# A simple game that has no solution

The fol­low­ing sim­ple game has one solu­tion that seems cor­rect, but isn’t. Can you figure out why?

The Game

Player One moves first. He must pick A, B, or C. If Player One picks A the game ends and Player Two does noth­ing. If Player One picks B or C, Player Two will be told that Player One picked B or C, but will not be told which of these two strate­gies Player One picked, Player Two must then pick X or Y, and then the game ends. The fol­low­ing shows the Play­ers’ pay­offs for each pos­si­ble out­come. Player One’s pay­off is listed first.

A 3,0 [And Player Two never got to move.]

B,X 2,0

B,Y 2,2

C,X 0,1

C,Y 6,0

The play­ers are ra­tio­nal, each player cares only about max­i­miz­ing his own pay­off, the play­ers can’t com­mu­ni­cate, they play the game only once, this game is all that will ever mat­ter to them, and all of this plus the pay­offs and the game struc­ture is com­mon knowl­edge.

Guess what will hap­pen. Imag­ine you are re­ally play­ing the game and de­cide what you would do as ei­ther Player One, or as Player Two if you have been told that you will get to move. To figure out what you would do you must for­mu­late a be­lief about what the other player has/​will do, and this will in part be based on your be­lief about his be­lief of what you have/​will do.

An In­cor­rect Ar­gu­ment for A

If Player One picks A he gets 3, whereas if he picks B he gets 2 re­gard­less of what Player Two does. Con­se­quently, Player One should never pick B. If Player One picks C he might get 0 or 6 so we can’t rule out Player One pick­ing C, at least with­out first figur­ing out what Player Two will do.

Player Two should as­sume that Player One will never pick B. Con­se­quently, if Player Two gets to move he should as­sume that C was played and there­fore Player Two should re­spond with X. If Player One be­lieves that Player Two will, if given the chance to move, pick X, then Player One is best off pick­ing A. In con­clu­sion, Player One will pick A and Player Two will never get to move.

Why the Game Has No Solution

I be­lieve that the above logic is wrong, and in­deed the game has no solu­tion. My rea­son­ing is given in rot13. (Copy what is be­low and paste at this link to con­vert to English.)

http://​​rot13.com/​​

Vs gur no­bir nanylfvf jrer pbeerpg Cyn­lre Gjb jb­hyq oryvrir ur jvyy ar­ire zbir. Fb jung unc­craf vs Cyn­lre Gjb qbrf trg gb zbir? Vs Cyn­lre Gjb trgf gb zbir jung fub­hyq uvf oryvrs or nobhg jung Cyn­lre Bar qvq tvira gung Cyn­lre Gjb xabjf Cyn­lre Bar qvq abg cvpx N? Cyn­lre Gjb pna’g nffhzr gung P jnf cyn­lrq. Vs vg jrer gehr gung vg’f pbzzba xab­jyrqtr gung Cyn­lre Bar jb­hyq ar­ire cynl O, gura vg fub­hyq or pbzzba xab­jyrqtr gung Cyn­lre Gjb jb­hyq ar­ire cynl L, ju­vpu jb­hyq zrna gung Cyn­lre Bar jb­hyq ar­ire cynl P, ohg pyr­neyl Cyn­lre Bar unf cvpxrq O be P fb fbzr­gu­vat vf je­bat.

Zber nof­genpgyl, vs V qrirybc n gurbel gung lbh jba’g gnxr npgvba Y, naq guvf arprffnevyl erfhygf va gur vz­cyvp­ngvba gung lbh jba’g qb npgvba Z, gura vs lbh unir pyr­neyl qbar rvgure Y be Z zl bevt­vany gurbel vf vainyvq. V’z abg nyy­b­jrq gb nffhzr gung lbh zhfg unir qbar Z whfg orpn­hfr zl vavgvny cebbs ubyq­vat gung lbh jba’g qb Y gbbx sr­jre fgrcf guna zl cebbs sbe jul lbh jba’g qb Z qvq.

Abar vs guvf jb­hyq or n ce­boyrz vs vg jrer veengvbany sbe Cyn­lre Bar gb abg cvpx N. Ns­gre nyy, V unir nffhzrq en­gvbanyvgl fb V’z abg nyy­b­jrq gb cbfghyngr gung Cyn­lre Bar jvyy qb fbzr­gu­vat veengvbany. Ohg vg’f veengvbany sbe Cyn­lre Bar gb Cvpx P bayl vs ur rfgvzn­grf gung gur ce­bonovyvgl bs Cyn­lre Gjb erfcbaq­vat jvgu L vf fhssvpvragyl ybj. Cyn­lre Gjb’f zbir jvyy qr­craq ba uvf oryvrsf bs jung Cyn­lre Bar unf qbar vs Cyn­lre Bar unf abg cvpxrq N. Pbafrdhragyl, jr pna bayl fnl vg vf veengvbany sbe Cyn­lre Bar gb abg cvpx N ns­gre jr unir svtherq bhg jung oryvrs Cyn­lre Gjb jb­hyq unir vs Cyn­lre Gjb trgf gb cynl. Naq guvf oryvrs bs Cyn­lre Gjb pna’g or on­frq ba gur nffhzcgvba gung Cyn­lre Bar jvyy ar­ire cvpx O orpn­hfr guvf erfhygf va Cyn­lre Gjb oryvri­vat gung Cyn­lre Bar jvyy ar­ire cvpx P rvgure, ohg pyr­neyl vs Cyn­lre Gjb trgf gb zbir rvgure O be P unf orra cvpxrq.

Va fhz, gb svaq n fby­hgvba sbe gur tnzr jr arrq gb xabj jung Cyn­lre Gjb jb­hyq qb vs ur trgf gb zbir, ohg gur bayl ern­fbanoyr pnaqvqngr fby­hgvba unf Cyn­lre Gjb ar­ire zbi­vat fb jr unir n pbagen­qvpgvba naq V unir ab vqrn jung gur ev­tug nafjre vf. Guvf vf n trareny ce­boyrz va tnzr gurbel ju­rer n fby­hgvba erd­hverf svthe­vat bhg jung n cyn­lre jb­hyq qb vs ur trgf gb zbir, ohg nyy gur ern­fbanoyr fby­hgvbaf unir guvf cyn­lre ar­ire zbi­vat.

Up­date: Emile has a great an­swer if you as­sume a “trem­bling hand.”

• Clas­si­cal game the­ory says that player 1 should chose A for ex­pected util­ity 3, as this is bet­ter than than the sub game of choos­ing be­tween B and C where the best player 1 can do against a clas­si­cally ra­tio­nal player 2 is to play B with prob­a­bil­ity 13 and C with prob­a­bil­ity 23 (and player 2 plays X with prob­a­bil­ity 23 and Y and with prob­a­bil­ity 13), for an ex­pected value of 2.

But, there are pareto im­prove­ments available. Player 1′s clas­si­cally op­ti­mal strat­egy gives player 1 ex­pected util­ity 3 and player 2 ex­pected util­ity 0. But sup­pose in­stead Player 1 plays C, and player 2 plays X with prob­a­bil­ity 13 and Y with prob­a­bil­ity 23. Then the ex­pected util­ity for player 1 is 4 and for player 2 it is 13. Of course, a clas­si­cally ra­tio­nal player 2 would want to play X with greater prob­a­bil­ity, to in­crease its own ex­pected util­ity at the ex­pense of player 1. It would want to in­crease the prob­a­bil­ity be­yond 12 which is the break even point for player 1, but then player 1 would rather just play A.

So, what would 2 TDT/​UDT play­ers do in this game? Would they man­age to find a point on the pareto fron­tier, and if so, which point?

• Two TDT play­ers have 3 plau­si­ble out­comes to me, it seems. This comes from my ad­mit­tedly in­ex­pe­rienced in­tu­itions, and not much rigor­ous math. The 1st two plau­si­ble points that oc­curred to me are 1)both play­ers choose C,Y, with cer­tainty, or 2)they sit at ex­actly the equil­ibrium for p1, giv­ing him an ex­pected pay­out of 3, and p2 an ex­pected pay­out of .5. Both of these im­prove on the global util­ity pay­out of 3 that’s got­ten if p1 just chooses A (giv­ing 6 and 3.5, re­spec­tively), which is a pos­i­tive thing, right?

The ar­gu­ment that sup­ports these pos­si­bil­ities isn’t un­fa­mil­iar to TDT. p2 does not ex­pect to be given a choice, ex­cept in the cases where p1 is us­ing TDT, there­fore she has the choice of Y, with a pay­out of 0, or not hav­ing been given a chance to chose at all. Both of these pos­si­bil­ities have no pay­out, so p2 is neu­tral about what choice to make, there­fore choos­ing Y makes some sense. Alter­na­tively, Y has to choose be­tween A for 3 or C for p(.5)*(6), which have the same pay­out. C, how­ever, gives p2 .5 more util­ity than she’d oth­er­wise get, so it makes some sense for p1 to pick C.

Alter­na­tively, and what oc­curred to me last, both these agents have some way to equally share their “profit” over Clas­si­cal De­ci­sion The­ory. For how­ever much more util­ity than 3 p1 gets, p2 gets the same amount. This pay­off point (p1-3=p2) does ex­ists, but I’m not sure where it is with­out do­ing more math. Is this a well for­mu­lated game the­o­retic con­cept? I don’t know, but it makes some sense to my idea of “fair­ness”, and the kind of point two well-for­mu­lated agents should con­verge on.

• “Clas­si­cal game the­ory says that player 1 should chose A for ex­pected util­ity 3, as this is bet­ter than than the sub game of choos­ing be­tween B and C ”

No since this is not a sub­game be­cause of the un­cer­tainty. From Wikipe­dia ” In game the­ory, a sub­game is any part (a sub­set) of a game that meets the fol­low­ing crite­ria...It has a sin­gle ini­tial node that is the only mem­ber of that node’s in­for­ma­tion set… ”

I’m un­cer­tain about what TDT/​UDT would say.

