Help with damage calculations DPR

[cheops]

Hi!

I'm thinking of designing a system and have a bit of strouble figuring this one out.

I have this idea of a fighting style that's a bit chaotic. MIght be fun for some people. But I'm not certain how to calculate the DPR (Damage Per Round) for this style.

The idea is that the character strikes once, contributing 1 dice of damage if he hits. The character then has the OPTION to roll a second time, and try to hit again. If he does, he can roll another die of damage. If he misses the second roll, he negates all damage done that round, including the damage from the first roll. So it's sort of a chancey, chaotic style, with the upshot of delivering relatively large amounts of damage if both strikes hit.

Let's say the chance to hit is 60% (9 or above on a d20). And the damage-die is a d8. How do I work out the average damage per turn/round for this style?

 

[Nagol]

If it were the case that the player always had to roll twice to hit then it's simple: DPR = (probability of hit) squared * 2 * expected hit damage (ignoring critical hit effects),

But it's not. The player has the option of rolling again. If we assume the player is somewhat rational, then if the probability of a successful hit is less than 50% then the option will be discounted since the effect on damage is "double or nothing" (ignoring edge cases where it may be more valuable to get a large damage result than just get a damage result) Then the DPR becomes probability of hit * expected damage. If the player expects to roll a second time then DPR calculation is the same as one in the first paragraph.

Even if the probability of a hit is greater than 50%, the player may opt to simply take the single effect as in many cases, the one set of damage will accomplish the goal (drop the enemy to 0, break the item, whatever).

 

[Blue]

Agree with [MENTION=23935]Nagol[/MENTION], there is some game theory here for "optimal" results, but who knows if the player will do it. Because if you roll a d8 and get an 8 you would likely stop there.

Hmm, at 60% chance to hit and 4.5 average damage on a d8, each roll can be expected to add 2.7 damage - except that a miss will negate all previous damage.

So you always take the first roll - there is no previous damage to negate.

The second roll. Let's call the first roll X. The second roll has a 60% chance of +4.5 (on average), and a 40% chance of -X (negating all damage). So the second roll adds 2.7-(0.4*x) on average. If you rolled a 7 or 8 the first time, it's a better call to stop then to try again at 60% chance to hit - you'll lose more then you gain. If you rolled a 6 or less it's better to roll again. But this doesn't really give you chaos and randomness - the roll of one dice doing well enough to stop or two dice when the first dice was worth risking aren't really all that far apart.

 

[Nagol]

Agree with @Nagol, there is some game theory here for "optimal" results, but who knows if the player will do it. Because if you roll a d8 and get an 8 you would likely stop there.

Hmm, at 60% chance to hit and 4.5 average damage on a d8, each roll can be expected to add 2.7 damage - except that a miss will negate all previous damage.

So you always take the first roll - there is no previous damage to negate.

The second roll. Let's call the first roll X. The second roll has a 60% chance of +4.5 (on average), and a 40% chance of -X (negating all damage). So the second roll adds 2.7-(0.4*x) on average. If you rolled a 7 or 8 the first time, it's a better call to stop then to try again at 60% chance to hit - you'll lose more then you gain. If you rolled a 6 or less it's better to roll again. But this doesn't really give you chaos and randomness - the roll of one dice doing well enough to stop or two dice when the first dice was worth risking aren't really all that far apart.

I didn't think of rolling the first damage die before choosing to roll again. If it works that way then the player has enough information to make an informed choice and to reasonably calculate the value of rolling again.

As you point out, the player would be foolish to roll again on a 7+. Before rolling any dice, the player is facing an expected damage of 3.285 compared to the normal expected damage of 2.7 if we assume the player will always attempt to maximise damage. The chance of doing no damage rises from 40% to 58%. The damage range goes to 2 - 14 on a lumpy bell curve with an expected result of 7.875.

 

[cheops]

Thanks guys! So, I'm trying to balance styles against eachother, and I'm a bit at loss. How would you proceed, I mean compared to say a style that always rolls 2 damage dies, and another that only do the normal thing 1 roll to thit and 1 damage die?

1) Style 1. Roll to hit + pick best one of 2 damage dies.
2) Style 2. Roll to hit + roll 1 damage die (normal)
3) Style 3. The one we're discussing above.

Would rolling all damage dies AFTER all the resolutions are done make it better you think? Or is there another way to think about how to balance these styles against eachother?

 

[Blue]

Rolling separate sets of dice isn't the fastest for play.

Maybe allow the player to pick their fighting style and they can roll one attack for one or three dice of damage, but if any of the dice match they do no damage and instead take as damage the amount of the matching dice.

Using the d8s you suggested, no matches occurs about 65% of the time. Any doubles is roughly 1/3 of the time, triples happen 1 in 64.