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I'm trying to solve this pure-strategy Nash equilibria of this game below:

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I highlighted the best pay off for player 1 and 2. But I don't get it when it comes to player 3. The correct answer is (A)

I tried to solve it as a gaming tree. But still I don't get it.

How shall I think?

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Three-player games are notoriously tricky to analyze. The ideal way to display them would be a three-dimensional array of cells, each containing three payoffs. But this is difficult to write down on two-dimensional paper. So typically an $n\times m\times l$-game is displayed as $l$ different $n\times m$-matrices. In your example you might think about it in this way:

  • Player 1 chooses the row (upper row, $A$, or lower row, $B$)

  • Player 2 chooses the column (left column, $A$, or right column, $B$)

  • Player 3 chooses the matrix (upper matrix, $A$, or lower matrix, $B$)

Therefore, to highlight the best payoff of player 3, for each of the 4 choices of players 1 and 2 (for each of the 4 cells in the matrices) you have to compare the player-3-payoffs between the upper and the lower matrix. So in the $(A,A)$-cell you highlight the $70$ in the upper matrix, since it is greater than the $60$ in the lower matrix. For the $(A,B)$-cell you highlight $23$, which is greater than $0$, and so on. In the end, the $(A,A)$-cell of the upper matrix (corresponding to the choice of $A$ by player 3) will have all three payoffs highlighted, and the same will be true for the $(B,B)$-cell of the upper matrix (again corresponding to the choice of $A$ by player 3). So the pure NE are $(A,A,A)$ and $(B,B,A)$.

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  • $\begingroup$ Thanks @VARulle, that was really good explanation . And it is written as (player1, player2, player3)? $\endgroup$ – soetirl13 Mar 4 at 12:57
  • $\begingroup$ @soetirl13: Yes, that's the convention. $\endgroup$ – VARulle Mar 4 at 13:51

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