• To see that it is in­deed a sub­game:

Rep­re­sent the whole game with a tree whose root node rep­re­sents player 1 choos­ing whether to play A (leads to leaf node), or to en­ter the sub­game at node S. Node S is the root of the sub­game, rep­re­sent­ing player 1′s choices to play B or C lead­ing to nodes rep­re­sent­ing player 2 choice to play X or Y in those re­spec­tive cases, each lead­ing to leaf nodes.

Node S is the only node in its in­for­ma­tion set. The sub­game con­tains all the de­scen­dants of S. The sub­game con­tains all nodes in the same in­for­ma­tion set as any node in the sub­game. It meets the crite­ria.

There is no un­cer­tainty that screws up my ar­gu­ment. The whole point of talk­ing about the sub­game was to stop think­ing about the pos­si­bil­ity that player 1 chose A, be­cause that had been ob­served not to hap­pen. (Of course, I also ar­gue that player 2 should be in­ter­ested in log­i­cally caus­ing player 1 not to have cho­sen A, but that gets be­yond clas­si­cal game the­ory.)

• I’m sorry but “sub­game” has a very spe­cific defi­ni­tion in game the­ory which you are not be­ing con­sis­tent with. Also, in­tu­itively when you are in a sub­game you can ig­nore ev­ery­thing out­side of the sub­game, play­ing as if it didn’t ex­ist. But when Player 2 moves he can’t ig­nore A be­cause the fact that Player 1 could have picked A but did not pro­vides in­sight into whether Player 1 picked B or C. I am a game the­o­rist.

• I’m sorry but “sub­game” has a very spe­cific defi­ni­tion in game the­ory which you are not be­ing con­sis­tent with.

I just ex­plained in de­tail how the sub­game I de­scribed meets the defi­ni­tion you linked to. If you are go­ing to dis­agree, you should be point­ing to some as­pect of the defi­ni­tion I am not meet­ing.

Also, in­tu­itively when you are in a sub­game you can ig­nore ev­ery­thing out­side of the sub­game, play­ing as if it didn’t ex­ist. But when Player 2 moves he can’t ig­nore A be­cause the fact that Player 1 could have picked A but did not pro­vides in­sight into whether Player 1 picked B or C.

If it is some­how the case that giv­ing player 2 info about player 1 is ad­van­ta­geous for player 1, then player 2 should just ig­nore the info, and ev­ery­thing still plays out as in my anal­y­sis. If it is ad­van­ta­geous for player 2, then it just strength­ens the case that player 1 should choose A.

I am a game the­o­rist.

I still think you are mak­ing a mis­take, and should pay more at­ten­tion to the ob­ject level dis­cus­sion.

• Let’s try to find the source of our dis­agree­ment. Would you agree with the fol­low­ing:

“You can only have a sub­game that ex­cludes A if the fact that Player 1 has not picked A pro­vides no use­ful in­for­ma­tion to Player 2 if Player 2 gets to move.”

• The defi­ni­tion you linked to doesn’t say any­thing about en­ter­ing sub­game not giv­ing the play­ers in­for­ma­tion, so no, I would not agree with that.

I would agree that if it gave player 2 use­ful in­for­ma­tion, that should in­fluence the anal­y­sis of the sub­game.

(I also don’t care very much whether we call this ob­ject within the game of how the strate­gies play out given that player 1 doesn’t choose A a “sub­game”. I did not in­tend that tech­ni­cal defi­ni­tion when I used the term, but it did seem to match when I checked care­fully when you ob­jected, think­ing that maybe there was a good mo­ti­va­tion for the defi­ni­tion so it could in­di­cated a prob­lem with my ar­gu­ment if it didn’t fit.)

I also dis­agree that player 1 not pick­ing A pro­vides use­ful in­for­ma­tion to player 2.

• “I also dis­agree that player 1 not pick­ing A pro­vides use­ful in­for­ma­tion to player 2.”

Player 1 gets 3 if he picks A and 2 if he picks B, so doesn’t know­ing that Player 1 did not pick A provide use­ful in­for­ma­tion as to whether he picked B?

• The rea­son player 1 would choose B is not be­cause it di­rectly has a higher pay­out but be­cause in­clud­ing B in a mixed strat­egy gives player 2 an in­cen­tive to in­clude Y in its own mixed strat­egy, in­creas­ing the ex­pected pay­off of C for player 1. The fact that A dom­i­nates B is ir­rele­vant. The fact that A has bet­ter ex­pected util­ity than the sub­game with B and C in­di­cates that player 1 not choos­ing A is some­how ir­ra­tional, but that doesn’t give a use­ful way for player 2 to ex­ploit this ir­ra­tional­ity. (And in or­der for this to make sense for player 1, player 1 would need a way to counter ex­ploit player 2′s ex­ploit, and for player 2 to try its ex­ploit de­spite this pos­si­bil­ity.)

• “The rea­son player 1 would choose B is not be­cause it di­rectly has a higher pay­out but be­cause in­clud­ing B in a mixed strat­egy gives player 2 an in­cen­tive to in­clude Y in its own mixed strat­egy, ”

No since Player 2 only ob­serves Player 1′s choice not what prob­a­bil­ities Player 1 used.

• Player 2 ob­serves “not A” as a choice. Doesn’t player 2 still need to es­ti­mate the rel­a­tive prob­a­bil­ities that B was cho­sen vs. that C was cho­sen?

Of course Player 2 doesn’t have ac­cess to Player 1′s source code, but that’s not an ex­cuse to set those prob­a­bil­ities in a com­pletely ar­bi­trary man­ner. Player 2 has to de­cide the prob­a­bil­ity of B in a ra­tio­nal way, given the available (albeit scarce) ev­i­dence, which is the pay­off ma­trix and the fact that A was not cho­sen.

It seems rea­son­able to imag­ine a space of strate­gies which would lead player 1 to not choose A, and as­sign prob­a­bil­ities to which strat­egy player 1 is us­ing. Player 1 is prob­a­bly mak­ing a shot for 6 points, mean­ing they are try­ing to tempt player 2 into choos­ing Y. Player 2 has to de­cide the prob­a­bil­ity that (Player 1 is us­ing a strat­egy which re­sults in [prob­a­bil­ity of B > 0]), in or­der to make that choice.

• Can you give an ex­am­ple a pair G1, G2 such that you con­sider G2 to be a “sub­game” of G1?

• Con­sider the Pri­soner’s Dilemma, mod­ified so that one per­son moves first and the other per­son gets to ob­serve their move be­fore choos­ing.

Ob­vi­ously the clas­si­cally cor­rect first move is to defect first. Thus the sec­ond player will never have to deal with a move of Co­op­er­ate.

There­fore if a move of Co­op­er­ate is made, the sec­ond player’s move is clas­si­cally un­defined (if one ac­cepts the logic of this post). And yet, if both play­ers play co­op­er­ate it’s bet­ter than (D,D), and so which move gets made first de­pends on the ac­tions of the sec­ond player if Co­op­er­ate is played first. There­fore, this pris­oner’s dilemma has no solu­tion.

(I con­sider this to be a re­duc­tio).

Viewed this way, there’s an ob­vi­ous re­la­tion­ship with the for­mal-agent prob­lem of “I can prove what op­tion is best, and I know I’ll take the best op­tion—there­fore, if I do some­thing else all log­i­cal state­ments are con­di­tion­ing on a false­hood, and so it’s true that I can get the best re­sults by do­ing noth­ing.” The solu­tion there is to not use log­i­cal con­di­tion­ing in­side the de­ci­sion-mak­ing pro­cess like that, and in­stead use causal insertion

Similarly, we might never ex­pect the sec­ond player in an or­dered Pri­soner’s Dilemma to have to deal with co­op­er­a­tion. But we can still talk about that coun­ter­fac­tual by declar­ing by fiat that the first move was co­op­er­a­tion and look­ing at the choices that re­sult. Note the similar­ity be­tween causal in­ser­tion and the trem­bling hand—al­most like this trem­bling hand stuff works for a deeper rea­son. If our sec­ond PD player is an or­di­nary clas­si­cal agent, they will choose Defect—prob­lem re­solved.

The game you pre­sent has an ex­tra di­men­sion, but upon learn­ing that B or C were cho­sen (again, via causal surgery, not log­i­cal con­di­tion­ing), a clas­si­cal agent with­out ad­di­tional in­for­ma­tion will just play the Nash equil­ibrium of the sub-game where only B or C are available—see JGWeiss­man’s com­ment for the cor­rect num­bers.

• P1: .5C .5B

P2: Y

It’s not a Nash equil­ibrium, but it could be a time­less one. Pos­si­bly more trust­wor­thy than usual for oneshots, since P2 knows that P1 was not a Nash agent as­sum­ing the other player was a Nash agent (clas­si­cal game the­o­rist) if P2 gets to move at all.

• I have no idea where those num­bers came from. Why not “P1: .3C .7B” to make “P2: Y” ra­tio­nal? Other­wise, why does P2 play Y at all? Why not “P1: C, P2: Y”, which max­i­mizes the sum of the two util­ities, and is the op­ti­mal pre­com­mit­ment un­der the Rawlian veil-of-ig­no­rance prior? Heck, why not just play the unique Nash equil­ibrium “P1: A”? Most im­por­tantly, if there’s no prin­ci­pled way to make these de­ci­sions, why as­sume your op­po­nent will time­lessly make them the same way?

• Why not “P1: C, P2: Y”, which max­i­mizes the sum of the two util­ities, and is the op­ti­mal pre­com­mit­ment un­der the Rawlian veil-of-ig­no­rance prior?

If we mul­ti­ply player 2′s util­ity func­tion by 100, that shouldn’t change any­thing be­cause it is an af­fine trans­for­ma­tion to a util­ity func­tion. But then “P1: B, P2: Y” would max­i­mize the sum. Ad­ding val­ues from differ­ent util­ity func­tions is a mean­ingless op­er­a­tion.

• You’re right. I’m not ac­tu­ally ad­vo­cat­ing this op­tion. Rather, I was com­par­ing EY’s seem­ingly ar­bi­trary strat­egy with other seem­ingly ar­bi­trary strate­gies. The only one I ac­tu­ally en­dorse is “P1: A”. It’s true that this spe­cific crite­rion is not in­var­i­ant un­der af­fine trans­for­ma­tions of util­ity func­tions, but how do I know EY’s pro­posed strat­egy wouldn’t change if we mul­ti­ply player 2′s util­ity func­tion by 100 as you pro­pose?

