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Lin Sherman
12-23-2002, 11:29 PM
Horatio Hornblower is accused of cheating at cards, and to preserve his honor, challenges his accuser to a duel. As the challenger, Hornblower gets to pick the weapons. His opponent is an expert at every sort of weaponry, while Hornblower is a bookish nerd, so Hornblower comes up with the following scheme to maximize his chances of survival: he decides the duel will be conducted with pistols, both of which will be loaded with powder, but only one with shot. This means his opponent, an expert marksman, will have a .5 probability of selecting the lethal pistol.

The pistols of the late 18th century, however, were wildly inaccurate, and even in an expert marksman using a highest quality pistol would have only a .8 probability of hitting his target. So Hornblower's actual probability of being killed is only .5 x .8 = .4

As it turns out, Hornblower (who throws down his pistol without taking a shot at his opponent) lives. What is the probability that Hornblower's opponent selected the pistol loaded with shot?

Lin

Bob T.
12-23-2002, 11:36 PM
Assuming he will die if hit, 5/6 chance that the opponent had the unloaded gun.

Good Luck,
Play Well,

Bob T.

Lin Sherman
12-24-2002, 03:11 AM
Yes, assume that Hornblower dies if he sustains any injury at all.

Zoe's Echo
12-24-2002, 01:14 PM
Mathmatical answer still 50% the result shouldn't change the original probability.

Non-mathmatical answer - I am not clear as to as if it was a rotation firing - marksman first than Hornblower or simultaneous.

If it was a staggered firing and Hornblower had the second shot after the marksman then the odds are 100% since if the Marksman fired first and came up without a bullet there is no doubt I would've fired back.

If they fired at the same time still 50%, Hornblower might never have fired because his chances of hitting the marksman were so small he might have assumed that his only chances to survive were to have the bullet or get missed.

Good Luck!

BruceZ
12-24-2002, 04:20 PM
Bayes' theorem problem. Hornblower lives 60% of the time, and his opponent draws the shot 50% of the time. The probability Hornblower drew the shot given that he lived is 50%/60% = 5/6, so the probability his opponent drew the shot is 1/6.

Here's another way.
P(opponent has shot GIVEN hornblower lives) = P(opponent has shot AND hornblower lives)/P(hornblower lives) = (.5)(.2)/.6 = 1/6.

akaLogic
12-25-2002, 02:56 PM
zero % given the actual conditions of the duel

MrRothstein
01-08-2003, 06:24 PM
What if you the question is rephrased:
"What is the % of time that the gun will have been loaded in this situation?"

I think its 10%
50% of the time the gun will not be loaded.
40% of the time the gun will be loaded and will kill.
That leaves 10% when the gun is loaded and will miss.

mobes
01-09-2003, 03:37 AM
50%..........the question only asks about selecting the gun with the shot, not hurting anyone.