Agent-Simulates-Predictor Variant of the Prisoner’s Dilemma

I don’t know enough math and I don’t know if this is im­por­tant, but in the hopes that it helps some­one figure some­thing out that they oth­er­wise might not, I’m post­ing it.

In Soares & Fallen­stein (2015), the au­thors de­scribe the fol­low­ing prob­lem:

Con­sider a sim­ple two-player game, de­scribed by Slep­nev (2011), played by a hu­man and an agent which is ca­pa­ble of fully simu­lat­ing the hu­man and which acts ac­cord­ing to the pre­scrip­tions of UDT. The game works as fol­lows: each player must write down an in­te­ger be­tween 0 and 10. If both num­bers sum to 10 or less, then each player is paid ac­cord­ing to the num­ber that they wrote down. Other­wise, they are paid noth­ing. For ex­am­ple, if one player writes down 4 and the other 3, then the former gets paid $4 while the lat­ter gets paid $3. But if both play­ers write down 6, then nei­ther player gets paid. Say the hu­man player rea­sons as fol­lows:

“I don’t quite know how UDT works, but I re­mem­ber hear­ing that it’s a very pow­er­ful pre­dic­tor. So if I de­cide to write down 9, then it will pre­dict this, and it will de­cide to write 1. There­fore, I can write down 9 with­out fear.”

The hu­man writes down 9, and UDT, pre­dict­ing this, pre­scribes writ­ing down 1. This re­sult is un­com­fortable, in that the agent with su­pe­rior pre­dic­tive power “loses” to the “dumber” agent. In this sce­nario, it is al­most as if the hu­man’s lack of abil­ity to pre­dict UDT (while us­ing cor­rect ab­stract rea­son­ing about the UDT al­gorithm) gives the hu­man an “epistemic high ground” or “first mover ad­van­tage.” It seems un­satis­fac­tory that in­creased pre­dic­tive power can harm an agent.

More pre­cisely: two agents A and B must choose in­te­gers m and n with 0 m, n 10, and if m + n 10, then A re­ceives a pay­off of m dol­lars and B re­ceives a pay­off of n dol­lars, and if m + n > 10, then each agent re­ceives a pay­off of zero dol­lars. B has perfect pre­dic­tive ac­cu­racy and A knows that B has perfect pre­dic­tive ac­cu­racy.

Con­sider a var­i­ant of the afore­men­tioned de­ci­sion prob­lem in which the same two agents A and B must choose in­te­gers m and n with 0 m, n 3; if m + n 3, then {A, B} re­ceives a pay­off of {m, n} dol­lars; if m + n > 3, then {A, B} re­ceives a pay­off of zero dol­lars. This var­i­ant is similar to a var­i­ant of the Pri­soner’s Dilemma with a slightly mod­ified pay­off ma­trix:

Like­wise, A rea­sons as fol­lows:

If I co­op­er­ate, then B will pre­dict that I will co­op­er­ate, and B will defect. If I defect, then B will pre­dict that I will defect, and B will co­op­er­ate. There­fore, I defect.

And B:

I pre­dict that A will defect. There­fore, I co­op­er­ate.

I figure it’s good to have mul­ti­ple takes on a prob­lem if pos­si­ble, and that this par­tic­u­lar take might be es­pe­cially valuable, what with all of the at­ten­tion that seems to get put on the Pri­soner’s Dilemma and its var­i­ants.