Curse

seekjy · May 19, 2008, 07:36 PM // 19:36

Your team has 16 copies of Arcane Thievery. All 16 copies are used on a single target in immediate succession. Assume the target has 8 spells.

What is the probability the target does not have his entire skillbar disabled?

poobert · May 19, 2008, 07:57 PM // 19:57

Depends how many silly mesmers are on your team who think it is useful to use [skill]simple thievery[/skill] every damn time.

I pwnd U · May 19, 2008, 08:05 PM // 20:05

The same as the probability of you attacking 1 target as 16 mesmers. >.>

Mr. G · May 19, 2008, 09:14 PM // 21:14

I fear [Arcane thievery] [Echo] [Arcane echo]

brrrrrrrrrr

or you know you could replace a team of echoing mesmers with useless skills with 1 single PD mes....

Turbobusa · May 19, 2008, 09:58 PM // 21:58

OK I'm not that good with maths but here's what I found:
Lets assume that each spell cast is one bit
1 = spell targeted now disabled
0 = spell targeted already disabled

Number of combinations = 2^16 = 65535
Number of possibilities of 8 bits being at one (found that on the net):

C^(p)_(n) = n! / ( p!.(n-p)!)
(source, in french: http://www.iut-bethune.univ-artois.f.../anal_com.html)

with p = combination of elements = 8
n = total number of elements = 16

That gives 12870 possibilities of 8 bits being at one.Now I only considered the cases where there is ONLY 8 bits at one, but that's also true for the cases where 9...16 bits are high also.
If I'm not mistaken the real answer would be given by (ok time for a proper formula):

I don't have any other way to calculate this than calculate each separatly and adding them together, so thats what I did, and it gave me 39203.
(for the records, thats:
8 ->12870
9 ->11440
10->8008
11->4368
12->1820
13->560
14->120
15->16
16->1)

In percents thats 39203/65535 = 59.82%

Now that is the number of possibilities of the whole skill bar to be disabled, so the opposite would be 40.18 % of not having the whole skill bar disabled.
That's more than what I expected, but again the more skills are disabled the harder it is to target a non disabled spell.

I'm not sure about the initial formula tho. Any comment on my maths are welcome, I was trying to apply what I found + some logic.

(Note: I love firefox. It crashed for X reason and I feared I lost this message, but it just saved it somewhere and restaured the session just as it was)

RedStar · May 19, 2008, 09:59 PM // 21:59

I think it's 16^8 = 1/4 294 967 296.

Edit : O_o what Turbobusa did seems so complicated xD.

Divinitys Creature · May 19, 2008, 10:50 PM // 22:50

If one of the enemy's 8 skills cannot have the disablement stack then it is certain the skill bar will be totally disabled but I haven't played in a while.

Owoc · May 19, 2008, 11:41 PM // 23:41

Correct me if I'm wrong, but it seems to me that, because the first spell will never fail, we actually end up with 15 digits, at least 7 of which must be 1 in order to black out the victim's bar entirely. Therefore, we need to find the value of

Code:

15          15!
 Σ       ----------
k=7       k!(15-k)!

which is apparently equal to 22819, so the chance to black out the skill bar completely is 22819/32768 ~ 0.69638.

Turbobusa · May 20, 2008, 07:40 AM // 07:40

You're right, my reasonning includes combination where it would fail, except it can't.
69.64% total disabling
30.36% one skill not disabled

EDIT:
After rethinking, it appears that this is just impossible to have the first bit +8...15 bits at one, seing as I said that 1 = skill disabled.
If 8 bits are already high, all the skills are disabled. Therefore the 8...15 high bits cases don't exist, the max combinations isn't 2^15, but rather:

with n=15. (note that it gives us 2^14, I'd like to know why)
Chances of 8 bits being at one is still:

which gives us 6435 over 16384 = 39.28% of total disabling (60.72% conversely)

**cosyfiep** · May 20, 2008, 07:56 AM // 07:56

how can you have 16 mesmers on your team????? (copies of AT that is....even with echo......)

and why would you want 16 copies of AT anyways????

seekjy · May 20, 2008, 03:36 PM // 15:36

What is the reasoning behind the "bits"? Also, bear in mind that it's possible for more than one skill to be missed by all copies of Arcane Thievery. Your calculation assumes that the one minus the probability of the entire bar being blacked out equals the probability that one skill is still active.

Turbobusa · May 20, 2008, 03:59 PM // 15:59

The bits represent the result of the casting of AT.
A high bit refers to "1 skill just got disabled", 0 is "spell disabled was already disabled".
You can also see the number of 1 in your 16 bit number as the number of spells disabled, which is the same.

The reasonning:
- the first spell will never fail
- we must find the chances of having 7 out of 15 AT copies that will disable a non-disabled spell.
- If 8 bits are high, all skills are disabled, therefore the rest of the AT copies will disable an already disabled spell (= 0)
- Hence, there can't be 9 or more bits active at the same time.
- Therefore the only cases where all skills will be disabled are the ones where 8 bits are at one.
- Since the first spell will always disable one spell, we must find the possibilities of having 7 out of the 15 copies of AT left that will disable another spell.
- we must count those cases, and make a ratio between that and the max number of possibilities

some exemples:

0000 0000 0000 0000 = nothing disabled (impossible)
0100 1100 1110 0110 = 8 skills disabled
1100 1100 1110 0110 = 9 skills disabled (impossible)

What I'm not too sure is the maximum number of combinations calculations.

RiKio · May 20, 2008, 04:27 PM // 16:27

Probability? Who cares about probabilities?

PD: My school doesnt :P