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.
A C G USA CHN GER
------------------
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.