(Along a similar vein, I don’t see how I can jus­tify my pro­posal of “P1: 310 C 710 B”. Where did the 10 come from? “P1: 27 C 57 B” works equally well. I only chose it be­cause it is con­ve­nient to write down in dec­i­mal.)

• Eliezer’s “ar­bi­trary” strat­egy has the nice prop­erty that it gives both play­ers more ex­pected util­ity than the Nash equil­ibrium. Of course there are other strate­gies with this prop­erty, and in­deed mul­ti­ple strate­gies that are not them­selves dom­i­nated in this way. It isn’t clear how ideally ra­tio­nal play­ers would se­lect one of these strate­gies or which one they would choose, but they should choose one of them.

• Eliezer, would your ideas from this post ap­ply here?

There could be many ac­cept­able ne­go­ti­at­ing equil­ibria be­tween what you think is the ‘fair’ point on the Pareto bound­ary, and the Nash equil­ibrium. So long as each step down in what you think is ‘fair­ness’ re­duces the to­tal pay­off to the other agent, even if it re­duces your own pay­off even more.

If I’m not too con­fused, the Nash equil­ibrium is [P1: A], and the Pareto bound­ary ex­tends from [P1: B, P2: Y] to [P1: C, P2: Y]. So the gains from trade give P1 1-3 ex­tra points, and P2 0-2 ex­tra points. As oth­ers have pointed out, a case could be made for [P1: C, P2: Y], as it max­i­mizes the to­tal gains from trade, but maybe, tak­ing your idea of differ­ent con­cepts of fair­ness from the linked post, P2 should hold P1 hostage by play­ing some kind of mixed X,Y strat­egy un­less P1 offers a “more fair” split.

Is that be­hav­ior by B the kind of thing that the rea­son­ing in the linked post en­dorses?

• I think this should get bet­ter and bet­ter for P1 the closer P1 gets to (2/​3)C (1/​3)B (with­out ac­tu­ally reach­ing it).

• Let’s say player 1 sub­mits a com­puter pro­gram that will re­ceive no in­put and print ei­ther A, B or C. Player 2 sub­mits a com­puter pro­gram that will re­ceive a sin­gle bit as in­put (tel­ling it whether P1′s pro­gram printed A), and print ei­ther X or Y. Both pro­grams also have ac­cess to a fair ran­dom num­ber gen­er­a­tor. That’s a si­mul­ta­neous move game where ev­ery Nash equil­ibrium leads to pay­offs (3,0). Hope­fully it’s not too much of a stretch to say that we should play the game in the same way that the best pro­gram would play it.

If ad­di­tion­ally each pro­gram re­ceives the other’s source code as in­put, many bet­ter Nash equil­ibria be­come achiev­able, like the out­come (4,1) pro­posed by Eliezer. In this case I think it’s a bar­gain­ing prob­lem. The Nash bar­gain­ing solu­tion pro­posed by Squark might be rele­vant, though I don’t know how to han­dle such prob­lems in gen­eral.

• Hope­fully it’s not too much of a stretch to say that we should play the game in the same way that the best pro­gram would play it.

Should we (hu­mans) play like the best pro­gram that don’t have ac­cess to each other’s source code play it, or play like the best pro­grams that do have ac­cess to each other’s source code play it? I mean, figu­ra­tively we have some in­for­ma­tion about the other player’s source code...

• I think that if we know about one an­other that we be­lieve in play­ing like pro­grams with ac­cess to each other’s source when play­ing against op­po­nents about which we know [QUINE], then we are jus­tified to play like pro­grams with ac­cess to each oth­ers source. :)

• I think the fol­low­ing is the unique proper equil­ibrium of this game:

Player One plays A with prob­a­bil­ity 1-ϵ, B with prob­a­bil­ity 13 ϵ, C with prob­a­bil­ity 23 ϵ. Player Two plays X with prob­a­bil­ity 23 and Y with prob­a­bil­ity 13.

JGWeiss­man had es­sen­tially the right idea, but used the wrong ter­minol­ogy.

ETA: I’ve changed my mind and no longer think the proper equil­ibrium solu­tion makes sense for this game. See later in this thread as well as this com­ment for the ex­pla­na­tion.

• Player Two plays X with prob­a­bil­ity 23 and Y with prob­a­bil­ity 13.

Should that be the other way round?

As writ­ten, player 1 ex­pects to score 3(1-ϵ) + 2(ϵ/​3) + (2ϵ/​3)(6/​3) = 3 − 3ϵ + 2ϵ/​3 + 4ϵ/​3 = 3-ϵ (as­sum­ing I haven’t made a dumb er­ror there), and so would do bet­ter by unilat­er­ally switch­ing to the pure strat­egy A.

But if player 2 plays Y with prob­a­bil­ity 23 and X with prob­a­bil­ity 13, player 1 can ex­pect to score 3(1-ϵ) + 2(ϵ/​3) + (2ϵ/​3)(12/​3) = 3 − 3ϵ + 2ϵ/​3 + 8ϵ/​3 = 3 + ϵ/​3, which beats the pure strat­egy A.

Edit: no, ig­nore me, I for­got that the whole point of proper equil­ibrium is that ϵ is an ar­bi­trary pa­ram­e­ter im­posed from out­side and as­sumed nonzero. Player 1 isn’t al­lowed to set it to zero. [Slaps self on fore­head.]

• You are prob­a­bly right but the proper equil­ibrium as­sump­tion that “more costly trem­bles are made with sig­nifi­cantly smaller prob­a­bil­ity than less costly ones” is a huge one.

• That as­sump­tion isn’t re­ally rele­vant here, since there aren’t ac­tu­ally any sec­ond level trem­bles with prob­a­bil­ity ϵ^2 in the solu­tion. (Maybe trem­bling hand perfect equil­ibrium already gives us this unique solu­tion and I didn’t re­ally need to in­voke proper equil­ibrium, but the lat­ter seems eas­ier to work with.) In­stead of talk­ing about tech­ni­cal­ities though, let’s just think about this in­tu­itively.

Sup­pose there is some definite prob­a­bil­ity of phys­i­cally mak­ing a mis­take when Player One in­tends to choose A. Let’s say when he in­tends to choose A, there’s prob­a­bil­ity 1100 for ac­ci­den­tally press­ing B and 1100 for ac­ci­den­tally press­ing C, and this is com­mon knowl­edge. Now we can just solve this mod­ified game us­ing stan­dard Nash equil­ibrium, and the only solu­tion is that Player One has to make his choices so that af­ter tak­ing both “trem­bling” and de­liber­ate choice into ac­count, the prob­a­bil­ities that Player Two face is that C is twice as likely as B. (I.e., Player One has to de­liber­ately choose C with prob­a­bil­ity about 1100.) That’s the only way that Player Two would be will­ing to choose a mixed strat­egy, which must be the case in or­der to have an equil­ibrium.

• Yes, I think you are right. I won­der if there is a way of chang­ing the pay­offs so there isn’t any trem­bling mixed equil­ibrium.

• Hmm, I just re­al­ized that Player Two’s strat­egy in the Nash equil­ibrium of this mod­ified game is differ­ent than in the proper equil­ibrium of the origi­nal game. Be­cause here Player Two has to make his choice so that Player One is in­differ­ent be­tween A and C, whereas in the proper equil­ibrium Player Two has to make his choices so that Player One is in­differ­ent be­tween B and C.

I think my “in­tu­itive anal­y­sis” does make sense, so I’m go­ing to change my mind and say that per­haps proper equil­ibrium isn’t the right solu­tion con­cept here...

• 20 Jul 2014 20:48 UTC
4 points

As­sume that each player’s hand may trem­ble with a small non-zero prob­a­bil­ity p, then take the limit as p ap­proaches zero from above.

• … Let’s do that!

Sim­ple model: A plays A, B and C with prob­a­bil­ities a, b, and c, with the con­straint that each must be above the trem­bling prob­a­bil­ity t (=p/​3 us­ing the p above). (Two doesn’t trem­ble for sim­plic­ity’s sake)

Two picks X with prob­a­bil­ity x and Y with prob­a­bil­ity (1-x).

So their ex­pected util­ities are:

One: 3a + 2b+6c(1-x)

Two: 2b(1-x) + cx = 2*b + (c − 2b) x

It seems pretty clear that One wants b to be as low as pos­si­ble (ei­ther a or c will always be bet­ter), so we can set b=t.

So One’s util­ity is (con­stant) − 3c+6c −6cx

So One wants c to max­i­mize (1-2x)c, and Two wants x to max­i­mize (c-2t)c

The Nash equil­ibrium is at 1-2x=0 and c-2t=0, so c=2t and x=0.5

So in other words, if One’s hand can trem­ble than he should also some­times de­liber­ately pick C to make it twice as likely as B, and Two should flip a coin.

(and as t con­verges to­wards 0, we do in­deed get One always pick­ing A)

• Ex­cel­lent! This does in­deed work given the as­sump­tion that Player 1 can not set the prob­a­bil­ity of him­self pick­ing B at zero. But if Player 1 can set the prob­a­bil­ity of him pick­ing B at zero, the game still has no solu­tion.

• Good think­ing, but I picked the pay­offs so this ap­proach wouldn’t give an easy solu­tion. Con­sider an equil­ibrium where Player 1 in­tends to pick A, but there is a small but equal chance he will pick B or C by mis­take. In this equil­ibrium Player 2 would pick Y if he got to move, but then Player 1 would always in­tend to Pick C, effec­tively pre­tend­ing he had made a mis­take.

• From Fu­den­berg & Ti­role (1995 edi­tion, chap­ter 8):

Sec­tion 8.4 then de­scribes a re­fine­ment of trem­bling-hand perfect equil­ibrium due to My­er­son(1978). A “proper equil­ibrium” re­quires that a player trem­ble less on strate­gies that are worse re­sponses.

