For B.M.
According to the Monster Manual, 3rd ed. (Gary Gygax; Lake Geneva, WI:TSR Hobbies, Inc., 1978), p. 14, centaurs are armor class 5 (leaders are AC 4) and have 4 hit dice. (For the record, they also have 2 attacks per melee round, doing 1-6 damage with their hooves with one attack and a variable amount of damage with the other attack, depending on the human weapon they’re wielding.) Importantly, the size is listed as Large.
So the expected value of the number of hit points of a centaur is E{HP} = E{4*1D8} = 4*E{1D8} = 4*4.5 = 18.
According to the Player’s Handbook (Gary Gygax; Lake Geneva, WI:TSR Hobbies, Inc., 1978), p. 38, a non-magical two-handed sword does 3-18 hit points of damage against large creatures, meaning the expected value (conditional on your having actually hit) is E{3*1D6} = 3*E{1D6} = 3*3.5 = 10.5. So on average you’d need to connect with a two-handed sword about twice before the average centaur buys it.
However, you’re not guaranteed to connect every time you swing. So let’s estimate how many times you’d need to swing that sword to defeat the centaur. On page 38 of the Player’s Handbook, it says that two-handed swords have a +2 Armor Class Adjustment against both AC 4 and 5. And on page 74 of the Dungeon Master’s Guide (Gary Gygax; Lake Geneva, WI:TSR Hobbies, Inc., 1979), it says that a first-level fighter needs a 15 (16) to hit AC 5 (4). With the +2 Armor Class Adjustment of the two-handed sword against AC 5 (4), this means that he/she’d need “only” a 13 (14). So every time he/she swings that sword, the expected damage against AC 5 is ((20-13+1)/20)*10.5 = 4.2 (3.675 for AC 4). This means that, on average, a first-level fighter would need to swing 18 / 4.2 ~ 4 times to kill an average non-leader centaur (and about 5 times for an average leader centaur).
But maybe it’s not fair to have a 1st-level fighter (“Veteran”) take on a centaur. Perhaps we should instead match a 4th-level fighter (“Hero”) against the centaur. OK, going through the same computations for a 4th-level fighter, I compute that the expected damage against AC 5 every time he/she swings that two-handed sword is ((20-11+1)/20)*10.5 = 5.25 (4.725 for AC 4), meaning he/she’d need to swing about 3 times (okay, 3.43) before an average non-leader centaur bites the dust (3.81, closer to 4 times for an average centaur leader).
Now, I don’t remember centaurs being quite this tough in Rogue. In Rogue, I thought that one or two hits would do. I think the rules were different in Rogue.
If I’ve made any mistakes in my computations, please let me know. But I have used the power of MATLAB (v. 7.4.0 R2007a, Student Version) to do these calculations, so they must be correct. 🙂
I knew there was a reason I’m studying to get a Master’s degree in biostatistics. If you studied statistics, you too could estimate fairly precisely whether you could take on that centaur. (Wait a sec, that word “precisely” bothers me. Maybe I should compute 95% confidence intervals…) You’d need to take your Panasonic Toughbook with you on your dungeon campaign, and be sure you’ve got MATLAB installed.
Future project: write a program that takes player character attributes (race, class, level, weapon, etc.) and monster attributes (hit dice, armor class, number of attacks, damage per attack, etc.) as input, and gives as output estimates for the outcome of combat (whether you’d win/lose, how many turns melee would last, damage dealt, etc.), with confidence intervals as appropriate. Actually, I bet somebody has already written this program.
I know. This was another really geeky post.
The Markov Thought Chain 