# cousin_it comments on Embedded Agency via Abstraction

• Hadn’t seen the dice ex­am­ple, is it from Jaynes? (I don’t yet see why you’re bet­ter off ran­domis­ing)

Well, one way to for­get the sum is to gen­er­ate ran­dom pairs of dice for each pos­si­ble sum and re­place one of them with your ac­tual pair. For ex­am­ple, if your dice came up (3 5), you can rewrite your mem­ory with some­thing like “the re­sult was one of (1 1) (2 1) (3 1) (4 1) (4 2) (2 5) (3 5) (4 5) (6 4) (6 5) (6 6)”. Is there a sim­pler way?

• Ob­vi­ously if you I the sum, I just want to know the die1-die2? The only prob­lem is that the signed differ­ence looks like a uniform dis­tri­bu­tion with width de­pen­dent on the sum—the signed differ­ence can range from 11 pos­si­bil­ities (-5 to 5) down to 1 (0).

So what I think you do is you put all the differ­ences onto the same scale by con­struct­ing a “unitless differ­ence,” which will ac­tu­ally be defined as a uniform dis­tri­bu­tion.

Rather than hav­ing the differ­ence be a sin­gle num­ber in a chunk of the num­ber line that changes in size, you con­struct a big set of or­dered points of fixed size equal to the least com­mon mul­ti­ple of the num­ber of pos­si­ble differ­ences for all sums. If you think of a differ­ence not as a num­ber, but as a uniform dis­tri­bu­tion on the set of pos­si­ble differ­ences, then you can just “scale up” this dis­tri­bu­tion from its set of vari­able into the big set of con­stant size, and sam­ple from this dis­tri­bu­tion to for­get the sum but re­mem­ber the most in­for­ma­tion about the differ­ence.

EDIT: I shouldn’t do math while tired.

• Note that the agent should rewrite its mem­ory with a dis­tri­bu­tion, not just a list of tu­ples—e.g. {(1 1): 136, (2 1): 236, …}. That way the “pos­te­rior” dis­tri­bu­tion on the sum will match the prior dis­tri­bu­tion on the sum.

That said, this is ba­si­cally cor­rect. It matches the an­swer(s) I got, and is more el­e­gant.

• Yeah. I guess I was as­sum­ing that the agent knows the list of tu­ples and also knows that they came from the pro­ce­dure I de­scribed; the dis­tri­bu­tion fol­lows from that :-)