Arcane Thievery Build?

Just wondering............if 8 people ran Arcane Thievery and hit 1 target...would it disable all 8 of that target's skills, effectively making this an uber-version of blackout.. or is this just wishful thinking :P

Please let me know if you've tried this. And if you're too scared to say it in public, PM me intead xD

A nice blackout for how much energy?? You are far better off to use echo+arcane echo+blackout instead...

Just because it'd be incredibly inefficient doesn't mean it wouldn't be funny as all get-out.

I'm guessing no-one's tried it then...

It doesn't work as you hoped it would.

Each A Thievery will target one of the spells, no matter if it has already been targetted or not by another A Thievery (which means that every spell can be "stolen" several times).

To give you an example, I tried an Echo-A Echo-A Thievery build and once stole THREE times the same spell (seed).

Result: you probably steal like 4 spells (remember it doesn't work on other skills). Not worth using 8 spots out of the team's 64.

Thanks for confirming, glad someone else tried it out

I guess if it did work like I thought it might, it would be used all over pvp. I mean, shutting someone down for 27 seconds is just insane...

It has been tested exhaustively; our guild tried running builds using it. Works well to some extent, but it is pointless to try to get all the spell a player has. Incredibly effective vs W/E hammer spikers, I've echoed it on seeing two W/E on the other team and had two aftershocks on my bar more often that I can count - pulls the fangs right out of that build.

Well; a bit of math for you guys. When you steal a spell it doesn't remove it from the pool of available spells (to steal), but it does alter the chance of the next attempt to steal a spell actually resulting in a steal (if it's the same as a previously stolen spell you get no advantage).

For simple cases it's easy:
1 spell on target, use 1 Arcane Thievery (AT) = 100% chance of hitting it.

2 spells on target; first AT automatically hits. the next AT is 50%hit/50%miss - this is nested of course - if you miss you can try with a 3rd etc...

With only 2 spells on the target, you have the probability of knocking N=2 spells out P(N)=P(2) depending on the number of trials (t) P(N|t); with the probability of 2 hits being 1-(0.5)^(t-1); that is
P(2|t) = 0.....0.5....0.75..0.875..0.9375
t =.........1.......2.......3.......4.......5

It gets more complex with more spells, as the odds begin to change with each success, so knowing how many trials (arcane thievery attempts) and the number of spells is necessary. Given that I posted a build for TA with 6 copies of AT avaiable, that's the first case I examined, though there are larger one.

Thus, trials = 6 for the following probabilities:

Number of spells on target = 8
probability of stealing N distinct spells = P(N)
expected value of 6 attempts = E(N) = sum(N*P(N))

N P(N)
1 3.05176E-05
2 0.006622314
3 0.115356445
4 0.416564941
5 0.384521484
6 0.076904297

E(N)=4.409637451

Number of spells = 7

N P(N)
1 5.9499E-05
2 0.011066817
3 0.160647349
4 0.464092342
5 0.321294699
6 0.042839293

E(N) = 4.224013804

Number of spells = 6

N P(N)
1 0.000128601
2 0.019933128
3 0.231481481
4 0.50154321
5 0.231481481
6 0.015432099

E(N) = 3.99061214

Number of spells = 5

N P(N)
1 0.00032
2 0.03968
3 0.3456
4 0.4992
5 0.1152
6 0

E(N) = 3.68928

Number of spells = 4

N P(N)
1 0.000976563
2 0.090820313
3 0.52734375
4 0.380859375
5 0
6 0


E(N) = 3.28808


Number of spells = 3

N P(N)
1 0.004115226
2 0.255144033
3 0.740740741
4 0
5 0
6 0


E(N) = 2.73662

Number of spells = 2

N P(N)
1 0.03125
2 0.96875
3 0
4 0
5 0
6 0

E(N) = 1.96875

Obviously, this is showing that we can expect roughly 4 slots stolen off a caster using 6-8 spells, and more like 3.5 slots stolen from a caster only using 5 spells, with 6 ATs cast. Denying 4 spells of 6 is great; follow it up with a diversion and you have likely reduced them to 1 spell. Diversion becomes very powerful in combo with the AT, as it takes from a restricted spell list, and lasts longer. Sure, the AT expires, but they've lost the spells used while AT was up for a minute, so they are still functioning at a disadvantage. Eliminating all spells is thus not necessary - if you can reliably reduce them to 1-2 spell slots a diversion or two will take care of the rest.

EDIT:

Table is started:

Number of Spells across top, number of AT attempts vertically - the readout is the expected value of the steal = thus, if you look on the table, for 4 arcane thievery attempts against a monk with 6 spells on his/her skill bar you get on average 3.11 spells stolen.

.....1___2___3___4___5___6___7___8
1 1.00 1.00 1.00 1.00 1.00 1.00 1.00 1.00
2 1.00 1.50 1.66 1.75 1.80 1.83 1.86 1.88
3 1.00 1.75 2.11 2.31 2.44 2.53 2.59 2.64
4 1.00 1.88 2.41 2.73 2.95 3.11 3.22 3.31
5 1.00 1.94 2.60 3.05 3.36 3.59 3.76 3.90
6 1.00 1.97 2.73 3.29 3.69 3.99 4.22 4.41
7 1.00 1.98 2.82 3.47 3.95 4.33 4.62 4.86

I haven't done the table past 7 arcane thieveries, as the payoffs aren't really that great beyond this point, except against casters with 8 spells perhaps.

...........my head exploded when you started assigning variables

Well, it is simply a cut+paste of a post in our forums; I probably overdid the analysis a bit, but I tend to do that, especially if doing in-depth stuff.

The table at the end is the bit that is most useful, as it tells you how many ATs to use to expect to eliminate however many spells from a skill bar with a given number of spells.

Heheh, well thanks for the info. I guess it's not too bad of a waste of skill slots as you can easily run it with echo, and that echo can be used on other stuff effectively as well.

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