2
2 of IDSDS, the game is as shown
2
After two rounds
in FIGURE SOL3.7.2.
consider x2 # 1. I want to show that this strategy is strictly dominated by
FIGURE SOL3.7.2
0. Note that regardless of x1, player 2 has the lower score whether he
chooses
Player
2
0 or x2 # 1. Since x2 # 0 gives a payoffSOLUTIONS
of 8 andMANUAL
x2 # 1 gives
a payoff
zero
w of 7,
yTHE
z
CHAPTER 3: ELIMINATING
IMPOSSIBLE:
SOLVING A GAME WHEN RATIONALITY IS COMMON KNOWLEDGE
strictly dominates effort of player 1. Now consider the remaining
strategy
b
1,3
5,3 of
2,0
2. Suppose x1 # 0. In that case, the payoff to player
2 from x % 1 is 8 " x6,2
2
Player 1 c 2 2,3 3,3
e he gets a B) and from x2 $ 2 is 10 " x2 (since he gets an d
A). His
payoff
is7,5
3,4
4,0
is some 5:50
strategy
for
player
1 such that x2 # 2
mized
at x2 # 2. Given there8/5/08
2976T_ch03_013-024.qxd
Page
3-11
payoff of at least
8PM
regardless
what player 2 does,
while x ! 2 yields a payoff of at
For8/5/08
player
d is best
player
2 uses z and
cannot
bewhen
strictly
dominated.
It b
is is best1 when player 2 uses y.
s the
maximum payoff,
then
x2 # 21,5:50
2976T_ch03_013-024.qxd
PM
Page
3-11
7. However,
most 8. cFurthermore,
zero effortby
does
strictly
betterofwhen
player
2 chooses was
effort
is
strictly
dominated
d.
Since
none
player
1’s strategies
uded that the strategies for player 2 that survive the first round of the IDSDS
eliminated
thestrategies
second round,
none1of
player
2’s strategies
can be
of 0 or 1.in
The
for player
that
survive
the first round
ofeliminated
the IDSDSinare
0,2}. Now move to round
2 ofthird
this iterative
process. The payoff matrix in FIGURE
round.
a. the
{0,1}. With
respect to player 2, it was previously shown that strategies 1, 3, 4, and
three The
rounds
IDSDS, in
thea game
as shown
.10.1 shows the surviving After
strategies.
firstofnumber
cell isisplayer
1’s in FIGURE SOL3.7.3.
5 are strictly dominated and, therefore, weakly dominated. Zero effort is not
ff.
FIGURE
weakly dominated by the
same SOL3.7.3
argument as in part (a). x ! 2 is weakly dominated
Player 2
2
SOLUTIONS
MANUAL to CHAPTER
3: ELIMINATING
THE IMPOSSIBLE:
A GAME WHEN
for reasons
analogous
those given
for player
1. The SOLVING
lone strategy
forRATIONALITY
player 2IS COMMON KNOWLEDGE
w
y
z
SOLUTIONS
MANUAL
CHAPTER
3: ELIMINATING
SOLVING A GAME WHEN RATIONALITY IS COMMON KNOWLEDGE
FIGURE
SOL3.10.1
! 0.
that survives
the first
round
of the
IDSDS isTHE
x1 IMPOSSIBLE:
x
b
1,3
5,3
2,0
Moving to2the second
round,
Player
1 x1 ! 0 weakly dominates x1 ! 1 since both result
d
3,4
4,0
2
in player 01’s getting
an A (as player 2 is choosing 7,5
0 effort). So the former yields a
payoff
of
10
and
the
latter awhat
payoff
of 9. 2The
IDSDS
that both
players
will
of
at
least
8
regardless
player
does,
whilepredicts
x1 ! 2 yields
a payoff
of at
payoff
0
10,8
8,8
payoff
at least
8 regardless
what
player
2 does,
while
x1 !
