SOLUTION MANUAL
Game Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12
1
,TABLE OF CONTENTS v v v
1 - Nim and Combinatorial Games
v v v v v
2 - Congestion Games
v v v
3 - Games in Strategic Form
v v v v v
4 - Game Trees with Perfect Information
v v v v v v
5 - Expected Utility
v v v
6 - Mixed Equilibrium
v v v
7 - Brouwer’s Fixed-Point Theorem
v v v v
8 - Zero-Sum Games
v v v
9 - Geometry of Equilibria in Bimatrix Games
v v v v v v v
10 - Game Trees with Imperfect Information
v v v v v v
11 - Bargaining
v v
12 - Correlated Equilibrium
v v v
2
,Game Theory Basics
v v
Solutions to Exercises
v v
©v BernhardvvonvStengelv2022
SolutionvtovExercisev1.1
(a) Letv≤vbevdefinedvbyv(1.7).v Tovshowvthatv≤visvtransitive,vconsidervx,vy,vzvwithvxv ≤vyvandvyv≤vz.vIfvxv=vyvthe
nvxv≤vz,vandvifvyv=vzvthenvalsovxv≤vz.vSovthevonlyvcasevleftvisvxv<vyvandvyv <vz,vwhichvimpliesvxv <vzvbecau
sev<visvtransitive,vandvhencevxv ≤vz.
Clearly,v≤visvreflexivevbecausevxv=vxvandvthereforevxv≤vx.
Tovshowvthatvvvvv≤isvantisymmetric,vconsidervxvandvyvwithvxvvvvvyvandvyvvvvv≤
x.vIfvwevhadvx≤v≠vyvthenvxv<vyva
ndvyv<vx,vandvbyvtransitivityvxv<vxvwhichvcontradictsv(1.38).vHencevxv=v y,vasvrequired.v Thisvshowsvt
hatv≤visvavpartialvorder.
Finally,vwevshowv(1.6),vsovwevhavevtovshowvthatvxv<vyvimpliesvxvvvyvandvxv≠vyvand ≤ vvicevversa.vLetvxv<vy,
vwhichvimpliesvxvyvbyv(1.7).vIfvwevhadvxv=vyvthenvxv<vx,vcontradictingv(1.38),vsovwevalsovhavevxv≠vy.v Co
≤
nversely,vxvvv yvandvxv≠vyvimplyvbyv(1.7)vxv <v yvorv xv =v yvwherevthevsecond≤vcasevisvexcluded,vhencev xv <v y,v
asvrequired.
(b) Considervavpartialvordervandv≤ assumev(1.6)vasvavdefinitionvofv<.vTovshowvthatv<visvtransitive,vsupp
osevxv<vy,vthatvis,vxvyvandvxv≠vy,vandvyv<vz,vthat ≤ vis,vyvzvandvyv≠vz.vBecausevvvvisvtransitive,≤
vxvvvvz.vIfvwevhadv
xv=vzvthenv≤ xvvvvvyvandvyvvvvvxvandvhence≤ vxv=vyvbyvantisymmetryvofvvvv,≤ vwhichv contradictsv xv ≠v y,vsov wevhav
≤
ev xvvvvzvandv xv ≠v z,vthatvis,vxv <v zvbyv(1.6),vasvrequired.
≤ ≤
Also,v<visvirreflexive,vbecausevxv<vxvwouldvbyvdefinitionvmeanvxvvvxvandvxv≠vx,vbut ≤ vthevlattervisvnotvtr
ue.
Finally,vwevshowv(1.7),vsovwevhavevtovshowvthatvxv ≤vyvimpliesvxv<vyvorvxv=vyvandvvicevversa,vgivenvtha
tv<visvdefinedvbyv(1.6).vLetvxv≤vy.vThenvifvxv=vy,vwevarevdone,votherwisevxv≠vyvandvthenvbyvdefinitionv
xv<vy.vHence,vxv≤vyvimpliesvxv<vyvorvxv=vy.vConversely,vsupposevxv <v yvorvxv=vy.v Ifvxv <v yvthenvxv ≤vyvbyv(
1.6),vandvifvxv=vyvthenvxv ≤v yvbecausev≤visvreflexive.v Thisvcompletesvthevproof.
