I spy on you from afar with a binoculars so I know what gun you have, then I bring a bigger and better one. Or maybe I just snipe you from the spying spot. Either way, I show up late and more well armed, just like the robot that looks at your hand and plays rock/paper/scissors.
I then lay my weapons down in peaceful protest. All of this violence is pretty unnecessary, I just brought this thing because I work in the ghetto..
I spy on you from afar with a binoculars so I know what gun you have, then I bring a bigger and better one. Or maybe I just snipe you from the spying spot. Either way, I show up late and more well armed, just like the robot that looks at your hand and plays rock/paper/scissors.
I then lay my weapons down in peaceful protest. All of this violence is pretty unnecessary, I just brought this thing because I work in the ghetto..
Reason? There are effects that increase crit range further. The last D&D character I played critted X2 on 17-20 at one point. ;^)
No. Since crit range and crit multiplier stack multiplicatively with one another, the imperative is to spread out over both of them. As such, if you can increase crit range further but cannot increase the multiplier, the multiplier is definitely better. For example, 17-20/x2 is equivalent to 19-20/x3.
Can someone who's actually good at stats check my work?
It's pretty badly wrong. First of all, there was no indication that this was about a direct combat between two equal opponents. However, if you do want to calculate the odds in a direct fight of Sword 1 vs Sword 2, you need to have some probability theory in there, and you need a better formulation of the problem. I've put up just such a formulation in the Math Puzzles/Problems thread.
Can someone who's actually good at stats check my work?
It's pretty badly wrong. First of all, there was no indication that this was about a direct combat between two equal opponents. However, if you do want to calculate the odds in a direct fight of Sword 1 vs Sword 2, you need to have some probability theory in there, and you need a better formulation of the problem. I've put up just such a formulation in the Math Puzzles/Problems thread.
Given that the only information we are giving when making the choice is that there are those two swords and their chance to hit and damage, I don't think you can assume anything else. Though I will cop to making the WILD assumption that it will be a contest between two equal parties.
I'll be interested to see how the math puzzle turns out.
I'm still going to say that Sword 2 is the better choice.
since a tie is a lose condition, I think that probably breaks sword 1 since you can only win if the opponent takes sword 2.
if swords are taken randomly then I think it might even out, but since you must Choose and sword 1 appears broken then Sword 2 becomes the superior choice.
Comments
As such, if you can increase crit range further but cannot increase the multiplier, the multiplier is definitely better. For example, 17-20/x2 is equivalent to 19-20/x3.
We'll assume true randomness on sword 2's swing so even of it hits 9 times in a row it still has a 50% chance to hit a 10th time.
We'll assume each combatant has 500hp since 10 strikes is a nice even number.
there can't be a draw since 0hp means death. So its either win/lose, lose/win or lose/lose
Sword 2 lucky = 6/10 hits Sword 2 unlucky means 4/10 hits Sword 2 even means 5/10 hits
Sword 1 Sword 2 lucky Sword 2 Unlucky Sword 2 even
Sword 1 L/L W/L L/W L/L
Sword 2 lucky L/W L/L L/W L/W
Sword 2 Unlucky W/L W/L L/L W/L
Sword 2 even L/L W/L L/W L/L
Sword 1: somewhere around 1/4
Sword 2: somewhere around1/3
So the only choice is sword 2.
Can someone who's actually good at stats check my work?
I'll be interested to see how the math puzzle turns out.
I'm still going to say that Sword 2 is the better choice.
if swords are taken randomly then I think it might even out, but since you must Choose and sword 1 appears broken then Sword 2 becomes the superior choice.