2 yields
a payoffeffort
of at
exert
zero
effort.
most
8.of
Furthermore,
zero effort
does
strictly
better
when
player
2 chooses
9,8 strategies
9,6 zero
most
8.1.Furthermore,
strictly
better
whenround
player
chooses
effort
The
for effort
playerdoes
1 that
survive
the first
of2the
IDSDS
are
10 or
b. of
(xd, z1) of 0 or
1. The
strategies
for 2,
player
1 that
surviveshown
the first
round
of the1,IDSDS
are
{0,1}.
With
respect
to
player
it
was
previously
that
strategies
3,
4,
and
2
8,8
8,6
11. {0,1}.
Groucho
Marx
once
said,
“I’ll
never
join
any
club
that
would
have
me
for
a
member.”
respect
to player
2, it
was previously
shown
that strategies
1, 3, is
4, not
and
are With
strictly
dominated
and,
therefore,
weakly
dominated.
Zero
effort
5Well,
is not interested
in joining your
investment
club,
but effort
Julie is.
5 are Groucho
strictly
dominated
and,argument
therefore,
dominated.
Zero
is Your
not club
dominated
weakly
dominated
by the same
asweakly
in part (a).
x2 ! 2 is weakly
has
members,
and
thesame
procedure
for admitting
a new
member
is
simple:
Each
per# 1 strictly dominates xweakly
# 10
2. Neither
of the
other
twoargument
strategies
for
player
!
2
is
weakly
dominated
dominated
by
the
as
in
part
(a).
x
1
2
8. for
(b1, bstrategies
2, banalogous
3) reasons
to those given for player 1. The lone
strategy for player 2
son2’s
receives
a ballot
that
has
two
options:
(1)
and
(2) dofor
not
admit2 Julie.
#
0TheJulie
dominated. None of player
is strictly
dominated
since
x2 admit
for
reasons
analogous
to
those
given
for
player
1.
lone
strategy
player
that survives the first round of the IDSDS is x1 ! 0.
#person
2 when
x1first
#
1 or
2 and
2 yields
same
s a higher payoff than does
x2 survives
Each
can
check
one
ofxthose
two options
or abstain by not submitting a ballot.
! 0.
the
round
IDSDS
isthe
x1dominates
2x #!
that
0 weakly
x1 ! 1 since both result
Moving
to the
second
round,the
1
#Moving
0.Julie
It is to
concluded
thatround,
the
that
survive
ff as does x2 # 0 when x1For
be
admitted,
shestrategies
must
receive
at
least
six votes
favor
of admittance.
!
0
weakly
dominates
xSo
1insince
both
result
to
the
second
x
1 2 is choosing 0 effort).
1 !the
in player 1’s getting an A (as player
former
yields
a
econd round of the IDSDS
for player
1 are
{0,1}
and
for player
2 are with
{0,2}.
Letting
m
be
the
number
of
ballots
submitted
option
1
checked,
assume
that
in
player
1’s
getting
an
A
(as
player
2
is
choosing
0
effort).
So
the
former
yields
a your
10. payoff of 10 and the latter a payoff of 9. The IDSDS predicts that both players will
ng to the third round,payoff
the resulting
payoff
matrix
is of
shown
FIGUREpredicts that both players will
of
10
andis
the latter
a payoff
9. TheinIDSDS
function
exert zero
effort.
.10.2.
a. exert
zero effort.
1 if m ! 6, 7, 8, 9, 10
11. Groucho
once said,
never join any club
FIGUREMarx
SOL3.10.2
. that would have me for a member.”
e “I’ll
11. Well,
Groucho
Marx
once
said,
never
foris.aYour
member.”
0 “I’ll
if m
0,join
1, 2,any
3, 4,club
5 that would
x
Groucho is2 not interested
in!joining
your
investment
club,have
but me
Julie
club
Well,
Groucho
is
not
interested
in
joining
your
investment
club,
but
Julie
is.
Your
club
has 10 members,
and
the procedure for admitting a new member is simple: Each per0
2
has
10
members,
and
the
procedure
for
admitting
a
new
member
is
simple:
Each
perProve that
checking
1 (admit
Julie)
is not a
dominant
sona. receives
a ballot
thatoption
has two
options:
(1) admit
Julie
and (2)strategy.
do not admit Julie.
