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  #11  
Old 09-13-2005, 03:39 PM
TorpedoBreath TorpedoBreath is offline
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Default Re: Countries Invading

I understand now.
Thanks for explaining.
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  #12  
Old 09-13-2005, 07:21 PM
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Default Re: Countries Invading

[ QUOTE ]
(Yes, I know there's no such thing as a three sided die)

[/ QUOTE ]

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  #13  
Old 09-13-2005, 07:32 PM
lastchance lastchance is offline
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Default Re: Countries Invading

You are a helluva lot better at math than I am. I have no idea what you just did to get the correct answer. (I took formal game theory.)
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  #14  
Old 09-14-2005, 02:13 AM
Jim T Jim T is offline
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Default Re: Countries Invading

Attack by air.
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  #15  
Old 09-14-2005, 04:42 AM
KidPokerX KidPokerX is offline
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Location: San Luis Obispo, California
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Default Answer

COUNTRIES INVADING

Solution: 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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