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According to the book Thinking Strategically by Avinash K. Dixit and Barry J. Nalebuff, a government can coerce every citizen to register for the military by threatening to punish only the first person whose name ranks highest in alphabetical order and does not register. In fact, it goes on to say, "Nothing is special about the alphabet. The key point is that the order of punishment is pre-specified. A randomly chosen and announced order of birthdates, or social security numbers, does just as well." It seems the assumption here is common knowledge of everyone's rationality, and they admit that this exercise would fail to incentivize total compliance in practice for this reason. However, everyone registering for the military is the Nash Equilibrium.

In one sense, I understand why the order must be pre-specified. With a uniformly random order, the probability of being punished would be the multiplicative inverse of the number of citizens who chose not to register. However, the following reasoning also seems correct.

Suppose we remodel the game (without changing the underlying physical problem) by treating the government as a player that chooses the order of citizens from which the first who has not registered will be punished, but instead of pre-specifying the order, the government and citizens choose their strategies at the same time.

The citizens could look ahead to each possible order the government could choose. For each case, the order is known by assumption, and the citizens' best responses collectively mirror the Nash Equilibrium of the above game where the order was actually announced ahead of time; everyone should register for the military. But given that these best responses are identical in each case, they should also constitute the best strategy overall in the simultaneous game.

What is wrong with this reasoning?

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  • $\begingroup$ I suppose there might be some (e.g. pacifist) people who prefer to not register (and accept the punishment instead joining the military). As long as at least one such person is before me in the line, I'm fine not registering. $\endgroup$ Aug 21, 2022 at 23:24

3 Answers 3

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Denote the cost of the punishment by $p$, the cost of registering by $r$.

In order for the plan to work, you need to have $p > r$.

If the order is known, then the person at the top will choose to register. Since rationality is common knowledge, the second person knows this, and thus knows that were he not to register he would be punished with probability 1. And so on, and so on.

If the order is unknown, or for example random, every person can believe that the probability of punishment is miniscule, 1 divided by the number of people. It can well be that the expected cost of punishment is less than the cost of registering.

This is why criminals in Gotham commit crimes: whoever meets Batman will get beaten up for sure, but since there are many crimes, only a few of them will meet Batman and they don't know in advance who it will be, so the expected costs may be lower than the gain from their crimes.

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    $\begingroup$ Excellent DC application ;) $\endgroup$
    – 1muflon1
    Aug 19, 2022 at 11:49
  • $\begingroup$ I understand everything said here, but it still seems to break the conclusiveness of argument by cases. In a prisoner's dilemma, we stratify the game across our opponent's possible strategies, and for each, our best response is defect, therefore we can conclude that our best response to the game as a whole when our opponent's strategy is unknown is defect. $\endgroup$
    – user10478
    Aug 19, 2022 at 16:16
  • $\begingroup$ Here, we stratify the game across the government's possible strategies, and for each, every citizen's best response is register, because a game where the government acts simultaneously with citizens but their strategy is stipulated is identical to a game where the government acts before citizens and announces that strategy. However, somehow we have the absurd result that the set of players' best responses is the same for each of the government's possible strategies, yet we cannot apply the same set of best responses to the game as a whole. $\endgroup$
    – user10478
    Aug 19, 2022 at 16:16
  • $\begingroup$ If the prisoner's dilemma were not so trivial, this result might make us question whether defect is our best response to each of the opponent's possible strategies, but not our best response to the game as a whole. $\endgroup$
    – user10478
    Aug 19, 2022 at 16:20
  • $\begingroup$ @user10478 Interesting counterargument. However, the argument in case of the prisoner's dilemma is that it is the same strategy that is always the best response, it is a strictly dominant strategy. You can show with similar expected utility calculation that no matter what a prisoner believes about the other's move, the strictly dominant strategy always yields the highest expected payoff. $\endgroup$
    – Giskard
    Aug 19, 2022 at 18:04
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A game can have multiple Nash equilibria.

You've correctly observed that, even if the punishment order isn't pre-specified, everyone registering is still a Nash equilibrium: each citizen knows that, as long as everyone else registers, they will be punished if they do not. Thus, unilaterally deviating from the Nash equilibrium strategy of registering is not a rational decision for any citizen.

However, if the citizens don't know the punishment order in advance, then nobody registering is also a Nash equilibrium (at least for the citizens, assuming that the risk of being randomly chosen to be punished is less than the cost of registering; I haven't read the book, so I don't know if the government is also considered a player in this game). This is also intuitively obvious: if nobody registers, so that the risk of being the one chosen for punishment is negligible, then there's no reason for any citizen to even consider registering.

