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I Created A Free Windows App That Helps You Enchant Gear/Failstack Efficiently- More Features Coming Soon!


81 posts in this topic

Posted (edited)

EDIT: Webapp Version:

weapons: http://deliddedtech.com/black-desert-failstack-calculator/

accessories: http://deliddedtech.com/black-desert-accessory-failstack-calculator/

 

Application is called Black Desert Tools:

https://www.microsoft.com/en-us/store/apps/black-desert-tools/9nblggh4wq2q

As of now, the app has 2 features:

Fail stack probability checker

The Fail stack probability checker tells you the probability that you will reach a given amount of failstacks without resetting to 0, based on the enchant level of gear you're using to failstack.

Fail stack optimizer

The fail stack optimizer determines the ideal amount of failstacks to use when enchanting an accessory or weapon. The amount of fail stacks recommended is based on a complex algorithm that involves the cost of the accessory you're enchanting(it's base price at +0), and the cost of obtaining failstacks via blackstones.

Ultimately, it figures out the point at which generating more failstacks provides less value in terms of increased enchant success rate than the average cost of reaching that many fail stacks.

The calculation is based on failstacking on +14 armor for low failstacks, then DUO armor for higher failstacks, and tells you when to use each.

 

Factors to consider:

  • the fail stack optimizer will never recommend more fail stacks than the cap for a given enchant(25 PRI, 35 DUO, etc)

  • If an item is considered priceless to you in that it isn't available on the market(IE Ogre Ring), I'd recommend using the maximum failstacks for that level.

  • Price of the accessory you enter should be what the accessory is worth at +0, not what its current worth is. The algorithm accounts for its true value when enchanted to the level before what you're going for.

Future Planned Additions:

  • Interactive map. Possibly have a node simulator, as well as the ability to mark the map with important points of interests you want to remember.

FAQ:

What data does the application use to determine probability?

http://i.imgur.com/G7Aw641.png

This graph is frequently referenced by the community and generally believed to be accurate, and is allegedly sourced from someone that data mined the client.

What is Black Desert Tools Programmed In?

The program is Written in C# using the UWP framework.

Why use a Windows 10 UWP format instead of EXE?

  • EXE requires a lot more work to achieve the same quality user interface. UI elements such as buttons need to be coded by hand unless I use a framework such as QT.

  • By distributing it as a UWP application on the Windows Store I can guarantee to users that it is virus free. Microsoft will not approve malware to their store. Most download sites like Sourceforge are filled with malware nowadays too, so distributing an exe file is difficult.

    If I made it as a standard C++ application, it would still be text based, which isn't very user friendly.

What if I use Windows 7 and don't want to upgrade for free to 10?

I'm planning on porting the application to Android, so if you have an Android phone, you can use that instead

I'll answer any questions in this topic as well as listen to any feedback!

Fallout_Prime likes this
Edited by Skilliard
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Posted

Awesome I'll check it out later today!

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Posted

Another user reported that they want the ability to see the chances of success when enchanting a particular item with a given amount of failstacks. Would anyone else be interested in this feature?

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Posted

Now that memory fragments are starting to become more expensive, and people are beginning to enchant stuff such as boss armor and Liverto weapons more, I am considering working on functionality to help you determine the optimal amount of failstacks for enchanting valuable weapons that require fragments to repair.

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Posted

awesome, seriously happy someone is taking some of their free time to do this.

You sir are a good human being.

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Posted

Wow, this is awesome! Make it Android soon! :) 

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Posted

Wow, this is awesome! Make it Android soon! :) 

I actually have an android version that's partially complete(only one of the features is done so far), I'm considering porting it to android as well!

Hoping to get some work done on other features this weekend! If all goes well there should be more features in 2-3 weeks.

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Posted

And For IOS if it´s not ask to much :)

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Posted (edited)

@Skilliard

I have a question regarding the numbers used.

If I put 12,500,000 MoS 200,000 blackstone the optimal enchantment for PRI is 21 according to your program.

 

We know on average it takes 124 stones to get 21 stacks using Reblath failstack method if we take 2.5% base 0.5% per failstack and 25 fs cap.

This means the cost of 21 failstack ignoring negligible armor repair is 24.8m

The average number of tries to get PRI from 21 stacks is 2.09 tries.

Therefore the expected cost is (12.5+12.5)*2.09 + 24.8 = 77.05m

 

If we say try at 18 stacks.

