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I started reading up on some Game Theory and started analysing the Battle of the Sexes game, using wikipedia as source (link). In the explanation they link to an article (link, reading section 4.5) that says that letting one person have the option of 'Burning money' and lowering their payout across the board can force rational players to go to the favored stategy of one of the players.

To me, the reasoning seems incorrect. To my understanding the BoS game is not sequential game. If it would have been sequential, a rational player would always choose their favorite choise causing the second rational player then always chosing that option as well.

Now if one player has the option of burning money, it is state that the second player observes the fact that player one plays 'Burn' or 'Do not burn' and then both players make their choise (to my understanding: simultaneously). If player oen plays 'Burn', the matrix has less payout for player one while player two receives the same payout. It's still the same Battle of the Sexes problem, but now player one just has less payout but still has the same optimal payout choises.

The linked article instead has a payout matrix that introduces 4 strategies for player one:

  1. 'Burn', then choose own favored option.
  2. 'Burn', then choose other player's favored option.
  3. 'Do not burn', then choose own favored option.
  4. 'Do not burn', then choose other player's favored option.

The listed strategies, according to the article, for player two: 1. Choose player one's favored option. 2. Choose own favored option. 3. Choose player one's favored option, unless player one burns. In that case, chose own favored option instead. 4. Choose own favored option, unless player one burns. In that case, chose player one's favored option instead.

I can't agree with the article that player two has strategies 3 and 4 available. Player two only has to observe the 2x2 payout matrix that exists after player one has made their choise to burn or not.

One could argue that introducing 'Burn' is similar to signaling your choise, like saying 'I'm going to play rock' in Rock-Paper-Scissors. In the Rock-Paper_scissors game, it's optimal to ignore such a signal. I would argue similarly to for player one playing or not playing the 'Burn' option. If could signal 'I'm lowering my highest possible payout to shown to make you favored choise more valuable compared to my favorable choise', but could also mean 'I want my option so badly, I'm even torching my gain from it to make it clear how much I want it'. Or perhaps some just want to world to burn, who knows. As such, I cannot see how player two 3 and 4 stategies are valid options at all.

Am I missing something here?

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It is assumed that Violetta can observe Alfredo's choice of burning money or not. The document you linked to explains everything very clearly. The conclusion is that if it is common knowledge that players will not play weakly dominated strategies (so we can find equilibria by iterated elimination of weakly dominated strategies), then only one one strategy profile is left: one where the outcome is that Alfredo does not burn money and they both choose his favoured option $g$. Note that this is not the only NE: there are NE where they play weakly dominated strategies ($(no,oo)$ for example).

The point of the example is to illustrate that iterated elimination of weakly dominated strategies may yield a unique NE which is implausible. As it says in the document in the link: "Of course, this result is not plausible. The fact that Alfredo has the capacity to do something bizarre, like burning money, should not lead rational players inexorably to choose an asymmetric equilibrium favoring Alfredo. The culprit here is the assumption of common knowledge of rationality, or the assumption that rational agents eliminate weakly dominated strategies, or both."

(Actually strictly speaking, the normal form in figure 4.3 is wrong, because a strategy of Alfredo should specify what he would choose after both burning money and not burning money, i.e. there should be three letters in each of his strategies: for example $nog$ would be the strategy where he does not burn money, and plays $o$ if he does not burn money and $g$ if he does burn money.)

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  • $\begingroup$ This I read in the article as well. But doesn't Violetta have to formulate a strategy after Alfredo choose to burn or not? Basically: why isn't the payout matrix 4x2 in the first place? $\endgroup$ – Xilconic Jul 29 '16 at 14:10
  • $\begingroup$ Yes, Violetta chooses $o$ or $g$ after observing whether or not Alfredo chose to burn money (this is made clear by the game tree). Thus her action can depend on Alfredo's choice of $n$ or $b$. Therefore each of Violetta's strategies is made up of two actions. $\endgroup$ – smcc Jul 29 '16 at 14:17
  • $\begingroup$ Okey, so I misunderstood the Game Theory definition of a strategy, being that you decide your whole 'decision tree' before the game has started. That explains the 4x4 matrix to me. What about solving the subgames first , as from what I understand it's good for eliminating noncredible threats (of which burning money is one). Both BoS subgames lead to the same equilibria and therefore the same strategies but the Burn-option is guaranteed to have the lowest payout for Alfredo. Correct? $\endgroup$ – Xilconic Aug 1 '16 at 6:49
  • $\begingroup$ There are quite a few subgame perfect Nash equilibria (which is what you get by using backward induction and requiring that players play a NE in all the subgames). $\endgroup$ – smcc Aug 1 '16 at 9:47
  • $\begingroup$ There is a SPNE where Alfredo burns money. In this SPNE the strategy profile is: Alfredo burns money, and would play $o$ if he did not burn money, and $g$ if he did burn money; Violetta plays $o$ if Alfredo did not burn money and $g$ if he did burn money. The outcome is that Alfredo burns money and then they play $g$ (go to the game), and payoffs are $(2,1)$, which is better for Alfredo than going to the opera. (Of course there are other SPNE where he does not burn money, just as in the usual BOS game there are multiple equilibria.) $\endgroup$ – smcc Aug 1 '16 at 9:47
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I spent a while staring at the article, the document, your question, and the previous answer, and I thought this might help clear up some remaining confusion...

  1. Choose player one's favored option, unless player one burns. In that case, chose own favored option instead.
  2. Choose own favored option, unless player one burns. In that case, chose player one's favored option instead.

I can't agree with the article that player two has strategies 3 and 4 available.

Player two only has to observe the 2 × 2 payout matrix that exists after player one has made their choice to burn or not.

The problem is, in a nutshell, that you need not be left with a 2 × 2 matrix (although you can be), so things may turn out differently than you anticipate in your last statement.

With iterative elimination of weakly dominated strategies (IEDWS), you can end up with different solutions, depending on the order in which you eliminate solutions. It is true that the most intuitive elimination would leave you with the first 2 × 2 matrix, the same one that you would have been left with if money-burning had not been an option. However, the key point is that money-burning can introduce additional solutions that were not already present, and that requires a rather counterintuitive elimination order.

In this case, instead of eliminating the burn branch in the first 2 iterations, they only eliminate half of it and use the other part to eliminate another part of the not-burn branch. Then they eliminate the burn branch. Of course, this means it's eliminated regardless, but its presence helped eliminate solutions that otherwise could not have been eliminated, giving us a paradoxical single solution.

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