Three world powers have just entered a nuclear war!

They have been waiting for this day for decades; everyone has their nuclear missiles ready to fire, and only need to choose a target.

  • The Chinese missiles hit their target 30% of the time.
  • The American missiles hit their target 60% of the time.
  • The German missiles hit their target 90% of the time.

This is similar to the Mexican Standoff, but instead, every round the three countries fire their missiles simultaneously at a chosen target. After everyone has fired their missile and found out which countries are left standing, the surviving nations begin a second round, firing at a chosen nation.

This continues until either the last World Power remains, or everyone has wiped each other out.

  1. What are the strategies for these three countries?
  2. What is the probability of each country winning if they follow their strategies?


  • Countries don't know where their neighbors fired the missiles until after the round is over.
  • Two countries can unknowingly hit and destroy the same target in the same round.
  • 2
    $\begingroup$ I missed the "one person firing at a time" thing; I've retracted my close vote (and +1!) $\endgroup$
    – WendiKidd
    Commented May 21, 2014 at 1:53
  • $\begingroup$ The brilliance and the trouble in this is that there's no simplifying "X should always target Y". In the Mexican Standoff, the best guy always wants to shoot the second best guy, to immediately eliminate the worst threat. Here, Not only is the probability not 100%, but since the attacks are simultaneous, your best choice might depend on what others are doing (e.g., if China is attacking America, there's a chance that attacking America would have been a waste). It's definitely interesting how making the attacks simultaneous makes such a drastic change. $\endgroup$ Commented Sep 12, 2014 at 20:25

3 Answers 3


We can see that as long as all the countries fire, at least two countries will be targeted. There are eight possible strategies to be played in the first round:

 Atkrs  P.O.D.
 C A A  93% 60%  0%  
 C A C  30% 96%  0%
 C G A  90% 60% 30%
 C G C   0% 96% 30%
 G A A  93%  0% 60%
 G A C  30% 90% 60%
 G G A  90%  0% 72%
 G G C   0% 90% 72%

If any one of them are destroyed, the strategies for any two remaining countries are just to fire at each other, in which case the probabilities become:

 A vs. G: A wins 6.25%, G wins 37.5%, they both get destroyed 56.25%.
 A vs. C: A wins 58.33%, C wins 16.67%, they both get destroyed 25%.
 C vs. G: G wins 67.74%, C wins 3.23%, they both get destroyed 29.03%.

These probabilities were calculated by dividing the overall probability of each event by the probability that not everybody survives. Because in that case, nothing happens, and they just keep on trying until something does happen, and the probabilities of the other events stay in the same proportion anyway.

If any two of them are destroyed, then the last one standing is the winner by default.

Since they all fire simultaneously, their strategies can be evaluated independently, since they can't depend on any other information. If America and Germany were to target each other, China wouldn't benefit at all by choosing not to fire, because just firing at Germany increases the chances that both of those countries will be destroyed, and China would emerge the sole survivor.

It seems obvious, given the symmetry of this problem, that each country would simply fire at the country that has the highest hit rate besides them, since by analyzing the two-country standoffs, eliminating that country will always maximize their chances in the resulting duel.

So in the first round, America and China both fire at Germany, and Germany fires at America. The probability table then looks as such:

America and Germany both get destroyed: 66.67%
Only America gets destroyed:            25.92%
Only Germany gets destroyed:             7.41%

In the first case, which has a 2/3 probability of happening under this naïve strategy, China is the sole survivor.

In the second case, the standoff between Germany and China has a 67.74% chance of Germany winning, which represents a 17.56% chance overall. And China still has a 3.23% chance of surviving, which is a 0.84% chance overall.

In the third case, the standoff between America and China has a 58.33% chance of America winning, which represents a 4.32% chance overall. And China has a 16.67% chance of winning, which represents a 1.23% chance overall.

So the total probabilities of surviving are as follows:

China:    68.74%
Germany:  17.56%
America:   4.32%
Nobody:    9.38%

An interesting quirk of this problem is that America and Germany are in a sort of prisoner's dilemma when it comes to deciding whether to target each other or China.

China will target Germany on the first round, no matter what. But if America and Germany decide to collude and both target China, then the probability matrix looks something like this:

China and Germany both get destroyed: 29.63%
Only China gets destroyed:            69.14%
Only Germany gets destroyed:           1.23%

And in the resulting standoff if only China gets destroyed, Germany has a 37.5% chance of winning, which is a total survival ratio of 25.92%. The resulting survival table looks like this:

Nobody:   39.20%
America:  34.67%
Germany:  25.92%
China:     0.21%

So America and Germany's survival ratios both go up if they decide to mutually target China. But since they can't contact each other to make this decision, they both "defect", try to target each other, and China ends up surviving over two-thirds of the time.

  • $\begingroup$ Wow! Great answer! But I'm thinking if it is really a prisoners dilemma... If America and Germany would collude to both attack China. Would it be better for each of them to hold their end of the bargain, or better to stab the ally in the back? - All under the assumption, if your ally survives the round and you attacked him, he will attack you next round... $\endgroup$
    – Falco
    Commented Aug 4, 2014 at 9:56
  • 1
    $\begingroup$ @Falco Revenge isn't logical. If you return to the same situation, you should replay the same. $\endgroup$
    – Florian F
    Commented Sep 26, 2014 at 22:08
  • 1
    $\begingroup$ @Falco If living through the war was the real goal, they would avoid nuclear war altogether. $\endgroup$
    – corsiKa
    Commented Nov 3, 2014 at 18:35

Joe Z gave an excellent answer in the spirit of the question, but what if one were to interpret the question more literally, like a national defense officer?

Technically, we need to take the amount of missiles of each country into account. Germany doesn't have any nuclear weapons (except like 10 or so warheads lent to them by the USA, which of course can't be fired without USA permission), so Germany is perfectly harmless to the USA. Also, China does not gain any benefits by firing at Germany, because Germany will be out of ammo after the first round anyway. So Germany fires at China, China fires at USA and USA fires at China. Considering the amount of missiles USA has, China will die pretty much 100% of the time, and USA will also go down most of the time at the hands of China, and Germany will go down depending on whether USA is feeling evil or not.


If the thinking is that America and China both fire at Germany, and Germany fires at America in the first round then both American and Germany realize that they should change their targets to better their odds. They see (without communication) that if they both change their target to China the probabilities go from (.69,.04,.18)(China,America,Germany) to (.002,.35,.26). But then Germany understands that Germany can now do better by retargeting back to America and we have (.12,.03,.58). But then America anticipates that and America can now retarget back to Germany and we're where we started (.69,.04,.18). There are other possibilities as well but there is no equilibrium.

However this can be avoided by employing a mixed strategy. For instance if America targeted China with a .75 probability and targeted Germany with a .25 probability then America would be guaranteed a .08 probability of survival.


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