neat! this was a fun one to think about. i think the program should two-box though. Let c be the program that’s the same as b, except it makes the opposite decision upon seeing (x, x). For simplicity, let’s assume the only two programs in P are b and c (and ill assume they two-box when not playing against themselves). Then, if b one-boxes against itself it makes (1/2) * 1m + (1/2) * 1k, wheras if it two-boxes against itself if makes (1/2) * 1k + (1/2) * (1m + 1k). Thus, it should two-box against itself. (I think the intuition that says to one-box is cheating a little—it assumes that by one-boxing you increase the number of one-boxers in P, which i agree would be awesome but not something that b has the ability to change)
neat! this was a fun one to think about. i think the program should two-box though. Let c be the program that’s the same as b, except it makes the opposite decision upon seeing (x, x). For simplicity, let’s assume the only two programs in P are b and c (and ill assume they two-box when not playing against themselves). Then, if b one-boxes against itself it makes (1/2) * 1m + (1/2) * 1k, wheras if it two-boxes against itself if makes (1/2) * 1k + (1/2) * (1m + 1k). Thus, it should two-box against itself. (I think the intuition that says to one-box is cheating a little—it assumes that by one-boxing you increase the number of one-boxers in P, which i agree would be awesome but not something that b has the ability to change)