Ye Olde Inn

Some odd mathematical questions I've always wanted opinions on:

What would you say is equal to a Bow that did One automatic Skull every time you fired it?(no roll required). Monster defends normally.

A) a Two CombatDice attack Bow? (roll)
B) a Three CombatDice attack Bow?(roll)
C) a One CombatDie attack Bow x three separate shots? (roll) Monster defends each separately.

Just trying to figure out some numbers for possible artifacts.
Thanks

According to probability, A. Two combat dice will roll an average of 1 skull.

C is not calculable since the defence would vary for each type of monster.

I was thinking two thigs about a bow:
2D attack
OR
2D attack but allow you to split all your attack dices between two targets before rolling them.

And was thinking about tweaking the Crossbow:
3D but can't attack in consecutive rounds (or attacks). But you can shot it point blank so you can attack adjacent or diagonal targets.

But for now those are just ideas, havent playtesyed anything.

Vorimir wrote:Crossbow:
3D but can't attack in consecutive rounds (or attacks). But you can shot it point blank so you can attack adjacent or diagonal targets.

Would be better to just pull out a long sword if the crossbow can't attack every round. Unless you have a penalty for switching weapons while in battle.
I personally like the idea of needing to reload a crossbow every other round do to its power.

Most of us like the splitting the bow attack, I do.

Another math question.

If (as a hero) you have 2Att & 2Def, Would you rather fight a creature with:

A# 2Att & 3Def...or

B# 2Att & 2Def but it rolls to defend as well?


And you both have the same BodyPoints.

If you picked A#, would a monster with 2Att & 4Def be to much defense to make things even?

Thanks!

According to my calculations, the damage dealt to the monsters per hero attack is:

monster A (3 black def dice): 0.666
monster B (2 white def dice): 0.556
monster C (4 black def dice): 0.579

So monster B is the toughest, but it is quite close to monster C, which makes sense because 2 white shields are the same number of successes on the dice as 4 black shields.

In practice, the hero can expect to kill any of these monsters with 2 attacks, so they are very similar enemies on the game board.

MonsterMotor wrote: 2 white shields are the same number of successes on the dice as 4 black shields.

Thank you! I needed to avoid making the enemy equal in defense to the hero but I wanted it to be close, so it will be as I thought for the monster: 2Att & 3Def .