PS1sol
The Saga of Big Monkey (BM) and Little Monkey (LM)
setup
Warifruit is worth kilocalories (Kc).
BM’s cost to climb is Kc; LM’s climbing cost is negligible.
If both climb, BM gets most of the fruit: BM = Kc, LM = Kc.
If only LM climbs, after LM returns BM has eaten Kc, LM gets Kc.
If BM climbs, LM can consume about Kc before BM climbs down and chases him away, leaving BM with Kc.
Actions: W = wait (don’t climb), C = climb.
a) Extensive form when BM decides first
BM at the root chooses between and .
If BM = , LM faces a decision node with actions or .
If BM = , LM faces a second decision node with actions or .
Terminal payoffs (BM, LM):
BM = , LM = :
BM = , LM = :
BM = , LM = :
BM = , LM = :
b) Strategies for each monkey
BM: ext{BM} o \{W, C}
LM: LM’s contingent plans across LM’s two decision nodes →
WW = wait after BM = W; wait after BM = C
WC = wait after BM = W; climb after BM = C
CW = climb after BM = W; wait after BM = C
CC = climb after BM = W; climb after BM = C
c) Normal-form (strategic form) representation
BM has 2 strategies:
LM has 4 strategies:
Payoff matrix (BM strategies as rows, LM strategies as columns):
BM = , LM = :
BM = , LM = :
BM = , LM = :
BM = , LM = :
BM = , LM = :
BM = , LM = :
BM = , LM = :
BM = , LM = :
d) How many Nash equilibria (NE) in this BM-first game? What are they?
There are 3 NE:
(W, CW)
(W, CC)
(C, WW)
In terms of strategies:
NE 1: BM = , LM =
NE 2: BM = , LM =
NE 3: BM = , LM =
e) Assume LM decides first. Redo a–d
Extensive form when LM decides first
LM’s choice at the root: or ; BM has contingent decisions after LM’s move across two BM decision nodes.
Strategies: LM ∈ \{W, C\}; BM ∈ \{WW, WC, CW, CC\}.
Payoff matrix (LM strategies as rows, BM strategies as columns):
If LM = , BM = :
LM = , BM = :
LM = , BM = :
LM = , BM = :
If LM = , BM = :
LM = , BM = :
LM = , BM = :
LM = , BM = :
NE (LM-first): 3 equilibria
(W, CW)
(W, CC)
(C, WW)
f) Assume the monkeys decide simultaneously. Redo a–d
Simultaneous-decision game with LM ∈ {W, C} and BM ∈ {W, C}.
Payoff matrix (LM rows, BM columns):
LM = W, BM = W: $(0,0)$
LM = W, BM = C: $(4,4)$
LM = C, BM = W: $(1,9)$
LM = C, BM = C: $(3,5)$
NE in this simultaneous game: two equilibria
(W, C)
(C, W)
g) Why not redo e when solving f?
The reason is that the extensive-form representation shown already captures the contingent move structure and the corresponding normal-form (strategic) representation is the same whether you derive it from the BM-first or the LM-first extensive form. The simultaneous-move game has a single normal-form representation that yields the same NE set, so redrawing the extensive form assuming the other player moves first would duplicate effort without changing the strategic (normal-form) outcomes.
Note on terminology used in this section: W = wait or don’t climb, C = climb; LM = Little Monkey; BM = Big Monkey.
Erste / Zweite (First and Second) game
a) Strategies
Erste (First mover): They have 3 decision nodes; thus a strategy specifies a choice at the first node and choices for the second two nodes. Strategies listed:
AGG, AGH, AHG, AHH, BGG, BGH, BHG, BHH
Zweite (Second mover): They have two decision nodes; thus a strategy lists a planned choice for each node. Strategies:
ce, cf, de, df
b) Normal-form representation
Strategy sets:
Erste: {AGG, AGH, AHG, AHH, BGG, BGH, BHG, BHH}
Zweite: {ce, cf, de, df}
Payoff table (Erste strategies as rows, Zweite strategies as columns). Example entries:
Erste AGG, Zweite ce: $(6,1)$
Erste AGG, Zweite cf: $(6,10)$
Erste AGG, Zweite de: $(9,3)$
Erste AGG, Zweite df: $(9,3)$
Erste AGH, Zweite ce: $(6,1)$
Erste AGH, Zweite cf: $(6,1)$
Erste AGH, Zweite de: $(2,5)$
Erste AGH, Zweite df: $(2,5)$
Erste AHG, Zweite ce: $(7,9)$
Erste AHG, Zweite cf: $(7,9)$
Erste AHG, Zweite de: $(9,3)$
Erste AHG, Zweite df: $(9,3)$
Erste AHH, Zweite ce: $(7,9)$
Erste AHH, Zweite cf: $(7,9)$
Erste AHH, Zweite de: $(2,5)$
Erste AHH, Zweite df: $(2,5)$
Erste BGG, Zweite ce: $(8,3)$