Jason's got a real noodle scratcher

February 5, 2000 9:05 PM

Jason's got a real noodle scratcher at his site. I'd say the first gunman and second gunman should take their first shots at the third gunman, but they're likely to miss, then the third gunman would kill number 2, the first one gets another shot, but it's again only 1/3 likely to hit #3. #3 then shoots #1 and it's over. I bet the correct answer is for #1 to shoot at himself or something, but that makes no sense either. What do you think?
posted by mathowie (8 comments total)

That seems like the right answer, if for no other reason than it would be funny. But if 1 shoots himself, he has a 66 percent chance of surviving. 2 then shoots 3 (because 1 has done him no wrong), and 1's CoS is still .66 and 3's CoS is .5. Then 3 fires back at 2 and 2 is dead. Then 1 fires at 3, who has a .66 CoS. I'm not sure how all that math adds up, but I think it gives 1 the best overall CoS.
posted by luke at 9:50 PM on February 5, 2000

Mark from RiotHero mentioned this possible solution as well. According to Jason's riddle the shootout continues only "until there is only one man standing". So when gunman number one shoots himself point blank in the foot, he falls down. At this point, gunman two has to shoot gunman thee, or else. If #2 hits #3, it's over. Otherwise, #3 will hit #2. And the first gunman lives.
posted by sixfoot6 at 10:17 PM on February 5, 2000

yeah, but does the 33% accuracy mean only when shooting at other objects? If a point-blank shot is definitely going to hit his foot, yeah, I think that's the correct solution.
posted by mathowie at 10:34 PM on February 5, 2000

". . .if someone gets shot, they're dead." Number one can't try to shoot himself in the foot and fall over.

I also think #1 should try to shoot himself. If he shoots himself he has a 66% CoS, if he shoots #2 we can assume #2 would shoot back with a 50% chance of killing #1, but #1 only has a 33% chance of killing #2. Then, if #1 does kill #2, #3 will kill #1 for sure.

On the other hand, if #1 does try killing #3, he has a 33% chance of hitting #1. But then #2 will shoot at #1 with a 50% chance of killing #1.

Now, if #1 tries shooting himself, there's no way #2 will go for #1 (if #1 survives his own shot). #2 will be doomed if he shoots at #1, so he'll obviously go for #3. I hope so, anyway. :D

Also, the answer could be that he doesn't shoot anybody, because then his chance of living increases a bit more. I don't know if that's a valid answer, though. . . .

Sorry, I just felt like rambling on about that. I thought it was amusing. . . .
posted by gleemax at 10:43 PM on February 5, 2000

Yup, I'm pretty sure about this:

"What should Gunman #1 do with his first shot so that he has his best chance of escaping with his life?"

He shouldn't use it.

(Kick me if I'm wrong!)
posted by gleemax at 10:46 PM on February 5, 2000

the key requirements:
"situation: if someone gets shot, they're dead."
"This continues until only one man is standing."
"What should Gunman #1 do with his first shot so that he has his best chance of escaping with his life?"

so we know that getting shot = death, the shooting continues until only 1 gunman is standing, and that we must define what the 1 gunman does with his first shot.

one of the major assumptions that everyone has made thus far is that if you shoot at someone that they'll shoot back at you; that's not a given, in the problem.

if this is a straight mathematical riddle, the correct answer would be to shoot at #3, as that's the way the math works out (semi-long equation that i'm not going to bore everyone with).

if this is a tricky little riddle (which i'm guessing it is), he should take a potshot at whomever he wishes, then lay down (so he's no longer standing).

if this were a clint eastwood movie, the answer would be shoot at #3, because you always take out the most dangerous gunman in the room, first.

and if it were a keanu reeves movie, the correct answer would be to shoot the hostage, of course.


posted by mmanning at 2:58 PM on February 6, 2000

I do believe I've been kicked. . . .
posted by gleemax at 4:19 PM on February 6, 2000

"What should Gunman #1 do with his first shot so that he has his best chance of escaping with his life?" He should line up the other two, and shoot them both at once.
posted by endquote at 1:33 PM on February 7, 2000