What's perhaps not so intuitively obvious is that announcing the punishment order before the game disrupts this second equilibrium: now the first citizen in the order, who knows for sure that they will be punished if they don't register, will indeed prefer to register no matter what anyone else does.

Furthermore, by announcing the full punishment order ahead of time, the government can ensure that there is no (pure or mixed) Nash equilibrium where some citizens would not register with 100% probability: if there was such an equilibrium, then the first such citizen in the punishment order would know that they'd definitely be the one getting punished if they did not register (since everyone else ahead of them in the order was already planning to register with 100% certainty), and thus they would prefer to switch strategy to also always registering.


ps. In the version of the game where the punishment order is unknown, there is (typically) also a third (mixed, unstable) Nash equilibrium where each of the $N$ citizens registers with probability $q = 1 - (P/R-1)/(N-1)$, where $R$ is the cost of registering and $P$ is the cost of getting chosen to be punished. At this equilibrium the expected payoffs for registering and not registering are equal for each citizen, so none of them can improve their expected payoff by deviating from the equilibrium. However, deviating from the equilibrium also doesn't reduce a citizen's expected payoff, so the equilibrium is only weak (as all mixed Nash equilibria are). And as soon as anyone does deviate from the equilibrium even slightly, others can improve their expected payoff by doing the same, so this equilibrium is unstable.

In effect, this unstable equilibrium marks the boundary between the domains of attraction for the two stable equilibria ("everybody registers" and "nobody registers") in the strategy space: if you believe that the average fraction of other citizens who will register is greater than $q$, then it's better for you to register as well; if it's less than $q$, you're better off not registering.

(Careful readers may note that, for this mixed equilibrium to exist, $q$ must be between zero and one, which implies that $R < P < NR$. But this was an implicit assumption of the game scenario anyway: if $P < R$, then the punishment would not be an effective deterrent even if it was administered to everyone who did not register; whereas if $P > NR$, then the punishment would be so severe, compared to the cost of registering, that it would always be better to register than to take even a one-in-$N$ risk of having to suffer the punishment.)

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  • $\begingroup$ Hello Dear Ilmari Karonen you're explained so clear. Can you please look at my question as well? I only want you to check for payoffs in the game. I am confused too much. Thank you economics.stackexchange.com/questions/52455/… $\endgroup$
    – studentp
    Aug 23, 2022 at 19:59
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What is wrong with this reasoning?

It assumes that the cost of punishment $p$ and the cost of registering $r$ are the same for everyone, therefore it is possible to find a punishment ensuring $p>r$, but this is not the case.

For some people, $r$ will be zero or even a gain: they want to register. For others, $r$ will be enormous, for example a conscientious objector will not want to go, almost no matter the cost. Thus there should be a probability distribution for $r$, which depends on current events (for example, if there is no war, $r$ is not that high, but if there is a war and registering means you will get sent to the front, put in a trench, and subjected to artillery bombardment, then the average $r$ would be pretty damn high).

Likewise the cost of punishment is not the same for everyone. If it is a fine, and the amount of revenue lost by registering and not working does not compensate the fine, then it is rational to pay the fine. If it is jail time, different people will have different perceptions of that. Even if the punishment was getting shot, you'd find people who'd prefer that over going to war.

My point being: across the population, for each individual there is a probability $Po$ they will object, refuse to go, and get punished willingly. This of course depends on the punishment.

Therefore, if you're first in line, $Po$ is of no concern to you. But if you're far back in the punishment order, knowing $p$ and $r$, it would be rational to try to estimate the average $Po$. If it is high enough, you could conclude the state can't jail or punish that many people without a revolt or economic collapse, so your turn will never come.

Even if $Po$ is impossible to estimate, it can be measured simply by watching how many people ahead of you in the line choose to get punished instead of enlisting.

So I think the optimum for the government would be to choose an order which puts those least likely to defect first. Good for public relations, as few will be punished, and it will cause the remaining people to underestimate Po.

However, this is not a Nash equilibrium: in this game, being the individual with the highest probability to defect is advantageous, as you get called last. So once this is known, everyone who doesn't want to go would loudly signal this fact.

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  • $\begingroup$ Hello Dear @bobflux your explanation is so good. Can you please look at my question as well? I only want you to check for payoffs in the game. I am confused too much. Thank you economics.stackexchange.com/questions/52455/… $\endgroup$
    – studentp
    Aug 23, 2022 at 19:57

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