• Hmm, is this not the cor­rect solu­tion for two su­per-ra­tio­nal play­ers:

Player One: Pick C with prob­a­bil­ity of 23 - e; pick B with prob­a­bil­ity of 13 + e, e be­ing some very small but not neg­ligible num­ber. Player Two: Pick Y

Ex­pected pay­off for Player One is 4 23 −4e; way bet­ter than play­ing A. For B is 23 + 2 e, a tiny bit bet­ter than play­ing X—so B will play Y, since he knows that A is to­tally ra­tio­nal and would have picked this very strat­egy.

• Purely the in­for­ma­tion that Player One be­haves ir­ra­tionally doesn’t give Player Two any more in­for­ma­tion on A’s be­havi­our than the fact that it is not ra­tio­nal. So other than know­ing Player One didn’t use the strat­egy “play A with 100% prob­a­bil­ity”, Player Two doesn’t know any­thing about Player One’s be­havi­our. What can Player Two do on that ba­sis? They can as­sume that any de­vi­a­tions from ra­tio­nal choice are small, which brings us to the trem­bling hand solu­tion. Or they can use a differ­ent model.

Which model of player One’s be­havi­our is the “cor­rect” one to as­sume in this situ­a­tion is not at all clear. Per­haps Player One can be mod­el­led as a RNG? Then pick Y. Per­haps One always mod­els its op­po­nents as RNG’s (which is, given no in­for­ma­tion about their ir­ra­tional­ity, ir­ra­tional)? Then pick X (since One is in­differ­ent be­tween A and C in this case). Just as re­versed stu­pidity is not in­tel­li­gence, [not in­tel­li­gence] doesn’t tell you any­thing about the kind of stu­pidity, un­less given more in­for­ma­tion.

Most Game The­ory Prob­lems in gen­eral can be said to have no solu­tion if one of the play­ers be­haves ir­ra­tionally. But that’s not a prob­lem for game the­ory be­cause the Ra­tional Choice as­sump­tion (per­haps al­low­ing small de­vi­a­tions) is perfectly fine in the real world, and re­ally the only sane way one can solve game-the­o­retic prob­lems, nar­row­ing down the space of pos­si­ble be­havi­ours tremen­dously!

And by the way, what use has a pay­off ma­trix if play­ers don’t do their best act­ing on that in­for­ma­tion?

• But that’s not a prob­lem for game the­ory be­cause the Ra­tional Choice as­sump­tion (per­haps al­low­ing small de­vi­a­tions) is perfectly fine in the real world

Not in the real world I’m fa­mil­iar with.

• Un­less you have more spe­cific in­for­ma­tion about the prob­lem in ques­tion, it’s the best con­cept to con­sider. At least in the limit of large stacks, long pon­der­ing times, and de­ci­sions jointly made by large or­ga­ni­za­tions, the as­sump­tion holds. Although think­ing about it, I’d re­ally like to see game the­ory for pre­dictably ir­ra­tional agents, suffer­ing from ex­actly those bi­ases un­trained hu­mans fall prey to.

• it’s the best con­cept to consider

I am not con­vinced about that at all.

Let’s con­sider the top head­lines of the mo­ment: the Rus­sian sep­a­ratists in the Ukraine shot down a pas­sen­ger jet and the IDF in­vaded Gaza. Both situ­a­tions (the sep­a­ratist move­ment and the Mid­dle Eastern con­flict) could be mod­eled in the game the­ory frame­work. Would you be com­fortable ap­ply­ing the “Ra­tional Choice as­sump­tion” to these situ­a­tions?

• I would at­tribute the shoot­ing of the pas­sen­ger jet to in­com­pe­tence; The IDF in­vad­ing Gaza yet again cer­tainly makes sense from their per­spec­tive.

Con­sid­er­ing the wide­spread false in­for­ma­tion in both cases, I’d ar­gue that by and large, the agents (mostly the larger ones like Rus­sia and Is­rael, less so the sep­a­ratists and the pales­tine fighters) act ra­tio­nally on the in­for­ma­tion they have. Take a look at Rus­sia, nei­ther ac­tively fight­ing the sep­a­ratists nor openly sup­port­ing them. I could ar­gue that this is the best strat­egy for ter­ri­to­rial ex­pan­sion, avoid­ing a UN mis­sion while strength­en­ing the sep­a­ratists. Spread­ing false in­for­ma­tion does its part.

I don’t know enough about the pales­tine fighters and the in­for­ma­tion they act on to eval­u­ate whether or not their be­havi­our makes sense.

I only con­sider in­stru­men­tal ra­tio­nal­ity here, not epistemic ra­tio­nal­ity.

• cer­tainly makes sense from their per­spec­tive.

That may well be so, but this is a rather differ­ent claim than the “Ra­tional Choice as­sump­tion”.

We know quite well that peo­ple are not ra­tio­nal. Why would you model them as ra­tio­nal agents in game the­ory?

• As I wrote above, in the limit of large stacks, long pon­der­ing times, and de­ci­sions jointly made by large or­ga­ni­za­tions, peo­ple do ac­tu­ally be­have ra­tio­nally. As an ex­am­ple: Bid­ding for oil drilling rights can be mod­el­led as auc­tions with in­com­plete and im­perfect in­for­ma­tion. Naïve bid­ding strate­gies fall prey to the win­ner’s curse. Game the­ory can model these situ­a­tions as Bayesian games and com­pute the emerg­ing Bayesian Nash Equil­ibria.

• in the limit of large stacks, long pon­der­ing times, and de­ci­sions jointly made by large or­ga­ni­za­tions, peo­ple do ac­tu­ally be­have ra­tio­nally.

I still don’t think so. To be a bit more pre­cise, cer­tainly peo­ple be­have ra­tio­nally some­times and I will agree that things like long de­liber­a­tions or joint de­ci­sions (given suffi­cient di­ver­sity of the de­cid­ing group) tend to in­crease the ra­tio­nal­ity. But I don’t think that even in the limit as­sum­ing ra­tio­nal­ity is a “safe” or a “fine” as­sump­tion.

Ex­am­ple: in­ter­na­tional poli­tics. Another ex­am­ple: or­ga­nized re­li­gions.

I also think that in an­a­lyz­ing this is­sue there is the dan­ger of con­struct­ing ra­tio­nal nar­ra­tives post-fac­tum via the claim of re­vealed prefer­ences. Let’s say en­tity A de­cides to do B. It’s very tempt­ing to say “Aha! It would be ra­tio­nal for A to de­cide to do B if A re­ally wants X, there­fore A wants X and be­haves ra­tio­nally”. And cer­tainly, that hap­pens like that on a reg­u­lar ba­sis. How­ever what also hap­pens is that A re­ally wants Y and de­cides to do B on non-ra­tio­nal grounds or just makes a mis­take. In this case our anal­y­sis of A’s ra­tio­nal­ity is false, but it’s hard for us to de­tect that with­out know­ing whether A re­ally wants X or Y.

• “They can as­sume that any de­vi­a­tions from ra­tio­nal choice are small,”

Not play­ing A might be ra­tio­nal de­pend­ing on Player Two’s be­liefs.

“And by the way, what use has a pay­off ma­trix if play­ers don’t do their best act­ing on that in­for­ma­tion?” The game’s un­cer­tainty and se­quen­tial moves mean you can’t use a stan­dard pay­off ma­trix.

• Not play­ing A might be ra­tio­nal de­pend­ing on Player Two’s be­liefs.

That, too, is the case in many games. The suc­cess in always as­sum­ing mu­tu­ally ra­tio­nal be­havi­our first lies in its nice prop­er­ties, like in­abil­ity to be ex­ploited, ex­is­tence of equil­ibrium, and a cer­tain re­sem­blance (albeit not a perfect one) to the real world.

The game’s un­cer­tainty and se­quen­tial moves mean you can’t use a stan­dard pay­off ma­trix.

Well, I mean, why spec­ify pay­offs at all if you then as­sume play­ers won’t care about them? If Player One cared about their pay­off and were mod­el­ling Two as a ra­tio­nal choice agent, they would’ve played A. So ei­ther they don’t model Two as a ra­tio­nal choice agent (which is stupid in its own right), or they sim­ply don’t care about their pay­off.

In any case, the fact that a game with ir­ra­tional play­ers doesn’t have a solu­tion, at least as long as the na­ture of play­ers’ ir­ra­tional­ity is not clear, doesn’t sur­prise me. Deal­ing with mad­men has never been easy.

• If Player One cared about their pay­off and were mod­el­ling Two as a ra­tio­nal choice agent, they would’ve played A

You can only prove this if you first tell me what Player two’s be­liefs would be if he got to move.

In any case, the fact that a game with ir­ra­tional play­ers doesn’t have a solu­tion, at least as long as the na­ture of play­ers’ ir­ra­tional­ity is not clear, doesn’t sur­prise me.

I agree, but I meant for the play­ers in the game to be ra­tio­nal.

• Player two could sim­ply play the equil­ibrium strat­egy for the 2x2-sub­game.

And to counter your re­sponse that it’s all one game, not two games, I can split the game by adding an ex­tra node with­out chang­ing the struc­ture of the game. Then, we ar­rive at the stan­dard sub­game perfect equil­ibrium, only that the 2x2 sub­game is in nor­mal form, which shouldn’t re­ally change things since we can just com­pute a Nash equil­ibrium for that.

After solv­ing the sub­game, we see that player one not play­ing A is not cred­ible, and we can elimi­nate that branch.

• I can split the game by adding an ex­tra node with­out chang­ing the struc­ture of the game

That ac­tu­ally does change the struc­ture of the game, if we as­sume that phys­i­cal mis­takes can hap­pen with some prob­a­bil­ity, which seems re­al­is­tic. (Think about play­ing this game in real life as player 2. If you do get to move, you must think there’s some non-zero prob­a­bil­ity that it was be­cause Player 1 just ac­ci­den­tally pressed the wrong but­ton, right?) With the ex­tra node, you get the oc­ca­sional “crap, I just ac­ci­den­tally pressed not-A, now I have to de­cide which of B or C to choose” which has no anal­ogy in the origi­nal game where you never get to choose be­tween B or C with­out A as an op­tion.

• Okay, I agree. But what do you think about the ex­ten­sive-form game in the image be­low? Is the struc­ture changed there?

