Country x is planning to attack country y, and country y is anticipating the attack.
Country x can either attack by land or by sea and country y can either prepare for a land defense or a sea defense.
Both countries must choose either an all land or all sea strategy, they may not divide their forces.
The following are the probabilities of a successful invasion according to both strategies used:

x attacks by y defends by prob. of success

sea sea 80%
sea land 100%
land land 60%
land sea 100%

1) What should the strategy of country x be, assuming the goal is to maximize the probability of a successful invasion? Assume the goal of country y to be to minimize the probability of a successful invasion.
2)What is the final probability of a successful invasion assuming both utilize an optimal strategy?
If country x has probability p1 of attacking by sea and of (1p1) of attacking by land, while country y has probability p2 of defending the sea and (1p2) of defending land, then the overall probability of a successful attack is given by:
p = .8 p1 p2 + p1 (1  p2) + p2 (1  p1) + .6 (1  p1)(1  p2)
= .6 p1 p2 + .4 p1 + .4 p2 + .6
We see that if country y makes its strategy to make its defense strategy of defending sea, p2 = 2/3, then no matter what country x does, the probability of successful invasion will be:
p = .4 p1 + .4 p1 + 8/30 + .6 = 13/15
But is this the best that it can do?
The equation for p is symmetric with regard to p1 and p2, so country x has it equally in its power to make the probability of success to equal 13/15, so country y can't do any better on its part, and likewise country x can't do any better on its part to guarantee a better positive outcome for its side.
So the final probability of a successful attack is 13/15.

Posted by Charlie
on 20090207 13:20:55 