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  1. #1
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    Enhancement Math

    Say you have a +6 weapon and want to make it +10. Assuming you are using enhancement runes, the following probability table should be true (based on the enhancement probabilities given by vindictusdb.com):

    Number of fails (x) P(fails less than x times) P(fails x times or less) P(fail more than x times)
    0 - 4% 96%
    1 4% 12.8% 87.2%
    2 12.8% ~24.9% ~75.1%
    3 ~24.9% ~38.3% ~61.7%
    4 ~38.3% ~51.2% ~48.8%

    Edit: This means, for example, there's a slightly above 50% probability you'll need not more than 5 enhancement runes to go from +6 to +10. Why 5? It's slightly above 50% to fail -less- than 5 times after all. So it's also the probability you'll consume not more than 4 runes.
    Well.. assuming you fail 4 times. If you have only 4 runes you'll have to do at least the last enhancement without rune. You could of course trust your luck, but.. yeah ..

    For those interested in the math (basically borish counting of combinations, please tell me if I made mistakes):

    From vindictusdb.com:
    Probability to fail upgrading to +6, +7 or to +8 (from one level below): 50% per try
    Probability to fail upgrading to +9 and to +10 (from one level below): 60% per try

    Assumptions:
    - start with +6 weapon
    - using runes

    Notation:
    - fail: F; x fails: xF
    - success: S; x successes: xS
    - fail at +x means fail when trying to enhance from +(x-1) to +x
    - fail between +x and +y means failing at either of the tries for any enhancement levels inbetween (inclusive) the two mentioned levels.
    Short: F (x-y)

    Probability to fail not at all:
    0.5 * 0.5 * 0.4 * 0.4 = 0.5² * 0.4² = 0.04 = 4%

    Probability to fail exactly once:

    - (1) fail at either +7 or +8: 0.5 (fail) * 0.5 * 0.5 + 0.5 * 0.5 (fail) * 0.5 = 0.5³ * 2 = 0.25
    Explanation: one fail and two successes. You can either fail at +7 or +8, so there are two possibilities, hence *2

    - (2) fail at either +9 or +10: 0.6 * 0.4² * 2 = 0.192

    - fail once at +7 or +8, then successes: 0.25 * 0.4² = 0.04 = 4%
    - fail once at +9 or +10, else successes: 0.5² * 0.192 = 0.048 = 4.8%

    => fail exactly once: 4% + 4.8% = 8.8%

    Probability to fail exactly twice:
    - (3) fail twice between +7 and +8: 0.5^4 * 3 = 0.1875
    => with rest successes: 0.1875 * 0.4² = 0.03 = 3%
    Explanation: two fails and two successes, three possible combinations: F F S S, F S F S, S F F S

    - (4) fail twice between +9 and +10: 0.6² * 0.4² * 3 = 0.1728
    => with rest successes: 0.5² * 0.1728 = 0.0432 = ~4.3%

    - fail once at +7 or +8 and once at +9 or +10 (see above: (1) and (2)): 0.25 * 0.192 = 0.48 = 4.8%

    => fail exactly twice: 3% + 4.32% + 4.8% = ~12.1%

    Probability to fail exactly three times:
    - (5) 3F between +7 and +8: 0.5^5 * 4 = 0.125
    => with rest successes: 0.125 * 0.4² = 0.02
    Explanation: four possible combinations: F F F S S, F F S F S, F S F F S, S F F F S. As you can see, there are always (number of fails) + 1 combinations, since you always quit with a success and the second success has that amount of possible positions to occur.

