10
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Two Pepperbox Pistols, a Lemat 1861 revolver, a Winchester Model 1873, a Colt Thuer Derringe, two Colt 1860 Army, a Smith&Weston Model 1 1/2, a Coach Gun, and a Double Barreled Shotgun.

10 dead men walking out of the saloon,

The mayor was in a hurry and was shot once, at noon.

9 dead men, flustered and jittery.

The cowboy was the culprit so they all shot him twice silly.

8 dead men, trying to calm down.

The barkeep was scared dead by the priest's lonely single round.

7 dead men slowly becoming weary,

The priest accused the soldier which 'back-fired' rather quickly.

6 dead men now hoping no one saw them.

The soldier could take them all, 5 bullet wounds solved the problem.

5 dead men still standing in the square.

The blacksmith wanted blood so the doctor gave him fair.

4 dead men now seeing the final picture.

The barber was the slowest, 3 rounds put him out for pasture.

3 dead men all grim and determined.

The doctor twitched too early so they both shot at him.

2 dead men still and silent and ready.

The sheriff was the quickest but came up empty.

1 dead man watching you stroll round the corner,

his weapon in hand, ready to do murder.

The list above totals 10 guns. If each man had one of the 10 weapons listed above so that no weapon remains, and no one reloaded, you've got to ask yourself: Do I feel lucky?

Well, do ya, punk?

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  • $\begingroup$ Hopefully I didn't mess anything up >_< . I don't think this is particularly hard but had quite a bit of fun making it up :P Hope you enjoy! $\endgroup$ – Spacemonkey May 27 '15 at 20:35
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    $\begingroup$ I tried to look for the number of rounds each of the weapons allow. I am note sure about Pepperbox pistols. Could you say how many rounds are supposed to have those? And wouldn't you matter to confirm that the numbers I got are correct? In order: ?, 9, 15, 1, 6, 5, 1, 2. I ask because I have no idea about guns, but this seems a quite funny riddle. Thanks a lot. $\endgroup$ – Masclins May 27 '15 at 21:23
  • $\begingroup$ en.wikipedia.org/wiki/Coach_gun this puzzle takes into account that you can fire both barrels separatly (That might be an inaccuracy on my part). As for the Pepperbox pistol indeed there seems to be more variants than I thought, for the sake of this puzzle, it has 4 rounds. Rest is about right :) $\endgroup$ – Spacemonkey May 27 '15 at 22:13
  • $\begingroup$ I'm trying to make out who started with what gun, and I feel like you made a mistake with it. Given the ammo capacity of each gun that we agreed on, its not possible for the priest to have a "lonely single round" unless he started with 3 rounds. So is this an oversight with the riddle, or we have the ammo capacities wrong? $\endgroup$ – MisterEman22 May 28 '15 at 1:08
  • $\begingroup$ I'll agree it's a bit ambiguous but I just meant that he fired a single round at the barkeep. $\endgroup$ – Spacemonkey May 28 '15 at 1:27
7
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I think I…

Feel lucky!

Here’s why: There are 10 guns, with the following numbers of rounds (based on what was stated in the comments): 4, 4, 9, 15, 1, 6, 6, 5, 2, 2.
There are 5 lines where everybody still alive shot a bullet, including 1 line where everybody still alive shot twice.

The cowboy was the culprit so they all shot him twice silly.
(6 people left) The soldier could take them all, 5 bullet wounds solved the problem.
(4 people left) The barber was the slowest, 3 rounds put him out for pasture.
(3 people left) The doctor twitched too early so they both shot at him.
(2 people left) The sheriff was the quickest but came up empty.

That means the survivor must have fired at least 6 times. This leaves 4 possible guns the survivor might have (9, 15, 6, 6). However, the soldier “could take them all” in the 5th stanza, which means that after firing at least twice, he still had enough bullets to kill 5 people. That means he had at least 7 bullets to start, so he had either the 9 or the 15. That means two of the remaining possible guns had 6 bullets, and one had (9 or 15). If the survivor started with 6 bullets, then he’s OUT OF BULLETS when you walk around the corner. 2/3 is pretty good odds, but I think we can also assume that since everybody shot the soldier because he had enough bullets to kill everybody, they would have done the same if another person had survived with that many bullets. Basically, if the other (9 or 15) gunman had survived past the time the soldier was shot, the others would have shot him as well. Thus, I’m going to assume that the other (9 or 15) gunman was killed earlier, meaning the survivor started with 6 bullets and now has none left. I FEEL LUCKY!

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  • $\begingroup$ So it seems I did mess up >_< I forgot to count the last round that killed the sheriff. My goal was to make the man with no name have one last bullet. Wouldve had to make the barber killed by 2 and clue to as which two fired. But oh well. :) Good Find! The trick was supposed to be that you know at the last two (Sheriff + survivor) They both had a gun with at least 5 or 6 rounds, but since the sheriff is out, he has the 5 which means the six-shooter is left (and then you count the rounds left) but with my mishap your answer is good ; ) $\endgroup$ – Spacemonkey May 28 '15 at 1:34

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