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X should attack by sea with probability 2/3 and by land with probability
1/3. The probability of a successful invasion is 13/15.
Both X and Y should choose their strategy randomly. Let x be the
probability that country X attacks by sea. Let y be the probability that
country Y defends by sea.
The probability of a successful invasion is :
f(x,y)=.8xy + x(1-y) + (1-x)y + .6(1-x)(1-y) = .8xy + x - xy + y - xy +
..6 - .6x - .6y + .6xy = -.6xy + .4x + .4y + .6 .
Y is obviously going to try to minimize the probability of a successful
attack. Taking the derivative of f(x,y) with respect to y yields:
-.6x + .4 = 0.
x=2/3.
Thus X should attack by sea with probability 2/3 and by land with
probability 1/3. Y should also defend by sea with probability 2/3 and by
land with probability 1/3. The probability of a successful invasion is
-.6*(2/3)*(2/3) + .4*(2/3) + .4*(2/3) + .6 = 13/15.
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| | Posted 9/14/2006 1:46 PM - 1 view - 0 comments
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