Stack cost: 14m

Tries 2.3

Exp cost: 71.5m

 

15 stacks:

Stack cost: 7.4m

Tries 2.53

Exp cost: 70.65m

 

5 stacks:

1.15m

3.83

Exp cost: 96.9m

 

I just sampled 3 random other stack values, and 2 of them came out cheaper than what your calculator suggested using? Am I overlooking something?

Edited by Featherine

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Posted (edited)

@Skilliard

I have a question regarding the numbers used.

If I put 12,500,000 MoS 200,000 blackstone the optimal enchantment for PRI is 21 according to your program.

 

We know on average it takes 124 stones to get 21 stacks using Reblath failstack method if we take 2.5% base 0.5% per failstack and 25 fs cap.

This means the cost of 21 failstack ignoring negligible armor repair is 24.8m

The average number of tries to get PRI from 21 stacks is 2.09 tries.

Therefore the expected cost is (12.5+12.5)*2.09 + 24.8 = 77.05m

 

If we say try at 18 stacks.

Stack cost: 14m

Tries 2.3

Exp cost: 71.5m

 

15 stacks:

Stack cost: 7.4m

Tries 2.53

Exp cost: 70.65m

 

5 stacks:

1.15m

3.83

Exp cost: 96.9m

 

I just sampled 3 random other stack values, and 2 of them came out cheaper than what your calculator suggested using? Am I overlooking something?

It takes a lot less than 124 blackstones on average to get 21 failstacks. It's much closer to 50-60. 

 

Using the reblath method at +14, it's about a 19.23% chance of reaching 21 failstacks without it succeeding and resetting to 0. On average, each attempt at generating an infinite amount of failstacks uses 11-12 blackstones before it resets. Therefore, the amount of blackstones needed on average to reach 21 failstacks is roughly 60.

Edited by Skilliard

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Posted (edited)

It takes a lot less than 124 blackstones on average to get 21 failstacks. It's much closer to 50-60. 

 

Using the reblath method at +14, it's about a 19.23% chance of reaching 21 failstacks without it succeeding and resetting to 0. On average, each attempt at generating an infinite amount of failstacks uses 11-12 blackstones before it resets. Therefore, the amount of blackstones needed on average to reach 21 failstacks is roughly 60.

That mathematical method is incorrect. The probability of failure is heavily skewed towards the higher stacks, and the probability of reaching 11 stacks is almost 60%. Cumulative probabilities cannot be divided in such way without considering conditional probabilities and combinatorics in this problem.

Taking your method would imply 35 stacks has a 1.8% chance of occurrence, and stacks 11-12 before it resets therefore 650~ stones average. However, a better method, such as Monte-Carlo simulation puts this at 1650 stones with 97.5% base 0.5% per stack up to 25 stacks.

 

A very simple dummy-proof way to illustrate the error in your methodology is simple. Suppose I want 3 stacks, the probability of reaching that is 91.3%. Given that on average you're saying it will generate 11 stacks before resetting (which is wrong) you imply that you will try 11 times at 91.3% probability, hence 12 average stones per attempt at 3 stacks. The answer is actually about 3.2.

Edited by Featherine

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Posted (edited)

That mathematical method is incorrect. The probability of failure is heavily skewed towards the higher stacks, and the probability of reaching 11 stacks is almost 60%. Cumulative probabilities cannot be divided in such way without considering conditional probabilities and combinatorics in this problem.

Taking your method would imply 35 stacks has a 1.8% chance of occurrence, and stacks 11-12 before it resets therefore 650~ stones average. However, a better method, such as Monte-Carlo simulation puts this at 1650 stones with 97.5% base 0.5% per stack up to 25 stacks.

 

A very simple dummy-proof way to illustrate the error in your methodology is simple. Suppose I want 3 stacks, the probability of reaching that is 91.3%. Given that on average you're saying it will generate 11 stacks before resetting (which is wrong) you imply that you will try 11 times at 91.3% probability, hence 12 average stones per attempt at 3 stacks. The answer is actually about 3.2.

Probability of reaching 11 stacks without resetting to 0 is ~56.8%.

35 fail stacks has a ~2.1% chance of success. Keep in mind higher stacks become increasingly rare due to the gear having a higher chance of success(all the way up to 25).

