Over-Enchantment Probabilities

Could someone (a statisticologist maybe ;)) post a table with the probabilities someone have to enchant succesfully an item up to maximum? Somewhere i read that the success rate up to +14 is 1.1%. Is that true? And if it is what are the % at each step?

% to succeed at ANY step is 66%. 1.1% to get +14 would be to go from 1-14 in one enchant. this is impossible.

% to succeed at ANY step is 66%. 1.1% to get +14 would be to go from 1-14 in one enchant. this is impossible.

LOL 1.1% chance is impossible...gg

If I remember my statistics right, overenchanting is a dependant compound event where each dependency has the same likelyhood of occuring (66%).

So, to determine the chance an overenchant will work 14 times in a row is calculated as:
0.66^14 = 0.002976 = 0.2976% chance of success

have to account for the first 3 as 100%, so its .66^11

It is quite literally 66% chance of success at each step as noted...

However, the tables you might have seen are the chance to succeed if you just take a +3 weapon and keep applying enchant after enchant to it wth the hopes it will reach a certain enchanted level.

Frankly, the tables aren't really that useful...
Basically every time you apply an enchant, you are "rolling" a 6-sided "die" and hoping not to roll 1 or 2.

P.S. 1.1% chance is improbable but not impossible.

oops, good catch there Spectres.

That makes a +14 overenchant a 1.035% change of success.
And it makes a +16 enchant a 0.45089% chance.

you are right, except the weapon can not break from 0 to 3.
0-14 is 1.16% chance ([censored])

Weapons enchantmant from +3 is always 66% ! ! ! That means u have always a 1/3 chance that u blow it ^^ Good luck

you could look @ [censored],,h :) then OE

***** all numbers. a good advice is, dont try +3-->14.

nope ^^ just +3:)

imo it costs 2 much money 2 play with weap :)

If I remember my statistics right, overenchanting is a dependant compound event where each dependency has the same likelyhood of occuring (66%).

So, to determine the chance an overenchant will work 14 times in a row is calculated as:
0.66^14 = 0.002976 = 0.2976% chance of success

0 -> +4 66.66%
0 -> +5 43.56%
0 -> +6 28.74%
0 -> +7 18.97%
0 -> +8 12.52%
0 -> +9 08.26%
0 -> +10 05.45%
0 -> +11 03.60%
0 -> +12 02.37%
0 -> +13 01.56%
0 -> +14 01.03%
0 -> +15 00.68%
0 -> +16 00.45%

Based on the theory you posted

yep, that's right, but I realized that the calculation should use a more accurate approximation of 2/3. If you use 0.66667 you get the 1.1% @ +14 and 0.51% @ +16