Question

I am playing a game of Dungeon and Dragons with my family. My son’s character must roll a critical hit, (19 or 20) on a 20 sided die. Without a critical hit, he cannot do enough damage to the troll. The troll will regenerate and kill me on the next turn. In addition to getting a high roll for hit chance, my son’s character must also get a high roll for damage. His weapon is a great sword and does 1D12 +3 damage (a 12 sided die +3). On a critical hit, his sword does double damage (2D12 +3 damage). My son’s character must deal a total of 16 damage to kill the troll. What is the probability that my son will roll a critical hit and do enough damage to kill the troll before the troll kills me?

Answer 1

The boy can only win if he rolls a critical hit.

If he doesn't roll a critical hit i.e. 19 or 20 then he could only damage a maximum of 12+3 = 15 (1D12 + 3) and 12 being the greatest side.

First let's take out the probability of getting a critical hit which we can get in only 2 cases i.e.

1. get a 19
2. get a 20

Favorable outcomes =2

sample space(total cases) =20

• P(rolling a critical hit)=2/20= 0.1

Next suppose a criticsl hit is rolled, a damage of 16 points has to be done, formula for damage now being 2D12+3

Damage would be more than 16 only when number comes out to be greater than or equal to 7 in the next roll (12 sided die).

For all numbers below seven the sum of 2D12+ 3 would be less than 16

> example if we get 6 the damage =2*6+3 = 15

> for 5 damage = 2*5+3= 13

Here favourable cases = 6 (numbers 7 to 12)

Total cases(12 sided die) =12

• P( doing a damage to kill) = 6/12 = 0.5

P(rolling a critical hit and doing a damage to kill) = P(rolling a critical hit) * P(doing damage to kill)

= 0.1 * 0.5

answer = 0.05