Countries Invading

[KidPokerX]

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. 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.

http://img193.imageshack.us/img193/9862/puzzle015xp.jpg (http://imageshack.us)

If x launches an attack by sea and y prepares a defense by sea the probability of a successful invasion is 80%.

If x launches an attack by sea and y prepares a defense by land the probability of a successful invasion is 100%.

If x launches an attack by land and y prepares a defense by land the probability of a successful invasion is 60%.

If x launches an attack by land and y prepares a defense by sea the probability of a successful invasion is 100%.

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. What is the final probability of a successful invasion assuming both utilize an optimal strategy?

[I will post the results after the answers are in]

[BluffTHIS!]

Game theory 101. First week. Dominant strategy available. Thus no need to randomize via min-max. Yawn.

[ QUOTE ]
I'm smarter than you.

[/ QUOTE ]
Isn't that what you really wanted to say?

[BluffTHIS!]

No it wasn't. Just that this isn't an interesting problem.

[udontknowmickey]

I never took formal game theory, but I did take micro econ (which the professor tried to teach us game theory) so here goes: Thoughts in white
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assigning probability to attack by land for X as q, and probability for defense of land for Y as p we have the EV of our attack set as

0.6*p*q+1*p*(1-q)+1*(1-p)*q+0.8*(1-p)*(1-q)

which comes out to 0.6*p*q+p-p*q-p*q+q+0.8-0.8*p-0.8*q+0.8*p*q

Which comes out to: -0.6*p*q+0.2*p+0.2*q+0.8

Now I'm treading in unknown waters. I was going to take a first derivative with respect to p but that doesn't seem to get me anywhere to solving for p in terms of q (rather it seems to give me q to minimize p!) So we take a guess and rearrange it as such:

0.2*p-0.6*q*p+0.2*q+0.8 = 0.2*p+q*(-0.6*p+0.2)+0.8

So thus I feel like I want to render Y's choice irrelevant which means setting -0.6*p+0.2 =0 or p=1/3 resulting in EV = 0.8+.2*1/3=0.866666666

To do the same for figuring out q:
0.2*q-0.6*q*p+0.2*p+0.8=
0.2*q+p(-0.6*q+0.2)+0.8

Setting -0.6*q+0.2 = 0 we get 1/3 for q as well

resulting in EV strategy 0.8666666

Optimal strategy for both is 1/3 land 2/3 sea, resulting in a victory probability of 0.8666666666 (or a failure probability of .1333333333)
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PS, cool graphic

[sam h]

[ QUOTE ]
Dominant strategy available.

[/ QUOTE ]

There is no dominant strategy that I can see for either player.

[ QUOTE ]
Game theory 101. First week. Dominant strategy available. Thus no need to randomize via min-max. Yawn.

[/ QUOTE ]

Just a suggestion: Try not to make incorrect observations while you are implying something is so easy that it's boring

[Darryl_P]

udontknow's solution looks right to me.

[TorpedoBreath]

"Both countries must choose either an all land or all sea strategy, they may not divide their forces."

Did udontknow's answer not violate this directive?

[udontknowmickey]

The strategy isn't to send 1/3 across land and 2/3 across sea, but it is to roll a three sided die, and if it comes up 1, send it across land, and any other number, across sea.

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

[TorpedoBreath]

I understand now.
Thanks for explaining.

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

[/ QUOTE ]

http://www.advancinghordes.com/images/gamescience/gms_43t3k.jpg

[lastchance]

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.)

[Jim T]

Attack by air.

[KidPokerX]

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.