Is it smarter to defend with one army or with two in the game Risk? I've recently played Risk with the following rules:


*

*One player can attack another player with at most 3 armies.

*The defender can defend with at most 2 armies. This is independent of the number of attacking armies.

*Each player (attacker and defender) gets to throw a die for each army they have. So the attacker throws at most 3 dice and the defender throws at most two dice.

*When an attack is launched and the dice are thrown, the question is which armies get defeated. To do so, the result of the attacker's dice $a_1, a_2, a_3$ and the defender's dice $d_1, d_2$ are ordered descending. They are compared pairwise ($a_1$ vs $d_1$; $a_2$ vs $d_2$; $a_3$ is ignored). The attacker wins if $a_i > d_i$, otherwise the defender wins. There will be a loss of exactly $k$ armies, where $k = \max(\text{number of attacking armies}, \text{number of defending armies})$. So $k \in \{1, 2\}$. Any combination is possible (the attacker loses 2 armies, the defender loses 2 armies, both lose 1 army for the case with $k=2$).


For the attacker, I think it always makes sense to attack with 3 armies if he can. However, suppose the attacker has $n$ armies and will keep attacking until the defender has lost or he has no armies to attack with any more. The interesting question is:
Should the defender always use two armies?
 A: Okay, I assume:
 (1) the defender chooses # of dice before seeing what the attacker rolls, 
 (2) the attacker and defender each have "lots" of armies, and 
 (3) the attacker always attacks with 3 armies.  
Then the defender can either defend with two or defend with one.  If the latter, we take the attacks in pairs.  The data are from my link in the comments.
Defending with two dice, the outcomes are  +2 (attacker loses 2) with probability $.29$, 0 (each loses one army) with probability $.34$, and -2 (defender loses 2) with probability $.37$.
Defending with one die twice, the outcomes are +2 with probability $.34^2=.12$, 0 with probability $.34\cdot .66+.66\cdot .34=.45$, and -2 with probability $.66^2=.44$.
Hence, it is generally better to defend with two dice.  Circumstances when it is better to defend with one are hard to construct.  One example is if you're trying to delay the game by causing lots of die rolls.  Another is if you're trying to lose on purpose (e.g. to an ally).
Some more mathy stuff on Risk can be found here.
A: According to
Lanchester's laws
(see https://en.wikipedia.org/wiki/Lanchester%27s_laws),
if
the forces fight
with firearms engaging each other directly 
with aimed shooting from a distance,
the effectiveness of a force
is proportional to
the square of its size.
Therefore
it would be better to
have the armies as large as possible,
so the number of armies should
be as small as possible
for a fixed number
of fighters.