    - (6) 3F between +9 and +10: 0.6³ * 0.4² * 4 = 0.13824
    => with rest successes: 0.5² * 0.13824 = 0.03456

    - 2F (7-8) and 1F (9-10) (see (3) and (2)): 0.1875 * 0.192 = 0.036
    - 1F (7-8) and 2F (9-10) (see (1) and (4)): 0.25 * 0.1728 = 0.0432

    => fail exactly three times: 0.02 + 0.03456 + 0.036 + 0.0432 = 0.13376 = ~13.4%

    Probability to fail exactly four times:
    - 4F (7-8): 0.5^6 * 5 = 0.078125
    => with rest successes: 0.078125 * 0.4² = 0.0125
    - 4F (9-10): 0.6^4 * 0.4^2 * 5 = 0.10368
    => with rest successes: 0.5² * 0.10368 = 0.02592

    - 3F (7-8) and 1F (9-10) (see (5) and (2)): 0.125 * 0.192 = 0.024
    - 2F (7-8) and 2F (9-10) (see (3) and (4)): 0.1875 * 0.1728 = 0.0324
    - 1F (7-8) and 3F (9-10) (see (1) and (6)): 0.25 * 0.13824 = 0.03456

    => fail exactly four times: 0.0125 + 0.02592 + 0.024 + 0.0324 + 0.03456 = 0.12938 = ~12.9%
    Last edited by Riyria; 25-01-2013 at 18:19. Reason: small formatting corrections
    MysticalEssence

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  2. #2
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    Re: Enhancement Math

    NezardZero disagrees with the calculations below, see discussion starting with this post.
    So take it with a grain of salt, please.

    Enhancement above +10

    Assuming you start with a +10 weapon.
    Probabilities of single steps again from vindictusdb.com:
    - 40% chance for +10->+11 and +11->+12
    - 33% chance for +12->+13, +13->+14 and +14->+15

    Enhancement level P(reach without fail) P(fail somewhere on the way)
    +11 40% 60%
    +12 16% 84%
    +13 ~5.3% ~94.7%
    +14 ~1.7% ~98.3%
    +15 ~0.6% ~99.4%

    Enhancement level expected number of tries (ENT) P(need less than half of ENT) P(need less than ENT) P(need less than twice of ENT) P(need twice of ENT or more)
    +11 2.5 40% 64% ~92.2% ~7.8%
    +12 6.25 ~50.2% ~70.5% ~89.6% ~10.4%
    +13 ~19 ~41.9% ~64.3% ~87.3% ~12.7%
    +14 ~57 ~39.9% ~63.3% ~86.5% ~13.5%
    +15 ~174 ~39.4% ~63.3% ~86.6% ~13.4%

    As you can see, there is always a low, but still noticable chance to need more than twice of the expected number of tries (and thus destroyed weapons) to get to the weapon of your dreams.

    Now, most people who successfully enhanced a +15 weapon will tell you they needed nowhere near 167 or more tries. But that's probably because people who are not lucky and get their weapon in fewer tries most likely will give up on making a +15 at some point.


    Some math again:

    (1) Probability to reach a certain enhancement level x without failing:
    multiply the success chances for each step needed with each other.

    Example: Probability of reaching +13 (starting with +10) without fail: 0.4 * 0.4 * 0.33 = 0.0528 = ~5.3%

    (2) Probability to reach a certain enhancement level failing exactly k times:

    S = Probability to reach +x without fail as above in (1)
    P(reach +x with exactly k fails) = (1 - S)k * S

    Let's make an example. How likely is it to enhance a weapon to +13 and failing exactly 3 times?
    1. Calculate the chance of success without failing as above which means S = 0.0528 = ~5.3% in our example.
    2. Calculate the chance to fail k times (in our example three times) by multiplying the probability of one failure (which is 1-S) with each other three times: 0.9472 * 0.9472 * 0.9472 = 0.9472³ = ~ 0.85 = ~85%
    3. Multiply these two probabilities with each other, basically calculating the chance to fail three times at reaching +13 from +10 and then succeed:


    => P(reach +13 with exactly 3 fails) = 0.9472³ * 0.528 = ~ 0.045 = ~ 4.5%

    Why is this less than getting there without fail? Because we have additional conditions (the failing 3 times before) which aren't 100% sure.
    What's more interesting is the following part.