• The struc­ture isn’t changed there, but with­out the ex­tra node, there is no sub­game. That ex­tra node is nec­es­sary in or­der to have a sub­game, be­cause only then can Player 2 think “the prob­a­bil­ities I’m fac­ing is the re­sult of Player 1′s choice be­tween just B and C” which al­lows them to solve that sub­game in­de­pen­dently of the rest of the game. Also, see this com­ment and its grand­child for why speci­fi­cally, given pos­si­bil­ity of ac­ci­den­tal presses, I don’t think Player 2′s strat­egy in the over­all game should be same as the equil­ibrium of the 2x2 “re­duced game”. In short, in the re­duced game, Player 2 has to make Player 1 in­differ­ent be­tween B and C, but in the over­all game with ac­ci­den­tal presses, Player 2 has to make Player 1 in­differ­ent be­tween A and C.

• In the 2x2 re­duced game, Player One’s strat­egy is 13 B, 23 C; Two’s strat­egy is 23 X, 13 Y. In the com­plete game with trem­bling hands, Player Two’s strat­egy re­mains un­changed, as you wrote in the starter of the linked thread, in­vok­ing proper equil­ibrium.

• Later on in the linked thread, I re­al­ized that the proper equil­ibrium solu­tion doesn’t make sense. Think about it: why does Player 1 “trem­ble” so that C is ex­actly twice the prob­a­bil­ity of B? Other than pure co­in­ci­dence, the only way that could hap­pen is if some of the but­ton presses of B and/​or C are ac­tu­ally de­liber­ate. Clearly Player 1 would never de­liber­ately press B while A is still an op­tion, so Player 1 must ac­tu­ally be play­ing a mixed strat­egy be­tween A and C, while also ac­ci­den­tally press­ing B and C with some small prob­a­bil­ity. But that im­plies Player 2 must be play­ing a mixed strat­egy that makes Player 1 in­differ­ent be­tween A and C, not be­tween B and C.

• After solv­ing the sub­game, we see that player one not play­ing A is not cred­ible, and we can elimi­nate that branch.

But Player 1 can make this perfectly cred­ible by ac­tu­ally not Play­ing A.

• But I just showed that this is ir­ra­tional as they would get less pay­off in that sub­game!

If that’s your at­ti­tude, then you have to aban­don the con­cept of sub­game perfect equil­ibrium en­tirely. Are you will­ing to do that?

• I think that adding the ex­tra node does change the struc­ture of the game. I also think that we have differ­ent views of what cred­i­bil­ity means.

• How does it change the struc­ture of the game? Of course, it was in nor­mal form be­fore, and is now in ex­ten­sive form, but re­ally, the way you set it up means it shouldn’t mat­ter which rep­re­sen­ta­tion we choose, since player two is get­ting ex­actly the same in­for­ma­tion.

Also, your ar­gu­ment about player two get­ting in­for­ma­tion about One’s be­havi­our can eas­ily be ap­plied to “nor­mal” ex­ten­sive form games. Re­gard­less of whether you in­tended to, if your ar­gu­ment were cor­rect, it would ren­der the con­cept of sub­game perfect equil­ibrium use­less.

I know that cred­i­bil­ity is nor­mally ap­plied as in “make cred­ible threats”. But if I change pay­offs to A: (3, 5) and add a few nodes above, then player 1′s threat to not play A (which in this case is a threat) is not cred­ible, and (3,5) car­ries over to the par­ent node.

By the logic of the ex­ten­sive form for­mu­la­tion, Two should sim­ply play the equil­ibrium strat­egy for the 2x2 sub­game.

Edit: Here is what the game looks like in ex­ten­sive form:

The dot­ted el­lipse in­di­cates that 2 can’t differ­en­ti­ate be­tween the two con­tained nodes. I don’t see how any of the play­ers has any more or less in­for­ma­tion or any more or less choices available.

• Yes this is the same game, but you can not cre­ate a sub­game that has B and C but not A.

• “as­sume both play­ers are ra­tio­nal. What should player 2 do when player 1 acts ir­ra­tionally?”

Player 2 should re­al­ize that his model is in­cor­rect and come up with a new the­ory for player 1′s mo­ti­va­tion. If one is hu­man-like, then two should guess that one was tempted by the shot at the 6 pay­off and chose C. Play X.

• But un­less you can prove that Player 2 would re­spond with X, you can’t tell me that Player 1 is ir­ra­tional for not pick­ing A.

• [The fol­low­ing ar­gu­ment is made with tongue some­what in cheek.]

Ra­tion­al­ity (with a cap­i­tal R) is sup­posed to be an ideal­ized al­gorithm that is uni­ver­sally valid for all agents. There­fore, this al­gorithm, as such, doesn’t know whether it will be in­stan­ti­ated in Player One (P1) or Player Two (P2). Yes, each player knows which one they are. But this knowl­edge is in­put that they feed to their Ra­tion­al­ity sub­rou­tine. The sub­rou­tine it­self doesn’t come with this knowl­edge built in.

Since Ra­tion­al­ity doesn’t know where it will end up, it doesn’t know which out­comes will max­i­mize its util­ity. Thus, the most ra­tio­nal thing for Ra­tion­al­ity (qua ideal­ized al­gorithm) to do is to pre­com­mit to strate­gies for P1 and P2 that max­i­mize its ex­pected util­ity given this un­cer­tainty.

That is, let EU₁(s, t) be the ex­pected util­ity to P1 if P1 im­ple­ments strat­egy s and P2 im­ple­ments strat­egy t. Define EU₂(s, t) similarly. Then Ra­tion­al­ity wants to pre­com­mit the play­ers to the re­spec­tive strate­gies s and t that max­i­mize

p EU₁(s, t) + q EU₂(s, t),

where p (re­spec­tively, q) is the mea­sure of the copies of Ra­tion­al­ity that end up in P1 (re­spec­tively, P2).

As­sum­ing that p = q, Ra­tion­al­ity will there­fore have P1 choose C (with prob­a­bil­ity 1) and have P2 choose Y (with prob­a­bil­ity 1).

That’s all well and good, but will the play­ers ac­tu­ally act that way? After all, they are stipu­lated to care only about them­selves, so P2 in par­tic­u­lar will not want to act ac­cord­ing to this self­less strat­egy.

Yes, but this self­ish de­sire on P2′s part is not built into P2′s Ra­tion­al­ity sub­rou­tine (call it R), be­cause Ra­tion­al­ity is uni­ver­sal. P2′s self­ish de­sires must be im­ple­mented el­se­where, out­side of R. To be sure, P2 is free to feed the in­for­ma­tion that it is P2 to R, but R won’t do any­thing with this in­for­ma­tion, be­cause R is already pre­com­mit­ted to a strat­egy for the rea­sons given above.

And since the play­ers are given to be ra­tio­nal, they are forced to act ac­cord­ing to the strate­gies pre-se­lected by their Ra­tion­al­ity sub­rou­tines, de­spite their wishes to the con­trary. There­fore, they will in fact act as Ra­tion­al­ity de­ter­mined.

[If this com­ment has any point, it is that there is a strong ten­sion, if not a con­tra­dic­tion, be­tween the idea of ra­tio­nal­ity as a uni­ver­sally valid mode of rea­son­ing, on the one hand, and the idea of ra­tio­nal agents whose re­vealed prefer­ences are self­ish, on the other.]

• Er­ror: Ad­ding val­ues from differ­ent util­ity func­tions.

See this com­ment.

• [Re­sum­ing my tongue-in-cheek ar­gu­ment...]

It is true that adding differ­ent util­ity func­tions is in gen­eral an er­ror. How­ever, for agents bound to fol­low Ra­tion­al­ity (and Ra­tion­al­ity alone), the differ­ent util­ity func­tions are best thought of as the same util­ity func­tion con­di­tioned on differ­ent hy­pothe­ses, where the differ­ent hy­pothe­ses look like “The util­ity to P2 turns out to be what re­ally mat­ters”.

After all, if the agents are mak­ing their de­ci­sions on the ba­sis of Ra­tion­al­ity alone, then Ra­tion­al­ity alone must have a util­ity func­tion. Since Ra­tion­al­ity is uni­ver­sal, the util­ity func­tion must be uni­ver­sal. What al­ter­na­tive does Ra­tion­al­ity have, given the con­straints of the prob­lem, other than a weighted sum of the util­ity func­tions of the differ­ent in­di­vi­d­u­als who might turn out to mat­ter?

• “Ra­tion­al­ity” seems to give differ­ent an­swer to the same prob­lem posed with differ­ent af­fine trans­for­ma­tions of the play­ers’ util­ity func­tions.

• [Still ar­gu­ing with tongue in cheek...]

That’s where the mea­sures p and q come in.

• I sus­pect that UDT play­ers always reach the Nash bar­gain­ing solu­tion al­though I have no proof.

For this game I proved that there is a lo­cal max­i­mum of the Nash product when 1 plays B with prob­a­bil­ity 38 and C with prob­a­bil­ity 58 and 2 plays Y. I’m not sure whether it’s global (can the Nash product have non-global lo­cal max­ima?)

• I thought the an­swer was Player One picks B with 1/​3+ϵ prob­a­bil­ity and C with 2/​3-ϵ prob­a­bil­ity. Player Two picks Y.

This gives Player One an ex­pected value of 2(1/​3+ϵ) + 6(2/​3-ϵ) = 14/​3-4ϵ and Player Two an ex­pected value of 2(1/​3+ϵ) = 2/​3+2ϵ.

If Player Two picked X, he’d have an ex­pected value of 2/​3-ϵ, and miss out on 3ϵ.

If Player One picked B with a higher prob­a­bil­ity, Player Two would still pick C, and Player One wouldn’t gain any­thing. If Player One picked C with a higher prob­a­bil­ity, Player Two would pick X, and Player One would get noth­ing. If Player One picked A, he’d only get 3, and miss out on 5/​3-4ϵ.

Did I mess up some­where?

• If Player One be­lieves that Player Two is go­ing to pick Y, then Player One will pick C, but of course this isn’t an equil­ibrium since Player Two would re­gret his strat­egy. All Player Two ever sees is Player One’s move, not the prob­a­bil­ities that Player 1 might have used so if C is played Player Two doesn’t know if it was be­cause Player One Played C with prob­a­bil­ity 1 or prob­a­bil­ity 2/​3-ϵ.

