 10%, and the problem isn’t interesting anymore.
That’s not too bad, actually. One of my ideas while thrashing about here was that an agent should have a “favourite” number in the set {1, 2} and pick that number with certainty. That way, Omega will definitely put the $1 million in Box 1 or Box 2 and each agent will have 50% chance that their favourite number disagrees with the simulated agent’s.
This won’t work if Omega describes the source-code of the simulation (or otherwise reveals the simulation’s favourite number) - since then any agent with that exact code knows it can’t choose deterministically, and its best chance is to pick each box with equal chance, as described in the original analysis.