SolutionvtovExercisev1.2
(a) Inv analysingv thev gamesv ofv threev Nimv heapsv wherev onev heapv hasv sizev one,v wev firstv lookvatvsomevexam
ples,vandvthenvusevmathematicalvinductionvtovprovevwhatvwevconjecturevtovbevthevlosingvpositions.vAvl
osingvpositionvisvonevwhereveveryvmovevisvtovavwinningvposition,vbecausevthenvthevopponentvwillvw
in.v Thevpointvofvthisvexercisevisvtovformulatevavprecisevstatementvtovbevproved,vandvthenvtovprovevit
.
First,vifvtherevarevonlyvtwovheapsvrecallvthatvtheyvarevlosingvifvandvonlyvifvthevheapsvarevofvequalvsi
ze.v Ifvtheyvarevofvunequalvsize,vthenvthevwinningvmovevisvtovreducevthevlargervheapvsovthatvbothvhe
apsvhavevequalvsize.
3
, Considervthreevheapsvofvsizesv1,vm,vn,vwherev1vvvvvmvvvvvn.vWe
≤ vobserve
≤ vthevfollowing:v1,v1,vmvisvwinni
ng,vbyvmovingvtov1,v1,v0.vSimilarly,v1,vm,vmvisvwinning,vbyvmovingvtov0,vm,vm.vNext,v1,v2,v3visvlosingv
(observedvearliervinvthevlecture),vandvhencev1,v2,vnvforvnv4visvwinning.v1,v3,vnvisvwinningvforvanyvnv
3vbyvmovingvtov1,v3,v2.vForv1,v4,v5,vreducingvanyvheapvproducesvavwinningvposition,vsovthisvisvlosi
≥ ≥
ng.
Thevgeneralvpatternvforvthevlosingvpositionsvthusvseemsvtovbe:v1,vm,vmv1,vforvevenvnumbers
+ vm.v Th
isvincludesvalsovthevcasevmv=v0,vwhichvwevcanvtakevasvthevbasevcasevforvanvinduction.v Wevnowvproce
edvtovprovevthisvformally.
Firstvwevshowvthatvifvthevpositionsvofvthevformv1,vm,vnvwithvmvvvvvvnvarevlosing ≤ vwhenvmvisvevenvandvnv
=vmv1,vthenvthesevarevthev+onlyvlosingvpositionsvbecausevanyvothervpositionv1,vm,vnv withvmv v nv isvwin
ning.v Namely,vifvmv =vnv then≤vavwinningvmovevfromv1,vm,vmvisvtov0,vm,vm,vsovwevcanvassumevmv<vn.v Ifvmvi
svevenvthenvnv>vmv v 1v(otherwisevwevwouldvbevinvthevpositionv1,vm,vmv v 1)vandvsovthevwinningvmovevis
+
vtov1,vm,v mv v 1.vIfv mvisvoddvthenvthevwinningvmovevisvtov1,vm,v mv1,vthevsamevasvpositionv1,vmv 1,vmv(thisv
+ +
wouldv alsov bev av winningv movev fromv 1,vm,vmv sov therev thev winningv movev isv notv unique).
– −
Second,vwevshowvthatvanyvmovevfromv1,vm,vmv+v1vwithvevenvmvisvtovavwinningvposition,vusingvasvinduc
tivevhypothesisvthatv1,vmJ,vmJv+v1vforvevenvmJvandvmJv<vmvisvavlosingvposition.vThevmovevtov0,vm,vmv+v
1vproducesvavwinningvpositionvwithvcounter-
movevtov0,vm,vm.vAvmovevtov1,vmJ,vmv+v1vforvmJv<vmvisvtovavwinningvpositionvwithvthevcounter-
movevtov1,vmJ,vmJv+v1vifvmJvisvevenvandvtov1,vmJ,vmJv−v1vifvmJvisvodd.vAvmovevtov1,vm,vmvisvtovavwinningv
positionvwithvcounter-
movevtov0,vm,vm.vAvmovevtov1,vm,vmJvwithv mJv<v mvisvalsovtovavwinningvpositionvwithvthevcounter-
movevtov1,vmJv−v1,vmJvifv mJvisvodd,vandvtov1,vmJv 1,vmJvifvmJvisvevenv(invwhichvcasevmJv 1v<vmvbecausevmvisv
even).vThisvconcludesvthevinductionvproof.
