Question

Consider the discrete Bertrand game. According to the rules of this game each student selects a number from the set {0,1,2, 3, 4, 5, 6, 7, 8, 9, 10} and is randomly matched with another student. Whoever has the lowest number wins that amount in dollars and whoever has the high number wins zero. In the event of ties, each student receives half their number in dollars. What number would you select if you played this game? Explain your reasoning.

Answer 1

IN order to win the game one would select thel lowest number possible .Also in the Bertrand price competition the firm will selects the price to be lowest possible equal to marginal cost. Even in this game one would select lowest possible number in order to win the game .

In this game, actions are restricted to the integers 0,...c−1,c,c + 1,c + 2,... Let’s consider the possible cases:

• p1 = p2 = c. Both fiems are making zero profit. If firm i lowered its price to below c, it would make a negative profit. If it raised its price to above c, it would not get any customers and hence make zero profit. As in the original game, there is no incentive to deviate, therefore (c,c) is a Nash equilibrium.
• pi < c for either firm. The firm with the lowest price is making negative profit, and can improve to zero profit by setting pi = c. Not a Nash equilibrium.
• pi = p2 = c + 1. Both firms are making positive profit. If firm i lowered its price, it would make zero or negative profit. If it raises its price, it makes zero profit. Therefore, (c + 1,c + 1) is also a Nash equilibrium.
• pi = c,pj > c. Firm i is making zero profit, can improve by raising its price to c + 1. Not a Nash equilibrium.
• pi > pj ≥ c+1. Firm i is making zero profit, can improve by lowering price to equal pj. Not a Nash equilibrium.

As compared to the original formulation of Bertrand duopoly, there is an additional Nash equilibrium because we have eliminated some possible actions that would give an incentive to deviate.