0 10,8
8,8 that has two options: (1) admit Julie and (2) do not admit Julie.
son receives
a ballot
Each
can check one of those two options or abstain by not submitting a ballot.
x1 person
Each
person
can check
of those two options or abstain by not submitting a ballot.
1 to 9,8
9,6the one
For ANSWER:
Julie
beIfadmitted,
she must
receive
least six
in favor
of admittance.
all
other
players
checkatoption
2, votes
then Julie
is not
admitted
Forregardless
Julie to be
admitted,
she
must
receive
at least
six votesabstain.
in favor
of admittance.
whether
you
check
option
1, check
Since
all
three
submitted
Letting m be the number of ballots
withoption
option2,1orchecked,
assume
that
your
Letting
m be yield
the number
of ballots
submitted
withnone
option
1 checked,
assume
that your
strategies
the
highest
payoff
in
that
case,
of
them
is
strictly
dominated
payoff
function
is
spection reveals that no
strategies
are strictly dominated. Thus, the IDSDS
b. (
0, 0
) payoff
is is no dominant strategy.
and function
thus
cts that player 1 will exert effort
of 0 there
or 1 and
player 2 will exert effort of 0 or 2.
1 if m ! 6, 7, 8, 9, 10
1 if
! 6, 7, dominated
8, 9, 10 . strategy.
edeletion
b. Prove
abstaining
is m
a weakly
ive the strategies that survive
thethat
iterative
stratee 0 if mof! weakly
0, 1, 2, dominated
3, 4, 5 .
11. 0
if
m
!
0,
1,
2,
3, 4, 5 dominated
. (The procedure works the same as the iterative deletion of strictly
ANSWER:
If at least
six of strategies
the otheratmembers
submit ballots with option 1
tegies, except that you eliminate
all weakly
dominated
each stage.)
a. Prove
thatthen
checking
option
1 (admit
Julie) of
is not
a you
dominant
strategy.
a. checked,
Julie
is
admitted
regardless
what
do; your
payoff is 1 with all
a. Prove that checking option 1 (admit Julie) is not a dominant
strategy.
strategies.
If
four
or
fewer
of
the
other
members
vote
in
favor
of option 1, then
WER: Now let us repeat the analysis when we are instead iteratively deleting
ANSWER:
If
all
the
other
players
check
option
2,
then
Julie
is
notisadmitted
Julie
is
denied
admittance
regardless
of
what
you
do;
your
payoff
0 with all
ly dominated strategies. ANSWER:
From part (a),
wethe
know
thatplayers
effort ofcheck
3, 4, and
5 are
If all
other
option
2, then Julie is not admitted
regardless
whether
you
check
option
1,
check
option
2,
or
abstain.
Since
all
strategies.
This
leaves
only
the
case
when
five
of
the
other
members
submit
ballots
rictly dominated and therefore
weakly
dominated.
Foroption
player1,1,check
efforts
of 0 2, or abstain. Since all three
regardless
whether
you check
option
three
strategies
yield
the
highest
payoff
in
that
case,
none
of them
is as
strictly
in
favor
of
admitting
Julie.
Abstaining
results
in
a payoff
of 0,
Julie dominated
ends up with
1 are not weakly dominated
for
the
same
reasons
given
in
part
(a):
there
is
a
strategies yield the highest payoff in that case, none of them is strictly dominated
and
thus
there
is strictly
no dominant
strategy.
only
five
supporting
votes.