• Per­haps Player One figures that Player Two knows enough about him to pre­dict him slightly more than ex­plain­able by chance, and vice versa.

• I think you can just com­pute the Nash Equil­ibria. For ex­am­ple, use this site: http://​​ba­nach.lse.ac.uk/​​

The an­swer ap­pears to be “always pick A”. Player 2 will never get to move.

• In the Nash equil­ibrium, what is Player 2′s be­lief if he gets to move? Also, the link you gave is for solv­ing si­mul­ta­neous move games, and the game I pre­sented is a se­quen­tial move game.

• That player one is not ra­tio­nal and he should aban­don clas­si­cal game the­ory.

• You are im­plic­itly us­ing cir­cu­lar rea­son­ing. Not pick­ing A is only ir­ra­tional for some but not all pos­si­ble be­liefs that Player 2 could have if Player 1 does not pick A. And even if we grant your as­sump­tion, what should Player 2 do if he gets to move, and if your an­swer al­lows for the pos­si­bil­ity that he picks Y how can you be sure that Player 1 is ir­ra­tional?

• I’m not us­ing cir­cu­lar rea­son­ing. The choice for player one be­tween A and ei­ther B or C is a choice be­tween a cer­tain pay­off of 3 and an as of yet un­cer­tain pay­off. If player A already chose to play ei­ther B or C, the game trans­forms into a game with a sim­ple 2x2 pay­off ma­trix. Writ­ing down the ma­trix we see that there is no pure dom­i­nant strat­egy for ei­ther player. We know though that there is a mixed strat­egy equil­ibrium as there always is one. Player one as­sumes that player two will play such that player one’s choice does not mat­ter and equal­ises his ex­pected pay­off to 2. Player two again as­sumes that player one plays in such a way that their choice does not mat­ter and equal­ises his ex­pected pay­off to 23. As the ex­pected pay­off for player one in the sec­ond game is lower than in the first game, at least one of the fol­low­ing as­sump­tions has to be false about player one:

1. Player one max­imises ex­pected util­ity in terms of game payoff

2. Player one cares only about his own util­ity, not the util­ity of player two

3. Player one as­sumes player two to not act ac­cord­ing to similar principles

• “If player [1] already chose to play ei­ther B or C, the game trans­forms into a game with a sim­ple 2x2 pay­off ma­trix.”

No be­cause Player 2 knows you did not pick A and this might give him in­sight into what you did pick. So even af­ter Player 1 picks B or C the ex­is­tence of strat­egy A might still effect the game be­cause of un­cer­tainty.

• Dis­tu­ingish be­tween the rea­son­ing and the out­come. Game the­o­retic rea­son­ing is mem­ory-less, the ex­act choice of ac­tion of one player does not mat­ter to the other one in the hy­po­thet­i­cal. As the rules are known, both play­ers come to the same con­clu­sion and can pre­dict how the game will play out. If in prac­tice this model is vi­o­lated by one player the other player im­me­di­ately knows that the first player is ir­ra­tional.

• “Game the­o­retic rea­son­ing is mem­ory-less”

No. Con­sider the tit-for-tat strat­egy in the in­finitely re­peated pris­on­ers’ dilemma game.

Why is it ir­ra­tional for Player 1 to not pick A? Your an­swer must in­clude be­liefs that Player 2 would have if he gets to move.

• Se­quen­tial move games are es­sen­tially a sub­set of si­mul­ta­neous move games. If two play­ers both write source code for pro­grams that will play a se­quen­tial move game, then the writ­ing of the code is a si­mul­ta­neous move game.

• I don’t think your game is se­quen­tial, if Player 2 doesn’t know Player 1′s move.

You re­ally have two games:

Game 1: Se­quen­tial game of Player 1 chooses A or B/​C, and de­ter­mines whether game 2 oc­curs.

Game 2: Si­mul­ta­neous game of Player 2 maybe choos­ing X or Y, against Player 1′s un­known se­lec­tion of B/​C.

edit: And the equil­ibrium case for Player 1 in the sec­ond game is an ex­pected pay­out of 2, so he should always choose A.

• I think you can just com­pute the Nash Equil­ibria.

Things don’t feel so sim­ple to me. (A, X) is a Nash equil­ibrium (and the only pure strat­egy NE for this game), but is nonethe­less un­satis­fac­tory to me; if player 1 com­pares that pure strat­egy against the mixed strat­egy pro­posed by Wei_Dai, they’ll choose to play Wei_Dai’s strat­egy in­stead. Nash equil­ibrium doesn’t seem to be a strong enough re­quire­ment (“solu­tion con­cept”) to force a plau­si­ble-look­ing solu­tion. [Edit: oops, dis­re­gard this para­graph. I mis­in­ter­preted Wei_Dai’s solu­tion so switch­ing to it from the NE pure equil­ibrium won’t ac­tu­ally get player A a bet­ter pay­off.]

(I also tried com­put­ing the mixed strat­egy NE by find­ing the player 1 move prob­a­bil­ities that max­i­mized their ex­pected re­turn, but ob­tained a con­tra­dic­tion! Maybe I screwed up the maths.)

• 21 Jul 2014 14:12 UTC
1 point

Hmm. The re­sults ap­pear quite differ­ent if you al­low com­mu­ni­ca­tion and re­peated plays. And they also in­tro­duce some­thing which seems slightly differ­ent than Trem­bling Hand (Per­haps Trem­bling Me­mory?)

With com­mu­ni­ca­tion and re­peated plays:

As­sume all po­ten­tial Player 1′s cred­ibly pre­com­mit to flip a fair coin and based on the toss, pick B half the time, and C half the time.

All po­ten­tial Player 2′s would know this, and as­sum­ing they ex­pect Player 1 to al­most always fol­low the pre­com­mit­ment, would pick Y, be­cause they would max­i­mize their ex­pected pay­out. (50% chance of 2 means ex­pected pay­out of 1, com­pared to pick­ing X, where 50% chance of 1 means ex­pected pay­out of 12.)

All po­ten­tial Player 1′s, based on fol­low­ing that pre­com­mit­ment uni­ver­sally, and hav­ing Player 2′s always pick Y, will get 2 50% of the time and 6 50% of the time, which would get an ex­pected per game pay­out of 4.

This seems bet­ter for ev­ery­one (by about one point per game) then Player 1 only choos­ing A.

With Trem­bling Me­mory:

As­sume fur­ther that some of the time, Player 2′s mem­ory trem­bles and he for­gets about the pre­com­mit­ments and the fact that this game is played re­peat­edly.

So If Player 2 sus­pects for in­stance, that they MIGHT be in the case above, but have for­got­ten im­por­tant facts (they are in­cor­rect about this be­ing a one-off game, they are in­cor­rectly as­sess­ing the state of com­mon knowl­edge, but they are cor­rectly as­sess­ing the cur­rent pay­off struc­ture of this par­tic­u­lar game) then fol­low­ing those sus­pi­cions, it would still make sense for them to choose Y, and it would also ex­plain why Player 1 chose some­thing that wasn’t A.

How­ever, noth­ing would seem to pre­vent Player 2 from sus­pect­ing other pos­si­bil­ities. (For in­stance, un­der the as­sump­tion that Player 1 hits player 2 with Am­ne­sia dust be­fore ev­ery game, know­ing that Player 2 will be forced into a mem­ory trem­bled and will be­lieve the above, Player 1 could play C ev­ery time, with player 2 draw­ing pre­dictably in­cor­rect con­clu­sions and play­ing Y at no benefit.)

I’m not sure how to model a situ­a­tion with trem­bling mem­ory, though, so I would not be sur­prised if I was miss­ing some­thing.

• I’m not overly fa­mil­iar with game the­ory, so for­give me if I’m mak­ing some el­e­men­tary mis­take, but surely the only pos­si­ble out­come is Player 1 always pick­ing A. Either other op­tion is es­sen­tially Player 1 choos­ing a smaller or no pay­off, which would vi­o­late the stated con­di­tion of both play­ers be­ing ra­tio­nal. A non­sen­si­cal game doesn’t have to make sense.

• To know that A gives you a higher pay­off than C you have to know what Player 2 would do if he got to move, but since Player 2 ex­pects to never move how do you figure this out?

• Right that makes sense, but wouldn’t Player 1 sim­ply re­al­ize that mak­ing an ac­cu­rate fore­cast of player 2′s ac­tions is func­tion­ally im­pos­si­ble, and still go with the cer­tain pay­out of A?

• By defi­ni­tion of ra­tio­nal­ity in game the­ory, Player 1 will max­i­mize his ex­pected pay­off and so need to have some be­lief as to the prob­a­bil­ities. If you can’t figure out a way of es­ti­mat­ing these prob­a­bil­ities the game has no solu­tion in clas­si­cal game the­ory land.

• Well, as i said I’m not fa­mil­iar with the math­e­mat­ics or rules of game the­ory so the game may well be un­solv­able in a math­e­mat­i­cal sense. How­ever, it still seems to me that Player 1 choos­ing A is the only ra­tio­nal choice. Hav­ing thought about it some more I would state my rea­son­ing as fol­lows. For Player 1, there is NO POSSIBLE way for him to max­i­mize util­ity by se­lect­ing B in a non-iter­ated game, it can­not ever be a ra­tio­nal choice, and you have stated the player is ra­tio­nal. Choos­ing C can con­ceiv­ably re­sult in greater util­ity, so it can’t be im­me­di­ately dis­carded as a ra­tio­nal choice. If Player 2 finds him­self with a move against a ra­tio­nal player, then the only pos­si­ble choice that player could have made is C, so a ra­tio­nal Player 2 must choose X. Both play­ers, be­ing ra­tio­nal can see this, and so Player 1 can­not pos­si­bly choose any­thing other than A with­out be­ing ir­ra­tional. Un­less you can jus­tify some sce­nario in which a ra­tio­nal player can max­i­mize util­ity by choos­ing B, then nei­ther player can con­sider that as a ra­tio­nal op­tion.

• Then please an­swer the ques­tion, “if Player 2 gets to move what should he be­lieve Player 1 has picked?”

