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How is Accuracy and Dodge calculated?

AuthorMessage
Ensign
Dec 20, 2010
4
Hi! I know that Accuracy increases the chance of you hitting an enemy, and Dodge increases the chance of an enemy missing you, but how exactly is it calculated? I used to think that if you had the same or more accuracy than the opponent had dodge, then you would always hit. My thinking was 100% is the base accuracy, the opponent has X dodge, and you have Y accuracy. So in my book, if I had 50 accuracy and the opponent had 60 dodge, then you would have a 90% chance of hitting (100+50=150-60=90%). However, it seems like my Hypothesis was wrong, because many people on Central think otherwise, but don't know how to calculate it, either. So how exactly is Accuracy and Dodge calculated? Any help is appreciated. Thanks!
Captain
Feb 27, 2009
505
Cedric Young on Apr 27, 2013 wrote:
Hi! I know that Accuracy increases the chance of you hitting an enemy, and Dodge increases the chance of an enemy missing you, but how exactly is it calculated? I used to think that if you had the same or more accuracy than the opponent had dodge, then you would always hit. My thinking was 100% is the base accuracy, the opponent has X dodge, and you have Y accuracy. So in my book, if I had 50 accuracy and the opponent had 60 dodge, then you would have a 90% chance of hitting (100+50=150-60=90%). However, it seems like my Hypothesis was wrong, because many people on Central think otherwise, but don't know how to calculate it, either. So how exactly is Accuracy and Dodge calculated? Any help is appreciated. Thanks!
It's like Wizard101's critical and the criticals will probably be the same as the accuracy and dodge if they ever add critical and block to Pirate101.
Gunner's Mate
Jan 06, 2011
228
I thought it was that way as well. But now I can see how it can't be because if you had more accuracy than your enemy had dodge, even just one percent higher, you would always hit. For example, if the attacker had 91% accuracy and the defender had 90% dodge you would have 100 + 91 - 90 = 101% chance to hit. Or if the attacker had 11% accuracy and the defender had 10% dodge, that too would be 101% chance to hit.
However, the formula cannot be simply accuracy - dodge, because then the opposite would occur (higher dodge would always cause misses).
It may be set to be around 50%, so that if accuracy and dodge are equal the chance to hit is 50%.

So, what the formula is exactly, I don't know. I have done game design and have actually had to deal with cases like this. They are very tricky to get balanced. Hopefully one of our benevolent developers will chime in to shed some light on the subject.
Developer
I saw the thread over on Central. Thank you for coming here for a definitive answer.

This poster (Emerald141) has it right (let's see if Jack will let me post a link):

(base chance of hitting target) + (attacker's accuracy) - (target's dodge) = (actual percent chance of hitting target)

And I'll confirm that the base chance is 75%.

However, there are a few more tricky bits involved. There's always a (small) chance that you will hit, and there's always a (small) chance that you will miss, no matter how much Accuracy or Dodge each combatant has.
Ensign
Dec 20, 2010
4
KiwiChickenz on Apr 28, 2013 wrote:
I thought it was that way as well. But now I can see how it can't be because if you had more accuracy than your enemy had dodge, even just one percent higher, you would always hit. For example, if the attacker had 91% accuracy and the defender had 90% dodge you would have 100 + 91 - 90 = 101% chance to hit. Or if the attacker had 11% accuracy and the defender had 10% dodge, that too would be 101% chance to hit.
However, the formula cannot be simply accuracy - dodge, because then the opposite would occur (higher dodge would always cause misses).
It may be set to be around 50%, so that if accuracy and dodge are equal the chance to hit is 50%.

So, what the formula is exactly, I don't know. I have done game design and have actually had to deal with cases like this. They are very tricky to get balanced. Hopefully one of our benevolent developers will chime in to shed some light on the subject.
Ya, I'm hoping that somebody from KI can shed some light on this subject, specifically Ratbeard. It would help, because the formula seems too complicated for most of us pirates to figure out lol.
Bosun
Oct 15, 2012
364
bonnie have 110 Accuracy scratch have 50 Dodge, 110 - 50 = 60

Bonnie has 60% chance of landing a blow on scratch,

i think, but, that seems like awfully little bit.... perhaps i have it wrong ....*shrugs*
Gunner's Mate
Jan 06, 2011
228
Thank you Ratbeard. That is actually quite simple, it was just that base chance that we didn't know. Would I be right in guessing that witchdoctors like Old Scratch have a slightly different formula since they don't take the targets dodge into account?
Ensign
Dec 20, 2010
4
Ratbeard on Apr 29, 2013 wrote:
I saw the thread over on Central. Thank you for coming here for a definitive answer.

This poster (Emerald141) has it right (let's see if Jack will let me post a link):

(base chance of hitting target) + (attacker's accuracy) - (target's dodge) = (actual percent chance of hitting target)

And I'll confirm that the base chance is 75%.

However, there are a few more tricky bits involved. There's always a (small) chance that you will hit, and there's always a (small) chance that you will miss, no matter how much Accuracy or Dodge each combatant has.
Thanks Ratbeard! This will definitely help me out with the guide.