This Riddler puzzle is about game theory… War or peace?

Two countries are eyeing each other’s gold. At the beginning of the game, the “strength” of each country’s army is drawn from a continuous uniform distribution and lies somewhere between 0 (very weak) and 1 (very strong). Each country knows its own strength but not that of its opponent. The countries observe their own strength and then simultaneously announce “peace” or “war.”

If both announce “peace,” then they each stay quietly in their own territory, with their own gold, which is worth \$1 trillion (so each “wins” \$1 trillion). If at least one announces “war,” then they go to war, and the country with the stronger army wins the other’s gold. (That is, the stronger country wins \$2 trillion, and the other wins \$0.)

What is the optimal strategy of each country (declaring “peace” or “war”) given its strength?

Extra credit: What if the countries don’t announce at the same time and instead one announces first and the other second? What if the value of winning the war were \$5 trillion rather than \$2 trillion?

Here is my solution for the first part, where both countries declare their intentions simultaneously.

[Show Solution]

### Pure, mixed, and threshold strategies

This problem describes what is known as a Bayesian game. Each country must make a decision based only on the strength of its army, which is a random variable. One country’s decision rule might look something like this: “If my army has a strength less than $a$, then declare peace. Otherwise, declare war”. This is a “threshold strategy” where the threshold value $a$ determines how likely the country is to go to war. The threshold strategy is an example of a pure strategy because the country behaves the same way every time it sees the same army strength. A more general strategy would be a mixed strategy, where the country decides on a *probability* of going to war that depends on the observed army strength.

Although neither country gets to know the other country’s strategy, we can still ask the question: suppose country A knows country B’s strategy, what would country A’s *best response* be? Similarly, we can ask what country B’s best response would be if they knew country A’s strategy. Suppose country A uses strategy $f_A$ and country B uses strategy $f_B$. If the best response to strategy $f_A$ is $f_B$ and the best response to strategy $f_B$ is $f_A$, then we have what is known as a Nash equilibrium. This means that neither country has an incentive to alter its strategy. This is what it means to have an *optimal strategy*.

### Threshold strategies are optimal

Let’s look for a Nash equilibrium for this game. Define the following:

- $x$ is the strength of the first country’s army. We’ll assume a mixed strategy defined by the function $p(x)$, which is the probability that the country declares peace given that its army has strength $x$.
- $y$ is the strength of the second country’s army. We’ll assume a mixed strategy defined by the function $q(y)$, which is the probability that the country declares peace given that its army has strength $y$.
- For the payoffs, we’ll assume mutual peace results in no change, while war results in $+W$ for the winner and $-L$ for the loser.

Given the above, the first country can expect to win:

\[

J_1 = \int_0^1 \int_0^1 (1-p(x)q(y))\, \text{war}(x-y)\, dy\,dx

\]where we defined:

\[

\text{war}(t) = \begin{cases}

+W&\text{if }t \ge 0\\

-L&\text{if }t < 0
\end{cases}
\]This follows because the probability of going to war is $(1-p(x)q(y))$, i.e. 1 minus the probability that both countries declare peace. Let's now ask ourselves what happens if we fix $q$ (the second country's strategy) and try to find $p$ in order to maximize $J_1$. Expanding, we obtain:
\begin{align}
J_1 &= \int_0^1 \int_0^1 (1 - p(x) q(y))\, \text{war}(x-y)\, dy\,dx \\
&= \tfrac{W-L}{2}-\int_0^1 p(x) \int_0^1 q(y)\, \text{war}(x-y)\, dy\,dx \\
&= \tfrac{W-L}{2}-\int_0^1 p(x) \left( W\!\int_0^x q(y)\,dy-L\!\int_x^1 q(y)\,dy\right) dx
\end{align}Inspecting the quantity in brackets, we notice that it is a nondecreasing function of $x$. When $x=0$, it's equal to $-L\!\int_0^1 q(y)dy$ and when $x=1$, it's equal to $W\!\int_0^1q(y)dy$. Therefore there is some value $x=a$ where the bracketed quantity is zero. In order to maximize $J_1$, we should set:
\[
p(x) = \begin{cases}
1 & \text{if }x < a \\
0 & \text{otherwise}
\end{cases}
\]In other words, the first country should use a threshold policy *no matter what the second country’s policy is*! By symmetry, the same argument holds if we fix the first country’s policy and optimize the second country’s policy; both countries should always use threshold policies.

### The solution is war!

We have established that both countries should use threshold strategies. So let’s suppose the second country uses the threshold $b$. What threshold should the first country use? We can compute this by solving the following equation for $a$:

\[

W\!\int_0^a q(y)\,dy = L\!\int_a^1 q(y)\,dy

\]Substituting the threshold policy for $q(y)$, we deduce that $a \le b$, since we must have a positive quantity on both sides of the equality. The result is that $a W = (b-a) L$. In other words:

\[

a = \left( \tfrac{L}{W+L} \right) b

\]So if $L=W=1$, the optimal threshold for the first country is half the optimal threshold for the second country. In fact, no matter what $L$ and $W$ are, the first country’s optimal strategy is to set a threshold that is *smaller* than the threshold of the other country. The exact same argument holds if we reverse the roles of the countries and we fix the first country’s strategy instead. We therefore conclude that each country’s best response is to continually lower its threshold. The only way to achieve a (Nash) equilibrium is if both countries set their thresholds at zero. Put another way, **both countries should always declare war**.

Here is my solution for the second part, where the countries declare their intentions sequentially.

[Show Solution]

If the countries declare their intentions sequentially, the second country gets to see the first country’s declaration before making its own, so it may use this information in its decision-making process. We can solve this version of the problem similarly to how we solved the first version, by looking for a Nash equilibrium.

### Threshold strategies are still optimal

Again, we assume the first country uses a mixed strategy $p(x)$. In principle, the second country can have two different strategies depending on what the first country declares, but if the first country declares war then nothing can be done to prevent war. So let $q(y)$ be the second country’s strategy *when the first country declares peace*.

The first country’s expected profit is the same as it was before, namely:

\[

J_1 = \int_0^1 \int_0^1 (1 – p(x) q(y))\, \text{war}(x-y)\, dy\,dx

\]Therefore we can draw the same conclusion as before: the first country should use a threshold strategy. The second country’s expected profit is different, since we must now condition on the first country declaring peace. This time, the probability of war is only $(1-q(y))$, and the density function of $x$ conditioned on the first country declaring peace is given by Bayes’ theorem:

\[

\mathbb{P}(x\,|\,\text{peace}) = \frac{\mathbb{P}(\text{peace}\,|\,x) \mathbb{P}(x) }{\int_0^1 \mathbb{P}(\text{peace}\,|\,x) \mathbb{P}(x) dx}

= \frac{p(x)}{\int_0^1 p(x) dx}

\]The expected profit for the second country is therefore:

\[

J_2 = \frac{\int_0^1 \int_0^1 (1 – q(y))\, \text{war}(y-x) p(x) \, dx\,dy}{\int_0^1 p(x)dx}

\]It’s not difficult to see that $J_2$ has the same optimal $q$ as it did before. Once again, a threshold strategy is optimal, and the threshold is the same as before.

### War is always the answer?

Since both countries use the same threshold strategies as in the original problem, the Nash equilibrium is the same; the first country should always declare war. Therefore, it doesn’t matter what the second country does; war is inevitable. This result is somewhat paradoxical because if the first country should always declare war, then what happens if it declares peace? The second country will be thoroughly confused!