    (3) Probability to reach a certain enhancement level x failing less than k times:

    This is actually pretty straight-forward since with each fail you have to start from the beginning again, and like in the rest of this post we assume the beginning to be a +10 weapon.

    So if you want to know how likely it is to enhance a weapon to +13 while failing less than 4 times (and thus at most destroying 3 weapons in the process) you do the following:
    1. Calculate the probability to reach the level without failing as above in (1)
    2. Now add up the probabilities to succeed without fail, with exactly one fail, with exactly two fails and with exactly three fails as calculated in (2). These are all possibilities you have when you want to succeed with less than 4 fails.


    => P(go to +13 from +10 with less than 4 fails) = 0.0528 + 0.0528 * 0.9472 + 0.0528 * 0.94722 + 0.0528 * 0.94723 = ~ 0.195 = ~19.5%

    In general, after a simplification:
    S = Probability to go to +x without fail
    P(go to +x from +10 with less than k fails) = S * (1 + (1-S) + (1-S)2 + ... + (1-S)k-1)

    You can also write this as follows:





    original post content:
    And all those probabilities don't prevent someone to fail 6 times in a row trying to go from +3 to +4 +cry
    Last edited by Riyria; 25-11-2013 at 21:34. Reason: Minor corrections to numbers (smoothing out some rounding errors); replaced "average number of tries" by "expected number of tries"
    MysticalEssence

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    Scissa - Vella
    Saede - Evie


  3. #3
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    Re: Enhancement Math

    Probability is just probability. Knowing my chances doesn't make me any more likely to succeed without burning a rune or any less likely to burn through 25 runes just to get up +1 rank =>




  4. #4
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    Re: Enhancement Math

    Quote Originally Posted by BladeSurge View Post
    Probability is just probability. Knowing my chances doesn't make me any more likely to succeed without burning a rune or any less likely to burn through 25 runes just to get up +1 rank =>
    I know that of course. I just thought it might be interesting to see some probabilities in a table like that.
    It's of course still perfectly possible to fail significantly more often.

    But when, according to the probability values given on vindictusdb, 6 fails from +3 to +4 in a row have a probability of around 0.02% (25% fail chance per try, 6 tries => 0.25^6) .. I still wonder
    Last edited by Riyria; 23-01-2013 at 02:21. Reason: typo
    MysticalEssence

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    Scissa - Vella
    Saede - Evie


  5. #5
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    Re: Enhancement Math

    in today's topic, we create conspiracy theories relating to nexon and money
    "Just... a young man!"
    ..
    "Caesar!"




  6. #6
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    Re: Enhancement Math

    This Math helps you to calculate more realistic value of enhanced Weapon/Armor.
    In average case, you must use at least 4 runes to get from +6 to +10, so average price should include rune costs

  7. #7
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    Re: Enhancement Math

    This is neat. Sweet stuff

  8. #8
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    Re: Enhancement Math

    Quote Originally Posted by whynotslave View Post
    This is neat. Sweet stuff
    Thank you.
    I'm thinking of making a section about enhancing above +10, too. I just need to find the time and to identify some metrics of interest.
    MysticalEssence

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    Scissa - Vella
    Saede - Evie


  9. #9
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    Re: Enhancement Math

    Quote Originally Posted by fewifewnewf View Post
    in today's topic, we create conspiracy theories relating to nexon and money
    Now that NEXON knows that we know about them, they will do drastic, stuff, thing!
    Quote Originally Posted by dooder41
    Looks like everyone in this forum has a quote from someone else in their signature. So I guess I have to put one too.
    Quote Originally Posted by Espy Rose
    "I'm sorry, did you want to have fun playing this game?"
    -Olimar.


  10. #10
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    Re: Enhancement Math

    My brain can't take so much math formulas..












 

 
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