Un­til you can an­swer this ques­tion you can not solve the game. If it is not pos­si­ble to an­swer the ques­tion, then the game can not be solved. I know that you want to say “Not pick­ing A would prove Player 1 is ir­ra­tional” but you can’t claim this un­til you tell me what Player 2 would do if he got to move, and you can’t an­swer this last ques­tion un­til you tell me what Player 2 would be­lieve Player 1 had done if Player 1 does not pick A.

• If Player 2 gets to move, then the only pos­si­ble choice for a ra­tio­nal Player 1 to have made is to pick C, be­cause B can­not pos­si­bly max­i­mize Player 1′s util­ity. The prob­a­bil­ity for a ra­tio­nal Player 1 to pick B is always 0, so the prob­a­bil­ity of pick­ing C has to be 1. For Player 1,there is no ra­tio­nal rea­son to ever pick B, and pick­ing C means that a ra­tio­nal Player 2 will always pick X, negat­ing Player 1′s util­ity. So a ra­tio­nal Player 1 must pick A.

• So are you say­ing that if Player 2 gets to move he will be­lieve that Player 1 picked C?

• Yes.

• But this does not make sense be­cause then player 1 will know that player 2 will play X, so Player 1 would have been bet­ter off play­ing A or B over C.

• You seem to be treat­ing the sub-prob­lem, “what would Player 2 be­lieve if he got a move” as if it is sep­a­rate from and un­in­formed by Player 1′s origi­nal choice. As­sum­ing Player 1 is a util­ity-max­i­mizer and Player 2 knows this, Player 2 im­me­di­ately knows that if he gets a move, then Player 1 be­lieved he could get greater util­ity from ei­ther op­tion B or C than he could get from op­tion A. As op­tion B can never offer greater util­ity than op­tion A, a ra­tio­nal Player 1 could never have se­lected it in prefer­ence to A. But of course that only leaves C as a pos­si­bil­ity for Player 1 to have se­lected and Player 2 will se­lect X and deny any util­ity to Player 1. So nei­ther op­tion B nor C can ever pro­duce more util­ity than op­tion A if both play­ers are ra­tio­nal.

• Ex­actly, but B is never a prefer­able op­tion over A, so the only ra­tio­nal op­tion is for Player 1 to have cho­sen A in the first place, so any cir­cum­stance in which Player 2 has a move ne­ces­si­tates an ir­ra­tional Player 1. The prob­a­bil­ity of Player 1 ever se­lect­ing B to max­i­mize util­ity is always 0.

• So what hap­pens if Player Two does get to move?

This is equiv­a­lent to “what if Omega can’t pre­dict what I would do?” im­plicit rea­son­ing done by two-box­ers in New­comb. Nei­ther pos­si­bil­ity is in the solu­tion do­main, pro­vided that one does not fight the hy­po­thet­i­cal (in your case “the play­ers are ra­tio­nal” in New­comb’s “Omega is a perfect pre­dic­tor”). Player two does not get to move, so there is no point con­sid­er­ing that. Omega knows ex­actly what you’d do, so no point con­sid­er­ing what to do if he is wrong.

• And this is the prob­lem I have with all the New­comb/​Omega busi­ness.

The hy­po­thet­i­cal should be fought. We should no more as­sign ab­solute cer­tainty to Omega’s pre­dic­tive power than we should as­sign ab­solute cer­tainty to CDT’s pre­dic­tive power.

In­stead of as­sign­ing 0 prob­a­bil­ity to the the­ory that Omega can make a mis­take, as­sign DeltaOmega. Similarly as­sign DeltaCDT to the prob­a­bil­ity that CDT anal­y­sis is wrong. I’m too lazy to ac­tu­ally do the math, but do you have any doubt that the right de­ci­sion will de­pend on the ra­tio of the two deltas?

There’s re­ally noth­ing to see here. .This is just an­other case of gen­er­at­ing para­doxes in prob­a­bil­ity the­ory when you don’t do a full anal­y­sis us­ing finite, non zero prob­a­bil­ity as­sign­ments.

This is a similar is­sue the OP has come up against. Propo­si­tion1 is that A obeys cer­tain game the­o­retic rules. Propo­si­tion2 is the re­port that im­pli­cates A vi­o­lat­ing those rules. When your propo­si­tions seem mu­tu­ally con­tra­dic­tory, be­cause you have lazily as­signed 0 prob­a­bil­ity to them, hilar­ity en­sues. As­sign finite val­ues, and the mys­ter­ies are re­solved.

• You are ba­si­cally us­ing the trem­bling hand equil­ibrium con­cept. I picked the pay­offs so this would not yield an easy solu­tion. Con­sider an equil­ibrium where Player 1 in­tends to pick A, but there is a small but equal chance he will pick B or C by mis­take. In this equil­ibrium Player 2 would pick Y if he got to move, but then Player 1 would always in­tend to Pick C, effec­tively pre­tend­ing he had made a mis­take.

• Player two does not get to move, so there is no point con­sid­er­ing that.

First, you are im­plic­itly us­ing cir­cu­lar rea­son­ing. You can not tell me that pick­ing B or C is ir­ra­tional un­til you tell me what be­liefs Player 2 would have if B or C were picked.

Also, imag­ine you are play­ing the game against some­one you think is ra­tio­nal. You are Player 2. You are told that A was not picked. What do you do?

• Also, imag­ine you are play­ing the game against some­one you think is ra­tio­nal. You are Player 2. You are told that A was not picked. What do you do?

If I think Player 1 is ra­tio­nal, I as­sume he must be mod­el­ing my de­ci­sion-mak­ing pro­cess some­how. If his model of my de­ci­sion-mak­ing pro­cess has pick­ing B or C seems ra­tio­nal, he must be mod­el­ing my choice of X and Y in a way that gives him a chance of a higher pay­off than he can get by choos­ing A. Since ev­ery com­bi­na­tion of (B,C) and (X,Y) is lower than his re­turn from A ex­cept [C,Y], no model of my de­ci­sion-mak­ing pro­cess would make B a good op­tion, while some mod­els (though in­ac­cu­rate) would recom­mend C as a po­ten­tially good op­tion. So while it’s un­cer­tain, it’s very likely I’m at C. In that case, I should pick X, and shake my head at my op­po­nent for dras­ti­cally dis­count­ing how ra­tio­nal I am, if he thought he could some­how go one level higher and get the big pay­off.

• imag­ine you are play­ing the game against some­one you think is ra­tio­nal. You are Player 2. You are told that A was not picked.

That’s the con­tra­dic­tion right there. If you are player 2 and get to move, Player 1 is not ra­tio­nal, be­cause you can always re­duce their pay­off by pick­ing X.

• Your be­hav­ior in im­pos­si­ble-in-re­al­ity but in some sense pos­si­ble-to-think-about situ­a­tions may well in­fluence oth­ers’ de­ci­sions, so it may be use­ful to de­cide what to do in im­pos­si­ble situ­a­tions if you ex­pect to be deal­ing with oth­ers who are moved by such con­sid­er­a­tions. Since de­ci­sions make their al­ter­na­tives im­pos­si­ble, but are based on eval­u­a­tion of those al­ter­na­tives, con­sid­er­ing situ­a­tions that even­tu­ally turn out to be im­pos­si­ble (as a re­sult of be­ing de­cided to be­come im­pos­si­ble) is a very nat­u­ral thing to do.

• But why is not pick­ing A “im­pos­si­ble-in-re­al­ity”? You can not an­swer un­til you tell me what Player 2′s be­liefs would be if A was not picked.

• I was mak­ing the more gen­eral point that im­pos­si­ble situ­a­tions (ab­stract ar­gu­ments that aren’t mod­eled by any of the “pos­si­ble” situ­a­tions be­ing con­sid­ered) can mat­ter, that im­pos­si­bil­ity is not nec­es­sar­ily sig­nifi­cant. Apart from that, I agree that we don’t ac­tu­ally have a good ar­gu­ment for im­pos­si­bil­ity of any given ac­tion by Player 1, if it de­pends on what Player 2 could be think­ing.

• Be­cause for Player 1 to in­crease his pay­off over pick­ing A, the only op­tion he can choose is C, based on an ac­cu­rate pre­dic­tion via some pro­cess of rea­son­ing that player 2 will pick X, thereby mak­ing a false pre­dic­tion about Player 1′s be­havi­our. You have stated both play­ers are ra­tio­nal, so I will as­sume they have equal pow­ers of rea­son, in which case if it is pos­si­ble for Player 2 to make a false pre­dic­tion based on their pow­ers of rea­son then Player 1 must be equally ca­pa­ble of mak­ing a wrong pre­dic­tion, mean­ing that Player 1 should avoid the un­cer­tainty and always go for the guaran­teed pay­off.

• To for­mu­late this math­e­mat­i­cally you would need to de­ter­mine the prob­a­bil­ity of mak­ing a false pre­dic­tion and fac­tor that into the odds, which I re­gret is be­yond my abil­ity.

• That’s the con­tra­dic­tion right there. If you are player 2 and get to move, Player 1 is not ra­tio­nal, be­cause you can always re­duce their pay­off by pick­ing X.

Note that “each player cares only about max­i­miz­ing his own pay­off”. By as­sump­tion, player 2 has only a self­ish prefer­ence, not a sadis­tic one, so they’ll only choose X (or be more likely to choose X) if they ex­pect that to im­prove their own ex­pected score. If player 1 can cred­ibly ex­pect player 2 to play Y of­ten enough when given the op­por­tu­nity, it is not ir­ra­tional for player 1 to give player 2 that op­por­tu­nity by play­ing B or C.

• Please an­swer the ques­tion, what would you do if you are player 2 and get to move? Might you pick Y? And if so, how can you con­clude that Player 1 was ir­ra­tional to not pick A?

• what would you do if you are player 2 and get to move?

I will re­al­ize that I was lied to, and the player 1 is not ra­tio­nal. Now, if you are ask­ing what player 2 should do in a situ­a­tion where Player 1 does not fol­low the best pos­si­ble strat­egy, I think Eliezer’s solu­tion above works in this case. Or Emile’s. It de­pends on how you model ir­ra­tional­ity.

