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%

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