Game tree: game tree Note that the information sets for player 1 are represented by red dotted lines, while the information sets for player 2 are represented by blue shaded tapes.

Described in words, the game involves the following sequence of steps:

  1. Player 1 chooses either to hurt the chance of player 2 being strong-type or do nothing. If player 1 decides to hurt, then he spends \$1 in exchange for decreasing the chance of player 2 being strong-type from $2/3$ to $1/3$. Player 2 does not know whether player 1 has chosen to hurt or do nothing.
  2. The game randomly decides whether player 2 is strong-type or weak-type with appropriate probabilities. If player 1 has decided to hurt in step 1, then player 2 is strong-type with probability $1/3$. Otherwise, player 2 is strong-type with probability $2/3$. Player 1 does not know the type of player 2, while player 2 knows his own type.
  3. Player 2 chooses either to stay in the game or quit the game early. If he decides to quit, then he walks away free if he is strong-type, but he has to spend \$1 to quit if he is weak-type. If he decides to stay, then the game proceeds to step 4.
  4. Player 1 chooses either to fight or run away. If player 1 runs away, then player 2 walks away free regardless of the type. If player 1 decides to fight, then the outcomes depend on the type of player 2. If player 2 is strong-type, then player 1 loses the fight and has to pay \$3 to player 2. If player 2 is weak-type, then player 1 wins the fight and he can get a payment of \$3 from player 2.
  5. In any case, the final payoff of player 1 takes into account whether he has spent the \$1 to hurt in step 1.


This is quite a complicated game. I don't even know how to start to tackle it. How should I analyze such a game?

  • $\begingroup$ Perhaps try this: gte.csc.liv.ac.uk/index $\endgroup$
    – afreelunch
    Nov 12, 2021 at 18:07
  • $\begingroup$ @afreelunch Actually I drew the game tree using the game theory explorer, but its solver was unable to do anything, saying that there were too many nodes... $\endgroup$
    – user141240
    Nov 12, 2021 at 18:17
  • 1
    $\begingroup$ I see... maybe try downloading Gambit? gambit-project.org $\endgroup$
    – afreelunch
    Nov 12, 2021 at 19:41

1 Answer 1


There is no guarantee that a random game can be solved by an easy trick. (For an example see Kuhn poker.)

Did you study something like this in class? If yes, was there a step-by-step method the teacher followed? Perhaps the steps were not designated as steps, but if you look back on it, you may realize that is what they were? If yes, you should probably follow that logic.

Otherwise you should

  1. try to eliminate strictly dominated strategies to make the game simpler (iterate).
  2. assign probability variables ($p_1,p_2,\dots$) to non-eliminated strategies.
  3. look for mixed equilibra that have all remaining strategies in the support sets (players should be indifferent between choosing these strategies given their conditional beliefs).
  4. if you have time left and feel up to it, look for mixed equilibria with smaller support sets.

That's is pretty much it.

"Spend to hurt" is very expensive for Player 1 relative to the expected gains. So unless I made a miscalculation somewhere, the solution is actually not that complicated.

$P_2$: Strong type should always Stay, this is her weakly dominant strategy. (Not a rigorous proof for always Staying.)

$P_1$: Following "Do nothing" should not fight, because $P$(type = strong $|$ Stay) $\geq 2/3$ and that means Fighting loses 3 more often than gains 3. (Can win at most $1/3$ of the time.)

$P_1$'s action following "Spend to hurt", $P_1$'s initial action and $P_2$'s weak type action remain to be determined.

Now assume $P_2$'s weak type action is Stay. Then following "Spend to hurt" $P_1$ is better off by Fighting than by Running ($0 > - 1$). In this case "Spend to hurt" is exactly as good as "Do nothing" ($0 = 0$).
$P_2$'s weak type action Stay is optimal iff $$ \frac{P(\text{Spend to hurt}) \cdot 2/3}{P(\text{Spend to hurt}) \cdot 2/3 + P(\text{Do nothing}) \cdot 1/3} \cdot (-3) + \frac{P(\text{Do nothing}) \cdot 1/3}{P(\text{Spend to hurt}) \cdot 2/3 + P(\text{Do nothing}) \cdot 1/3} \cdot 0 $$ is larger than or equal to $-1$. This is true when $P$(Spend to hurt) $\leq 1/5$.

Are there other equilibria? Any other weak type action $P_2$ chooses decreases the payoff of $P_1$ following "Spend to hurt", making it less than 0. This makes "Do nothing" and a consequent "Run" the better choice. This however would incentivize $P_2$ to always Stay, resulting in a contradiction.

  • $\begingroup$ I have learnt some basic game theory and I do know how to solve some simple bayesian games using perfect bayesian equilibrium. The problem with this game is, the information sets are interlaced with each other, making any equilibrium hard to find. It seems impossible to simplify the game by eliminating dominated strategies or solving proper subgames. $\endgroup$
    – user141240
    Nov 10, 2021 at 18:55
  • $\begingroup$ The game was described to me by someone else (for fun) and then I drew a game tree accordingly. I didn't see anything like this in class, and certainly no step-by-step method of solving such a game was taught in class. $\endgroup$
    – user141240
    Nov 10, 2021 at 19:04
  • $\begingroup$ @user141240 "...making any equilibrium hard to find." As I wrote: "There is no guarantee that a random game can be solved by an easy trick." $\endgroup$
    – Giskard
    Nov 11, 2021 at 7:27
  • 1
    $\begingroup$ @user141240 "It seems impossible to simplify the game by eliminating dominated strategies or solving proper subgames." There are no proper subgames, but there are definitely dominated strategies. E.g., a strong player 2 should clearly not quit, if there is any chance player 1 will run. The game reminds me of "Endgame poker", but with an added first step. $\endgroup$
    – Giskard
    Nov 11, 2021 at 7:34
  • $\begingroup$ Oh, indeed. Thank you for pointing it out. $\endgroup$
    – user141240
    Nov 11, 2021 at 7:57

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