The average amount of failstacks used when attemping to build failstacks infinitely is between 11 and 12. This was tested with a sample size of 1 million via simulation. Basically, as  F(which is the number of failstacks desired) approaches infinity, a, the average number of attempts, approaches somewhere around 12. 

For smaller amounts of failstacks being attempted(such as 15), the number of blackstones being used per attempt is actually less than 11.

"A very simple dummy-proof way to illustrate the error in your methodology is simple. Suppose I want 3 stacks, the probability of reaching that is 91.3%. Given that on average you're saying it will generate 11 stacks before resetting (which is wrong) you imply that you will try 11 times at 91.3% probability, hence 12 average stones per attempt at 3 stacks. The answer is actually about 3.2."

 

That is not how my algorithm works. A, the average number of attempts, is capped at either f(number of failstacks being attempted), or 11, whichever is lesser.

 

                            // need to change to cumulative probability, not single probability(prob)
                            // After a huge sample size test, average failstacks per attempt to stack plateaus at
                            // 11.Therefore, cost per failstack attempt is less than 12 blackstones.
                            if (current_failstacks < 11)
                            {
                                bstones_per_attempt = current_failstacks;
                            }
                            else
                            {
                                bstones_per_attempt = 11;
                            }

Obviously, this isn't perfect, it's only an estimation. But it isn't underestimating the amount of blackstones like you're saying, it's actually slightly overestimating, in cases of very low amounts of failstacks (<10). There is probably a better way to do this but I couldn't figure it out.

 

Here's how the probability of n failstacks is calculated:

 for (int passes = 0; passes < current_failstacks; passes++)
                            {
                                prob = 0.025 + (0.005 * passes);
                                cProb = cProb * (1 - prob);

}

It simulates the chance of reaching a given amount of failstacks by checking the probability after each attempt. It accounts for the gains in chances after each enchant. I don't see what's wrong with it.

EDIT: You are wrong, but your concerns made me look over my code again, and made me find a separate bug that caused the amount of failstacks recommended for DUO-PEN to be much lower than is optimal, especially TET/PEN. I will be putting out an update ASAP

Edited by Skilliard

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Posted (edited)

35 fail stacks has a ~2.1% chance of success. Keep in mind higher stacks become increasingly rare due to the gear having a higher chance of success(all the way up to 25).

 Yea, I accidentally did 36 stacks.

The average amount of failstacks used when attemping to build failstacks infinitely is between 11 and 12. This was tested with a sample size of 1 million via simulation. Basically, as  F(which is the number of failstacks desired) approaches infinity, a, the average number of attempts, approaches somewhere around 12. 

You don't need a simulation to work that out. The answer is:

(sum (product (0.975-0.005n), n=0 to k) * (0.025+0.005(k+1)) * (k+1), k=0 to 24) + (sum (0.106435 * (0.85^n) * 0.15 * (n+9)), n=1 to infinite) = 11.8.

This answer however is irrelevant, because the distribution is not symmetrical.

That is not how my algorithm works. A, the average number of attempts, is capped at either f(number of failstacks being attempted), or 11, whichever is lesser.

 

                            // need to change to cumulative probability, not single probability(prob)
                            // After a huge sample size test, average failstacks per attempt to stack plateaus at
                            // 11.Therefore, cost per failstack attempt is less than 12 blackstones.
                            if (current_failstacks < 11)
                            {
                                bstones_per_attempt = current_failstacks;
                            }
                            else
                            {
                                bstones_per_attempt = 11;
                            }

This method when compared to how I did my data set (a perfect Monte Carlo simulation which legitimately tests each try) showed a huge variation in results. Your values are about 2x lower near the 20~30 stack margin and the error only gets bigger and bigger.

Here's how the probability of n failstacks is calculated:

 for (int passes = 0; passes < current_failstacks; passes++)
                            {
                                prob = 0.025 + (0.005 * passes);
                                cProb = cProb * (1 - prob);

}

Hope you capped prob at 25 stacks.

Edited by Featherine

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Posted

Shame all these calculations are based on completely made up numbers

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Posted

Shame all these calculations are based on completely made up numbers

Shame that people who have zero knowledge in basic mathematics thinks their opinions matter here.

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Posted

Shame that people who have zero knowledge in basic mathematics thinks their opinions matter here.

Agreed, but that is why people like to make tools to help people. But no amount of mathematical rigor will work well on made up numbers.

