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The question asks for player B's equilibrium strategy, hence shouldn't player B choose the strategy that gives the highest payoff after player A's move? Why does player B choose 4 over 5 and 6 over 8 when he should choose the one with the highest payoff? I don't get why the answer should be B in this case...

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    $\begingroup$ I am also confused. While a,c,d are not even strategies (a strategy contains an action for every decision node), you can find a deviation for strategy b (as you said) and this is independent of whether "equilibrium" means NE or SPNE. $\endgroup$ – Bayesian May 20 at 18:18
  • $\begingroup$ Thanks for the reply. But any guesses for why b is chosen out of the 4? Was I right on the part where players will always choose the strategy with the highest payoff ? $\endgroup$ – onetwothree May 20 at 18:35
  • $\begingroup$ I suspect that the question may want you to think that A will certainly enter as all the outcomes for A entering dominate all those for A not entering: $\min(3,1) \gt \max(0,0)$. If so, option (a) is an equilibrium strategy for B, since $5 \gt 4$ $\endgroup$ – Henry May 20 at 23:12
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I think the question has a typo. Whoever wrote it was trying to make you think about the fact that a, c and d are not even properly defined strategies for player B. This is because a strategy must specify what player B would do in all possibly scenarios he might face.

However, the typo comes from the fact that b is not part of a NE. If the first payoff instead of being (3,5) it was (5,3) then b will be an equilibrium strategy for B.

Your intuition is partially right, for it to be an NE strategy it had to choose the highest payoffs on the path of play. If it was a SPNE it would have to choose the highest payoffs in all subgames, not only the ones that are reached in equilibrium.

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