I'm going to be the GM for the first time. One thing I want to get cleared up is concerning Monster hp and Hit Dice.
I understand the function of Hit Dice, but I need to know:
On the stat block for oh say, Goblin, it says hp: 6 (1d10 + 1). Does that mean that when I roll a d10 to determine hp, and I roll a 10, will the Goblins hp be (6+(10 + 1)) = 17, or will it be 10 + 1 = 11?
Or say I'm running the Tarrasque. Its stat block says hp: 525 (30d10 + 360). Say I roll 5's on every single one of those d10's (for the convenience of this thread). Would its hp end up being 525 + (150 + 360) or would it end up being just 150 + 360?
The 6 is the number of hit points it has. It has 6 hit points because monsters are assumed to have roll the average of their hd+con
The average of a d10 is actually 5.5, but the game rounds it down to a 5.
The 1 is coming from the constitution bonus
5+1=6
In short don't have to roll for the hit points. It is done for you. When you see (1d10+1) that is just the book showing you how it got the numbers it did.
The goblin's 6 hp is the average roll for the dice, rounded down, plus the goblin's con modifier. This is standard for all NPC's and Monsters.
Average for 1d10 = 5.5
For the Tarrasque:
(Average hit die roll * number of hit die) + (Con modifier * number of hit die) = Total HP
(1d10 * 30) + ([Constitution Score of 34] +12 Con Modifier * 30) = Total HP
(5.5 * 30 = 165) + (12 * 30 = 360) = 525 hp
And now everything makes sense! Thank you wraithstrike for showing me the light! Edit: And you too, Young Tully!
So having a group of five level 1 players fight 4 goblins with 11 hp apiece would be a disaster, correct?
It would not be a good idea. 1st level characters are fragile. I tend to be nice to players until somewhere between levels 3 and 5.
Not necessarily, though it would be a much more difficult encounter. Increasing a creature's HP is one of the quickest and easiest ways to increase the difficulty of a fight.
Since we're on the topic of nooby questions. One more thing I don't quite get (and won't have to worry about for a little while anyway). The way the book phrases it had me confused.
Will two of the same (or different) CR 2 monsters make a CR 4 encounter or a CR 3 encounter, and similarly, will 2 CR 20 monsters make for a CR 40 encounter or a CR 21?
Look at the second table for your answer. 2 CR 2 creatures = 2+2=4. 2 CR 20 creatures = 20+2=22.
Far from an exact science, though.
Look at the second table for your answer. 2 CR 2 creatures = 2+2=4. 2 CR 20 creatures = 20+2=22.
Far from an exact science, though.
That's good enough for me. Thank you Joana.