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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
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
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
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
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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