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Posted

Agreed, but that is why people like to make tools to help people. But no amount of mathematical rigor will work well on made up numbers.

You mean the numbers datamined from KR? If you don't believe it, just leave?

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Posted

Why so salty today?

My understanding is that these figures are made up or at very least inappropriately applied to different enhancement targets (weapons/armour/trinkets/green/blue/yellow)

Last I read of the data source it even stated this

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Posted

Why so salty today?

My understanding is that these figures are made up or at very least inappropriately applied to different enhancement targets (weapons/armour/trinkets/green/blue/yellow)

Last I read of the data source it even stated this

The numbers are datamined from KR. Anything else you heard about them are pure conjecture and rumors. I'm not salty, I'm pissed off that some idiot would come in here and spread more hearsay with nothing constructive to contribute to the discussion.

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Posted

Well, before you start calling people (who are better than you at mathematics) idiots:..

1) Quote your data source because many have show it has been made up

2) Fix the error is your stats thread where you fail to account for possibilities of getting greater that 1 drop.

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Posted

Well, before you start calling people (who are better than you at mathematics) idiots:..

1) Quote your data source because many have show it has been made up

2) Fix the error is your stats thread where you fail to account for possibilities of getting greater that 1 drop.

Before you even imply you're any good at maths:

1) Quote data source that "many" have shown it has been made up

2) Explain why you're talking about "drops" in a thread about fail stacks. 

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Posted

Can we get confirmation that this is safe @CM_Jouska

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Posted

Before you even imply you're any good at maths:

1) Quote data source that "many" have shown it has been made up

2) Explain why you're talking about "drops" in a thread about fail stacks. 

 

1) that's not how proof works, you claim the data is accurate, burden of proof is on you.

2) The clue is in here http://forum.blackdesertonline.com/index.php?/topic/93177-basic-lesson-on-probability/&do=findComment&comment=1241667

Your calculations are only good for 1 and only 1 drop, doesn't account for the chances of getting 2 or more drops out of n tries.

Additionally

3) You are the one calling people idiots, i didn't think that was like you.

4) You are being salty. Chill and use your brain.

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Posted (edited)

1) that's not how proof works, you claim the data is accurate, burden of proof is on you.

KR table is the only piece of evidence we have. The data is presumed accurate in the model, so in order to disprove the validity of the model, you must disprove the validity of the data. Failure to do so can only cast doubt about absolute certainty of the model, it does not disprove it.

You can go roll dices for fail stack optimization, but people like me prefer to use even limited evidence over no evidence at all.

2) The clue is in here http://forum.blackdesertonline.com/index.php?/topic/93177-basic-lesson-on-probability/&do=findComment&comment=1241667

Your calculations are only good for 1 and only 1 drop, doesn't account for the chances of getting 2 or more drops out of n tries.

I'm glad you're salty enough to pull a red herring and try use it as an argument here, I will address it anyway. Please read the OP in that thread again, the question answered in that thread is the probability of getting at least ONE drop after n kills. Your question here only demonstrates how bad you are at reading.

3) You are the one calling people idiots, i didn't think that was like you.

4) You are being salty. Chill and use your brain.

Given here you've demonstrated that you're also a fan of using ad hom attacks, I think you're too hypocritical if you continue this line of reasoning. 

Edited by Featherine

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Posted

Well I can't disprove your data source if you won't tell me what it is. However I assume you mean the standard table of data that all these things use e.g.

http://www.blackd.de/failstacks/?mwl=0&mwp=90000&swl=0&swp=24000&ahl=0&ahp=92000&aal=0&aap=104000&agl=0&agp=80000&abl=0&abp=80000&wbs=340000&abs=220000&mf=280000&cap=35000

If you read the bottom disclaimer:

Data Source

All calculations are based on the following data / assumptions. Please be aware that there is no proof that these numbers are correct. But it is the closest we can get at the moment to make educated guesses about the optimal way to enhance gear.

Now I am happy to accept you have access to confirmed reliable data, but I haven't seen it and you seem reluctant to post it.

I'm sorry if you think I have been guilty of an ad hominem attack, (please, which part) certainly wasn't my intention. Just wanted to respond to you calling me an idiot who doesn't know anything about maths (you know nothing about me). I can assure you I intend only to enlighten with accuracy as are you but your attack on me was unworthy and I thought unlike you (hence I assumed you were just in a bad mood).

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