Thursday, April 13, 2006
Pirate shootout
Incensed at the imbalanced distribution of doubloons, Darrrian and Edwarrrd challenge Arrrthur to a duel, winner take all. They will stand at the corners of a triangle on the beach and take turns shooting. Each turn will consist of a pirate shooting at anyone he chooses, and the last pirate standing takes the gold. They draw straws to determine shooting order. If Arrrthur always kills his target, Darrrian kills his target 80% of the time, and Edwarrrd kills his target 50% of the time, what strategy should each pirate adopt? Who has the best odds of winning?
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3 comments:
It seems Arrrthur let his success go to his head - this deal is not so good for him.
Here's my reasoning so far:
Arrrthur's best strategy is to shoot Darrrian, and then he will obviously have a 50% chance of life.
If Edwarrrd shoots and kills Arrrthur with his first shot, then he has an 80% chance of dying with Darrrian's first shot. If he shoots and kills Darrrian with his first shot, then Arrrthur will definitely kill him.
So Edwarrrd's best strategy is to shoot the ground. Then Arrrthur will shoot Darrrian (since this is Arrrthur's best strategy) and Edwarrrd will have a 50% chance of life (he kills Arrrthur on his next turn with 50% probability, and dies if he misses once).
Darrrian's best strategy is to shoot Arrrthur with his first shot. Then he has a 40% chance of being alive after one round (.8 he kills Arrrthur * .5 Edwarrrd misses him). So his overall chances of winning if he goes first are less than 40%.
Let's assume the straw drawing is fair and consider each pirate's chances of winning given the 6 possible shooting orders (ADE, AED, DAE, DEA, EAD, EDA):
Arrrthur
ADE - 1/6 * 1/2 = 1/12
AED - 1/6 * 1/2 = 1/12
DAE - 1/6 * 2/10 * 1/2 = 2/120 = 1/60
DEA - 1/6 * 2/10 * 1/2 = 2/120 = 1/60
EAD - 1/6 * 1/2 = 1/12
EDA - 1/6 * 2/10 * 1/2 = 2/120 = 1/60
So Arrrthur's overall chance of living is 1/12 + 1/12 + 1/60 + 1/60 + 1/12 + 1/60 = 18/60 = 30%
The others are messier.
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