• I don’t agree since you can’t prove that not pick­ing A is ir­ra­tional un­til you tell me what player 2 would do if he gets to move and we can’t an­swer this last ques­tion.

• Semi-plau­si­ble in­ter­pre­ta­tion of the game:

Player One and Two are in a war. Player One can send a mes­sage to his team, and Player Two can in­ter­cept it. There is a price for send­ing the mes­sage, and a price for in­ter­cept­ing it.

A is “don’t send any mes­sage”. B is “send a use­less (blank, or ran­dom) mes­sage”. (Uses up 1 util­ity. Also gives the other player 2 free util­ity points.) C is “send a use­ful mes­sage”. (Uses up 1 util­ity, but gains 4 if not in­ter­cepted, and loses ex­tra 1 if in­ter­cepted.) X is “in­ter­cept”, which costs 2 util­ity. How­ever, in­ter­cept­ing a use­ful mes­sage gains you 3 util­ity (af­ter pay­ing 2). Y is “don’t in­ter­cept”.

(I think I got the num­bers right, if they don’t match up some­where, tell me and I’ll ad­just them.)

The ana­log to the prob­lem is this: send­ing a use­less mes­sage can never beat do­ing noth­ing. (We’re as­sum­ing you don’t gain from the op­po­nent wast­ing en­ergy. It’s pos­si­ble if the en­ergy is small enough to be out­weighed by the cost of send­ing the mes­sage.) So if Player Two sees a mes­sage, he should always as­sume it is use­ful. There­fore Two will in­ter­cept all mes­sages. There­fore Player One should never send any mes­sages.

How­ever, if One does, in fact, send a mes­sage, then Two can’t rely on his ra­tio­nal­ity, and can’t as­sume the mes­sage has value to in­ter­cept.

• This can be an­a­lyzed as a reg­u­lar 2-player game with pay­off matrix

-- X Y

A 3,0 3,0

B 2,0 2,2

C 0,1 6,0

Player 2′s in­differ­ence be­tween X and Y when player 1 plays A means that player 2 only con­sid­ers whether player 1 plays B or C.

• Your model doesn’t in­cor­po­rate the un­cer­tainty of my game. Even if Player 2 knows that Player 1 didn’t play A, the fact that he could have im­pacts his es­ti­mate of whether Player 1 picked B or C.

• I’m say­ing that Player 2′s re­ward is strictly con­trol­led by what­ever frac­tion of the time player 1 plays B or C, since if player 1 plays A player 2′s re­ward is guaran­teed to be zero, and diminishes ex­pected re­ward from X and Y in the same pro­por­tion.

If player 1 moves and when they pick ei­ther A,B or C player 2 is told “player 1 picked A, B or C” then player 2 can re­duce it to only con­sid­er­ing the pos­si­bil­ity of B and C be­cause even though A strictly dom­i­nates B, player 2′s re­ward is only non-zero in the case where B or C are played.

This anal­y­sis would change if A,X were 3,0.5 or even 3,0.01

• We look at game the­ory in differ­ent ways. By my anal­y­sis it is ir­rele­vant what Player 2 would get if A were played, it could be \$1 trillion or -\$1 trillion and it would have no im­pact on the game as I see it. But then I don’t use time­less de­ci­sion the­ory, and you might be. This could be the source of our dis­agree­ment.

• No, I’m just say­ing that since in your par­tic­u­lar ex­am­ple Player 2 is in­differ­ent when Player 1 chooses A, the fact that they don’t get a de­ci­sion doesn’t mat­ter. Noth­ing to do with TDT.

• What is this game sup­posed to analo­gize to in re­al­ity? I usu­ally don’t like to fight the hy­po­thet­i­cal but in this sort of situ­a­tion I feel like as player 2 I would as­sume they are in­clud­ing con­sid­er­a­tions from out­side the game’s ra­tio­nal­ity like a sense of fair­ness or gen­eros­ity.

• As far as I know the game has no di­rect real life anal­ogy. But I be­lieve it shows a weak­ness of game the­ory, and shows that some­times to know what you are go­ing to do you need to take into ac­count what an­other per­son would do if that per­son were in a situ­a­tion they would never ex­pect to be in.

• This game is ex­actly equiv­a­lent to the stan­dard one where player one chooses from (A,B,C) and player two chooses from (X,Y), with the pay­off for (A,X) and for (A,Y) equal to (3,0). When choos­ing what choice to make, player two can ig­nore the case where player one chooses A, since the pay­offs are the same in that case.

And as oth­ers have said, the pure strat­egy (A,X) is a Nash equil­ibrium.

• It’s not equiv­a­lent be­cause of the un­cer­tainty.

Also, even if it were, lots of games have Nash equil­ibra that are not rea­son­able solu­tions so say­ing “this is a Nash equil­ibrium” doesn’t mean you have found a good solu­tion. For ex­am­ple, con­sider the si­mul­ta­neous move game where we each pick A or B. If we both pick B we both get 1. If any­thing else hap­pens we both get 0. Both of us pick­ing A is a Nash equil­ibrium, but is also clearly un­rea­son­able.

• It’s not equiv­a­lent be­cause of the un­cer­tainty.

Could you ex­plain what you mean? What un­cer­tainty is there?

Also, even if it were, lots of games have Nash equil­ibra that are not rea­son­able solu­tions so say­ing “this is a Nash equil­ibrium” doesn’t mean you have found a good solu­tion.

For ex­am­ple, con­sider the si­mul­ta­neous move game where we each pick A or B. If we both pick B we both get 1. If any­thing else hap­pens we both get 0. Both of us pick­ing A is a Nash equil­ibrium, but is also clearly un­rea­son­able.

This game has two equil­ibria: a bad one at (A,A) and good one at (B,B). The game from this post also has two equil­ibria, but both in­volve player one pick­ing A, in which case it doesn’t mat­ter what player two does (or in your ver­sion, he doesn’t get to do any­thing).

• Could you ex­plain what you mean? What un­cer­tainty is there?

If Player 2 gets to move he is un­cer­tain as to what Player 1 did. He might have a differ­ent prob­a­bil­ity es­ti­mate in the game I gave than one in which strat­egy A did not ex­ist, or one in which he is told what Player 1 did.

I’m not con­vinced that the game has any equil­ibrium un­less you al­low for trem­bling hands. For A,A to be an equil­ibrium you have to tell me what be­lief Player 2 would have if he got to move, or tell me that Player 1′s be­lief about Player 2′s be­lief can’t effect the game.

• If Player 2 gets to move he is un­cer­tain as to what Player 1 did. He might have a differ­ent prob­a­bil­ity es­ti­mate in the game I gave than one in which strat­egy A did not ex­ist, or one in which he is told what Player 1 did.

In a clas­si­cal game all the play­ers move si­mul­ta­neously. So to re­peat, your game is:

• player 1 chooses A, B or C

• then, player 2 is told whether player 1 chose B or C, and in that case he chooses X or Y

• pay­offs are (A,-) → (3,0); (B,X) → (2,0); (B,Y) → (2,2); (C,X) → (0,1); (C,Y) → (6,0)

The clas­si­cal game equiv­a­lent is

• player 1 chooses A, B or C

• with­out be­ing told the choice of player 1, player 2 chooses X or Y

• pay­offs are as be­fore, with (A,X) → (3,0); (A,Y) → (3,0).

I hope you agree that the fact that player 2 gets to make a (use­less) move in the case that player 1 chooses A doesn’t change the fun­da­men­tals of the game.

In this clas­sic game player 2 also has less in­for­ma­tion be­fore mak­ing his move. In par­tic­u­lar, player 2 is not told whether or not player 1 choose A. But this in­for­ma­tion is com­pletely ir­rele­vant for player 2′s strat­egy, since if player 1 chooses A there is noth­ing that player 2 can do with that in­for­ma­tion.

I’m not con­vinced that the game has any equil­ibrium un­less you al­low for trem­bling hands.

If the play­ers choose (A,X), then the pay­off is (4,0). Chang­ing his choice to B or C will not im­prove the pay­off for player 1, and switch­ing to Y doesn’t im­prove the pay­off for B. There­fore this is a Nash equil­ibrium. It is not sta­ble, since player 2 can switch to Y with­out get­ting a worse pay­off.

• In a clas­si­cal game all the play­ers move si­mul­ta­neously.

I’m not sure what you mean by “clas­si­cal game” but my game is not a si­mul­ta­neous move game. Many se­quen­tial move games do not have equiv­a­lent si­mul­ta­neous move ver­sions.

“I hope you agree that the fact that player 2 gets to make a (use­less) move in the case that player 1 chooses A doesn’t change the fun­da­men­tals of the game.”

I do not agree. Con­sider these pay­offs for the same game:

A 3,0 [And Player Two never got to move.]

B,X 2,10000

B,Y 2,2

C,X 0,1

C,Y 4,4

Now al­though Player 1 will never pick A, its ex­is­tence is re­ally im­por­tant to the out­come by con­vinc­ing Player 2 that if he moves C has been played.

• I do not agree. Con­sider these pay­offs for the same game: …

Differ­ent pay­offs im­ply a differ­ent game. But even in this differ­ent game, the si­mul­ta­neous move ver­sion would be equiv­a­lent. With re­gards to choos­ing be­tween X and Y, the ex­is­tence of choice A still doesn’t mat­ter, be­cause if player 1 chose A X and Y have the same pay­off. The only differ­ence is how much player 2 knows about what player 1 did, and there­fore how much player 2 knows about the pay­off he can ex­pect. But that doesn’t af­fect his strat­egy or the pay­off that he gets in the end.

• There’s no math­e­mat­i­cal solu­tion for sin­gle-player, non-zero sum games of any sort. All these con­structs lead to is ar­gu­ments about “what is ra­tio­nal”. If you a full math model of a “ra­tio­nal en­tity”, then you could get a math­e­mat­i­cally defined solu­tion.

This is why I pre­fer evolu­tion­ary game the­ory to clas­si­cal game the­ory. Evolu­tion­ary game the­ory gen­er­ally has mod­els of its ac­tors and thus guaran­tees a solu­tion to the prob­lems it posits. One can ar­gue with the mod­els and I would say that’s where such ar­gu­ments most fruit­fully should be.