+ +
ThisvresultvisvinvagreementvwithvthevtheoremvonvNimvheapvsizesvrepresentedvasvsumsvofvpowersvofv2:v
1v v mv v nvisvlosing 0
vifvandvonlyvif,vexceptvforv2 ,vthevpowersvofv2vmakingvupvmvandvnvcomevinvpairs.vSovthe
∗v +∗ +∗
sevmustvbevthevsamevpowersvofv2,vexceptvforv1v=v20,vwhichvoccursvinvonlyvmvorvn,vwherevwevhavevassum
edvthatvnvisvthevlargervnumber,vsov1vappearsvinv thev representationv ofv n:v Wev havev mv =v 2avvvvvv2bvvvvvv2c
forv av >v bv >v cv >vvvvvvvv 1,vsov mv isv
a b c + + +v ·v ·v · ·v ·v·v ≥
even,v and,v withv thev samev a,vb,vc,v.v.v.,v nv =v 2 v v v
2 v v v
2 1v =v mvvvv 1.v Then
+ + +v ·v ·v ·v + +
∗1v vvvvvv
+v ∗mvvvvv+nv ∗vvvvvv 0.
≡vv∗Thev followingv isv anv examplev usingvthev bitvrepresentationv where
mv=v12v(whichvdeterminesvthevbitvpatternv1100,vwhichvofvcoursevdependsvonvm):
1 = 0001
12 = 1100
13 = 1101
Nim-sum 0 = 0000
(b) Wevusev(a).vClearly,v1,v2,v3visvlosingvasvshownvinv(1.2),vandvbecausevthevNim-
sumvofvthevbinaryvrepresentationsv01,v10,v11visv00.vExamplesvshowvthatvanyvothervpositionvisvwin
ning.vThevthreevnumbersvarevn,vnv 1,vnv v 2.vIf+
vnvisvevenvthenvreducingvthevheapvofvsizevnv2vtov1vcreatesv
+
thevpositionvn,vnv 1,v1vwhichvisvlosingvasvshownvinv(a).vIfvnvisvodd,vthenvnv 1visvevenvandvnvvv2v=v nvvv1
+ +
vvv1vsovbyvthevsamevargument,vavwinningvmovevisvtovreducevthevNimvheapvofvsizevnvtov1v(whichvonl
+ + (v +v )v+
yvworksvifvnv >v1).
4
Game Theory Basics 1st Edition
By Bernhard von Stengel. Chapters 1 - 12
1
,TABLE OF CONTENTS v v v
1 - Nim and Combinatorial Games
v v v v v
2 - Congestion Games
v v v
3 - Games in Strategic Form
v v v v v
4 - Game Trees with Perfect Information
v v v v v v
5 - Expected Utility
v v v
6 - Mixed Equilibrium
v v v
7 - Brouwer’s Fixed-Point Theorem
v v v v
8 - Zero-Sum Games
v v v
9 - Geometry of Equilibria in Bimatrix Games
v v v v v v v
10 - Game Trees with Imperfect Information
v v v v v v
11 - Bargaining
v v
12 - Correlated Equilibrium
v v v
2
,Game Theory Basics
v v
Solutions to Exercises
v v
©v BernhardvvonvStengelv2022
SolutionvtovExercisev1.1
(a) Letv≤vbevdefinedvbyv(1.7).v Tovshowvthatv≤visvtransitive,vconsidervx,vy,vzvwithvxv ≤vyvandvyv≤vz.vIfvxv=vyvthe
nvxv≤vz,vandvifvyv=vzvthenvalsovxv≤vz.vSovthevonlyvcasevleftvisvxv<vyvandvyv <vz,vwhichvimpliesvxv <vzvbecau
sev<visvtransitive,vandvhencevxv ≤vz.
Clearly,v≤visvreflexivevbecausevxv=vxvandvthereforevxv≤vx.
Tovshowvthatvvvvv≤isvantisymmetric,vconsidervxvandvyvwithvxvvvvvyvandvyvvvvv≤
x.vIfvwevhadvx≤v≠vyvthenvxv<vyva
ndvyv<vx,vandvbyvtransitivityvxv<vxvwhichvcontradictsv(1.38).vHencevxv=v y,vasvrequired.v Thisvshowsvt
hatv≤visvavpartialvorder.
