Rolling the results at the same time as the check

Question

I was in a roll playing game last night. In combat we throw 3 six sided dice when we attack and roll some amount of dice if we hit our target by rolling high enough on the attack (for example the dice sum to at least 10, or 13, or something like that).

The number the dice sum to is called the target number. So if the attack succeeds you roll another batch of dice (simplifying assumption that they be another 3 six sided dice), the sum of which is the damage. If the first row is too low then the attack misses and does 0 damage.

The person running the game suggested that we roll the attack and damage together, basically checking the sum of the dice against the target number and then dealing it as damage. I told him my intuition is that this would change the amount of damage dealt in the game, and that it would increase on average, over all target numbers, with the hardest to hit target numbers getting the biggest damage boost (because the damage cannot be less than the target number).

My question is, is there a formula (it can be an estimate) of the percent of damage increase using this rule? It would be a function of the target number.

(Fyi, the system we are using is the Dragon Age tabletop RPG)

Answer 1

By eliminating lower-valued damage amounts, the expected (i.e., average) damage is indeed increased. You should be able to work out the increase yourself by computing weighted averages. There’s a nice table here that gives the number of ways of rolling each value for various numbers of d6 that you can use for this.

The average amount of damage per successful attack would normally be: $$ (1\cdot3+3\cdot4+6\cdot5+\cdots+3\cdot17+1\cdot18)/(1+3+6+\cdots+3+1) = 10.5 $$ If you only take rolls of 5 or greater, say, this becomes $$ (6\cdot5+10\cdot6+\cdots+3\cdot17+1\cdot18)/(6+10+\cdots+3+1) \approx 10.63, $$ which isn’t a very large increase. On the other hand, for a target number of 15, the average becomes 15.75, an over 50% increase. It’s probably easiest to do this in a spreadsheet.

This doesn’t give you the expected damage per attack, however, since not all attacks are successful. You have to multiply that average damage by the probability of the attack landing. For that same target number of 15, the average amount of damage per attack is about 0.97, while the suggested method ups that to 1.46 or so. It’s still the same large increase in percentage terms, but less than half a point in absolute terms.

It’s hard to say how much of an effect this will actually have on the game. That depends on other factors such as the distribution of target numbers for the attacks that are made, how enemy hit points compare to target number, and so on. What you can say for sure is that distribution of damage will change drastically at higher target numbers: higher-difficulty attacks will either do no damage at all, or a lot of damage all at once.

If you’re looking to speed up the game, why not simply roll two differently-colored sets of dice at once?

Answer 2

Let $p$ be the probability of scoring a hit.

Let $\mathbb{E}[X]$ be the expected value of the sum of the dice.

The expected total damage per attack would be $p\cdot \mathbb{E}[X]$

Compare this to the simplified rules that the dice are not rerolled:

The expected total damage per attack would be $p\cdot \mathbb{E}[X\mid (X\geq\text{target})]>p\cdot \mathbb{E}[X]$, where $\mathbb{E}[X\mid (X\geq\text{target})]$ represents the expected value of the sum of the dice given that the total is greater than or equal to the target.

Without specific numbers, it is difficult to say exactly, but for two fair d6, target value of 8 or greater to hit, and two fair d6 for damage, under the original rules the expected damage per turn would be $\frac{15}{36}\cdot 7\approx 2.91$, whereas under the simplified rules, the expected damage per turn would be $\frac{15}{36}\cdot \frac{8\cdot 5+9\cdot 4+10\cdot 3+11\cdot 2+12\cdot 1}{15} = \frac{35}{9}\approx 3.89$

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Calculating $p$ for a specific number of dice: Rolling $n$ fair six-sided dice, consider the generating function $(x+x^2+x^3+x^4+x^5+x^6)^n(1+x+x^2+\dots)$. The coefficient of $x^{\text{target-1}}$ will be the number of ways in which you can achieve a roll of strictly less than the target value. Dividing by $6^n$ gives you the probability of rolling strictly less than the target value. Subtracting this value from one gives you the probability of rolling equal to or greater than the target value.

Calculating $\mathbb{E}[X]$ is simply $n\cdot 3.5$ for $n$ fair d6.

Calculating $\mathbb{E}[X\mid (X\geq \text{target})]$ becomes trickier. Consider the generating function $(x+x^2+x^3+x^4+x^5+x^6)^n$. For each value of $k\geq \text{target}$, calculate $k$ times the coefficient of $x^k$ and add the results. Divide this result by the sum of the coefficients from $x^\text{target}$ to $x^{6n}$.