#Power Dice 10/12 add the same "pity chance" as regular Power Dice do

1 messages · Page 1 of 1 (latest)

small gull
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Summary: Failed attempts with PD10/12 give the same cumulative chance in the pity system as PD4/6/8 do.

Server: all (in this case Totemia)
Character: all (in this case Eluva)
Position: all (in this case Flaris at my homeboy Boboku boboku )

Visual proof: attached

Expected result: Failed attempts with the rare PD10/12 should increase the chance for the next attempt accordingly.

Actual result: Failed attempts with the rare PD10/12 only increase the chance with the same factor as regular PDs do.

Additional comments: I am unsure whether this is intended or if it's an actual bug, but as of right now it is a waste using PD10/12 on the first bunch of attempts as the chance is just too low.

worn ice
small gull
# worn ice I think PD10 increase by 20% the chance. If you have 1% chance with PD8 => 1.2%...

Well as you can see in the screenshots when the PD10 fails it only adds the default 0.1508% chance to the next attempt in my case.
I'd have thought that a failed attempt with a PD10 would increase the chance for the next attempt by the same factor, let's say like you said it's 1.2 times

So it would be ~0.181% chance added to the next attempt when using a PD10.

That means if I have some PD10 from the battle pass and use them on a fresh item that has like 0.0x% chance to succed, I'll only add some normal failed attempts and when my PD10 are gone and I have to go back to PD8 I have absolutely no benefit from having wasted my PD10 in the beginning.

Thinking about it though I'm sure there's some technicalities that would make the way I thought it works hard to implement. They'd have to add a 2nd value "Piercing attempts PD8" and "Piercing attempts PD10" so that progress is properly saved.

I'd still love a quick answer from a developer if this is indeed intended design.

Answer to below: @worn ice
I don't think you understand what I mean. In the scenario I mentioned where I used my PD10 in the beginning and I am only left with PD8 it will be the exact same chance as if I had used PD8 only from the beginning. So using PD10 is an absolute waste on the early attempts that are very very low chance.

worn ice
rare rain
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I think this behaviour is intended. As you stated correctly, using PD10/12 really early or really late isnt worth it. I devised a formula to calculate the optimal attempt to use better powerdices:

0.75 * sqrt(1 / initialChance)

rounded up.

Aurae also incorporated this feature into his upgrade-material-cost-calculator

slow spire
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Pretty sure that the way it appears to work is that it has an integer counter tracking failed attempts, and just multiplies the chance modifier of the dice against this number to produce the current probability. Which means if you have 20 fails, switching from PD12/10 to PD6/8 will drop your success rate, and increase it going the other direction, without having made any attempts.

They need to also track the probability as a float and peg the current probability to that number rather than reconstruct it (badly).

So rather than a forumula like dice * failed count * probability modifier you get previous probability + dice modifier at current fail count.

This is such a basic, common programming pattern I'm surprised they didn't start with this.

@rare rain Then they need to not have two separate tracks of probability. Because the solution above to maximize your gains using a mix of PD8/10 and PD6/12 is also complicated and opaque.

And why should you benefit from past failed attempts using PD10s/12s? Because imagine burning 30 PD12s and then having to continue with PD6s and your success rate is back down significantly below where it was 10 seconds ago. If that's intended behavior, then they need to re-evaluate.

Storing the probability you were at and accounting for the types of dice used is fine, and should be how it is handled. Your average success rate does go up if you use a dice that has a higher probability than one that doesn't. The overall average being higher because you chose to employ those dice for some of your attempts should be the case. That's what happens when you take the average of two different probabilities.

As it is you get no value for having used PD10/12s if you don't succeed with them, because the only thing that is stored is the attempt as integer.