Game Development Reference

In-Depth Information

⎨

s
1
(100
− s
1
)if
s
1
<s
2

s
1
(100

−

s
1
)

p
1
(
s
1
,s
2
):=

if
s
1
=
s
2

⎩

2

0

if
s
1
>s
2

⎧

⎨

⎩

s
2
(100

−

s
2
)if
s
2
<s
1

s
2
(100

−

s
2
)

p
2
(
s
1
,s
2
):=

if
s
1
=
s
2

2

0

if
s
2
>s
1
.

Consider now each player's best responses to the strategies of the opponent.

Since
s
1
= 50 maximises the value of
s
1
(100

s
1
) in the interval (0
,
100],

the strategy 50 is the unique best response of the first player to any strategy

s
2
>
50 of the second player. Further, no strategy is a best response to

a strategy
s
2
≤

−

50. By symmetry the same holds for the strategies of the

second player.

So the elimination of never best responses leaves each player with a single

strategy, 50. In the second round we need to consider the best responses

to these two strategies in the
original
game
G
.In
G
the strategy
s
1
=49

is a better response to
s
2
= 50 than
s
1
= 50 and symmetrically for the

second player. So in the second round of elimination both strategies 50 are

eliminated and we reach the restriction with the empty strategy sets. By

Theorem 1.22 we conclude that the original game
G
has no Nash equilibrium.

Note that if we defined
S
i
in the definition of the operator

using

the restriction
R
instead of the original game
G
, the iteration would stop in

the above example after the first round. Such a modified definition of the

RAT

RAT

operator is actually an instance of the IENBR (iterated elimination

of never best responses) in which at each stage all never best responses are

eliminated. So for the above game
G
we can then conclude by the IENBR

Theorem 1.10(
i
) that it has at most one equilibrium, namely (50
,
50), and

then check separately that in fact it is not a Nash equilibrium.

1.5.4 A comparison between the introduced notions

We introduced so far the notions of strict dominance, weak dominance,

and a best response, and related them to the notion of a Nash equilibrium.

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