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No Equilibrium in Pure Strategy:

Each  of  the games we  have considered  so  far  has  had  at least  one  Nash equilibrium  in  pure  strategies. But there could be  games  in which  there  is not a single pure  strategy Nash equilibrium. Consider the following  example of constant sum game (the constant sum of payoffs being 100).  

930_No Equilibrium in Pure Strategy.png

We will  try  to  find  the  solution  (Nash  equilibrium) of the game  by  trial and error method. We will  take each  possible  combination  of  strategies and investigate whether that survives the condition of Nash equilibrium, as there  is no strictly dominated strategy.

Case  I:  (S1,  T1)-> If player  1 adopts the strategy S1,  player 2 will choose TI (this  is because when  player  1 chooses S1,  playing T1  fetches her 50, whereas playing T2  fetches 20 only). Thus, given the equilibrium  strategy of player  1  to be  S1,  player  2  is  doing  her best.  But if  player  2 chooses  T1, player  1 will choose S2 (why?). Thus,  the strategy vector  (S1,  T1)  is not  self-enforcing.  To be  specific, player 1  has no  incentive  to stick  to this equilibrium. Therefore, (S1,  T1) cannot serve as a Nash equilibrium.

Similarly, we can  show that none of (S1,  T1),  (S1,  T2), (S2,  T1)  can  serve as a Nash equilibrium of the game.

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