Finally,vwevshowv(1.6),vsovwevhavevtovshowvthatvxv<vyvimpliesvxvvvyvandvxv≠vyvand ≤ vvicevversa.vLetvxv<vy,
vwhichvimpliesvxvyvbyv(1.7).vIfvwevhadvxv=vyvthenvxv<vx,vcontradictingv(1.38),vsovwevalsovhavevxv≠vy.v Co
≤
nversely,vxvvv yvandvxv≠vyvimplyvbyv(1.7)vxv <v yvorv xv =v yvwherevthevsecond≤vcasevisvexcluded,vhencev xv <v y,v
asvrequired.
(b) Considervavpartialvordervandv≤ assumev(1.6)vasvavdefinitionvofv<.vTovshowvthatv<visvtransitive,vsupp
osevxv<vy,vthatvis,vxvyvandvxv≠vy,vandvyv<vz,vthat ≤ vis,vyvzvandvyv≠vz.vBecausevvvvisvtransitive,≤
vxvvvvz.vIfvwevhadv
xv=vzvthenv≤ xvvvvvyvandvyvvvvvxvandvhence≤ vxv=vyvbyvantisymmetryvofvvvv,≤ vwhichv contradictsv xv ≠v y,vsov wevhav
≤
ev xvvvvzvandv xv ≠v z,vthatvis,vxv <v zvbyv(1.6),vasvrequired.
≤ ≤
Also,v<visvirreflexive,vbecausevxv<vxvwouldvbyvdefinitionvmeanvxvvvxvandvxv≠vx,vbut ≤ vthevlattervisvnotvtr
ue.
Finally,vwevshowv(1.7),vsovwevhavevtovshowvthatvxv ≤vyvimpliesvxv<vyvorvxv=vyvandvvicevversa,vgivenvtha
tv<visvdefinedvbyv(1.6).vLetvxv≤vy.vThenvifvxv=vy,vwevarevdone,votherwisevxv≠vyvandvthenvbyvdefinitionv
xv<vy.vHence,vxv≤vyvimpliesvxv<vyvorvxv=vy.vConversely,vsupposevxv <v yvorvxv=vy.v Ifvxv <v yvthenvxv ≤vyvbyv(
1.6),vandvifvxv=vyvthenvxv ≤v yvbecausev≤visvreflexive.v Thisvcompletesvthevproof.
SolutionvtovExercisev1.2
(a) Inv analysingv thev gamesv ofv threev Nimv heapsv wherev onev heapv hasv sizev one,v wev firstv lookvatvsomevexam
ples,vandvthenvusevmathematicalvinductionvtovprovevwhatvwevconjecturevtovbevthevlosingvpositions.vAvl
osingvpositionvisvonevwhereveveryvmovevisvtovavwinningvposition,vbecausevthenvthevopponentvwillvw
in.v Thevpointvofvthisvexercisevisvtovformulatevavprecisevstatementvtovbevproved,vandvthenvtovprovevit
.
First,vifvtherevarevonlyvtwovheapsvrecallvthatvtheyvarevlosingvifvandvonlyvifvthevheapsvarevofvequalvsi
ze.v Ifvtheyvarevofvunequalvsize,vthenvthevwinningvmovevisvtovreducevthevlargervheapvsovthatvbothvhe
apsvhavevequalvsize.
3
, Considervthreevheapsvofvsizesv1,vm,vn,vwherev1vvvvvmvvvvvn.vWe
≤ vobserve
≤ vthevfollowing:v1,v1,vmvisvwinni
ng,vbyvmovingvtov1,v1,v0.vSimilarly,v1,vm,vmvisvwinning,vbyvmovingvtov0,vm,vm.vNext,v1,v2,v3visvlosingv
(observedvearliervinvthevlecture),vandvhencev1,v2,vnvforvnv4visvwinning.v1,v3,vnvisvwinningvforvanyvnv
3vbyvmovingvtov1,v3,v2.vForv1,v4,v5,vreducingvanyvheapvproducesvavwinningvposition,vsovthisvisvlosi
≥ ≥
ng.
Thevgeneralvpatternvforvthevlosingvpositionsvthusvseemsvtovbe:v1,vm,vmv1,vforvevenvnumbers
+ vm.v Th
isvincludesvalsovthevcasevmv=v0,vwhichvwevcanvtakevasvthevbasevcasevforvanvinduction.v Wevnowvproce
edvtovprovevthisvformally.