Voting
and While
checking
1 results in her admittance
2
egy for player 2 such that
each
yields
the
highest
payoff.
x1 #option
and
thus
there
is no dominant
strategy.
thusdominated
a payoff ofby
1. Hence,
weakly
The latter is
yields
a dominated by voting in favor
t strictly dominated, it is and
weakly
x1 # 0. abstaining
of Julie.
b. Prove
that abstaining is a weakly dominated strategy.
b. b. Prove that abstaining is a weakly dominated strategy.
c. Now suppose
you’re
at the
end members
of the daysubmit
and so ballots
it is costly
you to1 attend
ANSWER:
If at least
sixtired
of the
other
withfor
option
ANSWER:
If at meeting
least sixtoofvote.
the By
other
submit
ballots from
with the
option
the evening’s
not members
showing up,
you abstain
vote. 1This is
checked, then Julie is admitted regardless of what you do; your payoff is 1 with all
checked,
then
Julie
is admitted
regardless
ofthe
what
you do; your payoff is 1 with all
reflected
your
function
having
form
strategies.
If in
four
or payoff
fewer of
the other
members
vote in favor of option 1, then
strategies. If four or fewer of the other members vote in favor of option 1, then
Julie is denied admittance regardless of what you do; your payoff is 0 with all
Julie is denied admittance regardless of what you do; your payoff is 0 with all
strategies. This leaves
case
when
the abstained
other members submit ballots
1 only
if mthe
! 6,
7, 8,
9, 10five
andof
strategies. This leaves
only
the
case
when
five
ofyou
the other members submit ballots
in favor of admitting
Julie. Abstaining results in a payoff of 0, as Julie ends up with
1
in favor of admitting
Julie.
Abstaining
results
a payoff
of 0, as Julie ends up with
if mVoting
! 6, 7,and
8, 9,checking
10 andinyou
voted
2 votes.
only five supporting
option
1 results in her admittance
µ
only five supporting
votes. Voting and checking option 1 results in her admittance
0 1.ifHence,
m ! 0,abstaining
1, 2, 3, 4, 5is and
you dominated
abstained by voting in favor
and thus a payoff of
weakly
and thus a payoff 1of 1. Hence, abstaining is weakly dominated by voting in favor
of Julie.
#
if
m
!
0,
1,
2,
3,
4,
5
and
you
voted
of Julie.
2
c. Now suppose you’re tired at the end of the day and so it is costly for you to attend
c. Now
suppose
you’re tired not
at the
end ofdominated
the day and
so it is costly for you to attend
Prove
that abstaining
a weakly
strategy.
evening’s
meeting toisvote.
By
not showing
up, you
abstain from the vote. This is
c. the
the evening’s meeting to vote. By not showing up, you abstain from the vote. This is
reflected in your payoff function having the form
reflected
payoffmembers
function vote
having
the form
ANSWER:inIfyour
the other
so that
m " 6, then the payoff from showing
up and voting is 12, while it’s 1 from not showing up and thus abstaining. Since
1
if m ! 6, 7, 8, 9, 10 and you abstained
there are strategies
other
whereby
abstention is the unique optimal
1
iffor
mthe
! 6,
7, 8,players
9, 10 and
you abstained
1
if m ! cannot
6, 7, 8, be
9, 10
and you
voted
strategy, then abstention
weakly
dominated.
21
µµ 2 if m ! 6, 7, 8, 9, 10 and you voted
0 if m ! 0, 1, 2, 3, 4, 5 and you abstained
0 if m ! 0, 1, 2, 3, 4, 5 and you abstained
#121 if m ! 0, 1, 2, 3, 4, 5 and you voted
#2 if m ! 0, 1, 2, 3, 4, 5 and you voted
Prove
Prove that
that abstaining
abstaining is
is not
not aa weakly
weakly dominated
dominated strategy.
strategy.
ANSWER:
members
ANSWER: IfIf the
the other
other
members vote
vote so
so that
that m
m"
"6,
6, then
then the
the payoff
payoff from
from showing
showing
1
up
and
voting
is
,
while
it’s
1
from
not
showing
up
and
thus
abstaining.
1
up and voting is 22, while it’s 1 from not showing up and thus abstaining. Since
Since
there
thereare
arestrategies
strategiesfor
forthe
theother
otherplayers
playerswhereby
wherebyabstention
abstentionis
isthe
theunique
uniqueoptimal
optimal
3-11
3-11
3-11