Re best of three: if sufficiently precise and 100% free randomization is allowed, I think the optimal policy is ratifiable by CDT. (Though not paying is also ratifiable, and more robustly so.)
Or if your knowledge of the environment does helpful randomization for you (if you’re not >99% sure your two copies will take the same action), CDT’ll at least press the button. But yeah, interesting problem.
Is the correct policy an equilibrium? Suppose the payoff was 5$, not 1000$. If you all press with probability P, you get: (1-P)^3 of 0, 3P(1-P)^2 of −1, 3P^2(1-P) of 3, and P^3 of 2. Optimal P is 0.8873 for payoff of 2.162.
Now suppose you know your two copies are pressing the button with P=0.8873. You press with probability Q. You get (1-P)^2(1-Q) of 0, 2P(1-P)(1-Q) + (1-P)^2Q of −1, 2P(1-P)Q + P^2(1-Q) of 3, and P^2Q of 2. Optimal Q is 0. If you never press the button, you get 2*0.8873*(1-0.8873) of −1 and 0.8873^2 of 3, which is 2.262.
So if you know your copies are playing the optimal policy for three, you shouldn’t press the button :D
Re best of three: if sufficiently precise and 100% free randomization is allowed, I think the optimal policy is ratifiable by CDT. (Though not paying is also ratifiable, and more robustly so.)
I did say “suppose you are deterministic”. That said, can you spell out how CDT ratifies the optimal policy if randomization is allowed?
I believe it follows from this proof: https://www.alignmentforum.org/posts/5bd75cc58225bf06703751b2/in-memoryless-cartesian-environments-every-udt-policy-is-a
Or if your knowledge of the environment does helpful randomization for you (if you’re not >99% sure your two copies will take the same action), CDT’ll at least press the button. But yeah, interesting problem.
Is the correct policy an equilibrium? Suppose the payoff was 5$, not 1000$. If you all press with probability P, you get: (1-P)^3 of 0, 3P(1-P)^2 of −1, 3P^2(1-P) of 3, and P^3 of 2. Optimal P is 0.8873 for payoff of 2.162.
Now suppose you know your two copies are pressing the button with P=0.8873. You press with probability Q. You get (1-P)^2(1-Q) of 0, 2P(1-P)(1-Q) + (1-P)^2Q of −1, 2P(1-P)Q + P^2(1-Q) of 3, and P^2Q of 2. Optimal Q is 0. If you never press the button, you get 2*0.8873*(1-0.8873) of −1 and 0.8873^2 of 3, which is 2.262.
So if you know your copies are playing the optimal policy for three, you shouldn’t press the button :D
I think if others play with probability P, every value of Q is equally good.
Not sure if this is a typo, but I get 2*0.8873*(1-0.8873)(-1)+0.8873^{2}(3) = 2.162
Which is the same as if you play Q=P. Which supports the claim that every value of Q is equally good.
I can’t check today, but whoops, sorry if I typoed the equation at some step.