Firstvwevshowvthatvifvthevpositionsvofvthevformv1,vm,vnvwithvmvvvvvvnvarevlosing ≤ vwhenvmvisvevenvandvnv
=vmv1,vthenvthesevarevthev+onlyvlosingvpositionsvbecausevanyvothervpositionv1,vm,vnv withvmv v nv isvwin
ning.v Namely,vifvmv =vnv then≤vavwinningvmovevfromv1,vm,vmvisvtov0,vm,vm,vsovwevcanvassumevmv<vn.v Ifvmvi
svevenvthenvnv>vmv v 1v(otherwisevwevwouldvbevinvthevpositionv1,vm,vmv v 1)vandvsovthevwinningvmovevis
+
vtov1,vm,v mv v 1.vIfv mvisvoddvthenvthevwinningvmovevisvtov1,vm,v mv1,vthevsamevasvpositionv1,vmv 1,vmv(thisv
+ +
wouldv alsov bev av winningv movev fromv 1,vm,vmv sov therev thev winningv movev isv notv unique).
– −
Second,vwevshowvthatvanyvmovevfromv1,vm,vmv+v1vwithvevenvmvisvtovavwinningvposition,vusingvasvinduc
tivevhypothesisvthatv1,vmJ,vmJv+v1vforvevenvmJvandvmJv<vmvisvavlosingvposition.vThevmovevtov0,vm,vmv+v
1vproducesvavwinningvpositionvwithvcounter-
movevtov0,vm,vm.vAvmovevtov1,vmJ,vmv+v1vforvmJv<vmvisvtovavwinningvpositionvwithvthevcounter-
movevtov1,vmJ,vmJv+v1vifvmJvisvevenvandvtov1,vmJ,vmJv−v1vifvmJvisvodd.vAvmovevtov1,vm,vmvisvtovavwinningv
positionvwithvcounter-
movevtov0,vm,vm.vAvmovevtov1,vm,vmJvwithv mJv<v mvisvalsovtovavwinningvpositionvwithvthevcounter-
movevtov1,vmJv−v1,vmJvifv mJvisvodd,vandvtov1,vmJv 1,vmJvifvmJvisvevenv(invwhichvcasevmJv 1v<vmvbecausevmvisv
even).vThisvconcludesvthevinductionvproof.
+ +
ThisvresultvisvinvagreementvwithvthevtheoremvonvNimvheapvsizesvrepresentedvasvsumsvofvpowersvofv2:v
1v v mv v nvisvlosing 0
vifvandvonlyvif,vexceptvforv2 ,vthevpowersvofv2vmakingvupvmvandvnvcomevinvpairs.vSovthe
∗v +∗ +∗
sevmustvbevthevsamevpowersvofv2,vexceptvforv1v=v20,vwhichvoccursvinvonlyvmvorvn,vwherevwevhavevassum
edvthatvnvisvthevlargervnumber,vsov1vappearsvinv thev representationv ofv n:v Wev havev mv =v 2avvvvvv2bvvvvvv2c
forv av >v bv >v cv >vvvvvvvv 1,vsov mv isv
a b c + + +v ·v ·v · ·v ·v·v ≥
even,v and,v withv thev samev a,vb,vc,v.v.v.,v nv =v 2 v v v
2 v v v
2 1v =v mvvvv 1.v Then
+ + +v ·v ·v ·v + +
∗1v vvvvvv
+v ∗mvvvvv+nv ∗vvvvvv 0.
≡vv∗Thev followingv isv anv examplev usingvthev bitvrepresentationv where
mv=v12v(whichvdeterminesvthevbitvpatternv1100,vwhichvofvcoursevdependsvonvm):
1 = 0001
12 = 1100
13 = 1101
Nim-sum 0 = 0000
(b) Wevusev(a).vClearly,v1,v2,v3visvlosingvasvshownvinv(1.2),vandvbecausevthevNim-
sumvofvthevbinaryvrepresentationsv01,v10,v11visv00.vExamplesvshowvthatvanyvothervpositionvisvwin
ning.vThevthreevnumbersvarevn,vnv 1,vnv v 2.vIf+
vnvisvevenvthenvreducingvthevheapvofvsizevnv2vtov1vcreatesv
+
thevpositionvn,vnv 1,v1vwhichvisvlosingvasvshownvinv(a).vIfvnvisvodd,vthenvnv 1visvevenvandvnvvv2v=v nvvv1
+ +
vvv1vsovbyvthevsamevargument,vavwinningvmovevisvtovreducevthevNimvheapvofvsizevnvtov1v(whichvonl
+ + (v +v )v+
yvworksvifvnv >v1).
4
