Mathematical proof that Mboxing is not unfair

[Ðeceased]

Hi there everyone,

was goingonver this rly annoying post where despite most mboxers and some supporers destroying the arguments of those saying its illegal/unfair.. I came up with mathematical proof that this is not the case

pls correct my workings


assuming we are talking a 5 boxer

A: 1 mboxer =/= 5 players - correct
B: 1 mboxer = 1 player - correct
C: 5 characters = 5 character - correct
D: 5 characters =/= 1 character - correct

E: 1 mboxer = 5 characters - correct
F: 1 player = 1 character - correct
G: 5 players = 5 characters - correct

E = G ?

1mboxer = 5 players ?

we just saw from before they are not equal to one another either so that statement is incorrect!

E = F ?

5characters = 1 character?

this is also incorrect

conclusion:

1 mboxer does not equal one individual player, and an mboxer does not eqaul 5 players! the argument can never be about player number as i have just proven mathematically

the argument can only be made in number of characters.. and the statement 5v1 characters is always been one everyone agrees on, ie outnumbering guarantees an upper hand.. thus saying mboxing gives an unfair advantage is saying grouping and teamwork gives an unfair advantage

PS: soz about the spelling mistakes.. was in a rush

http://forums.wow-europe.com/thread....sid=1&pageNo=1

[Tehtsuo]

Erm, don't mean to be a hardass, but that's not really math...

[Sun]

well...I prefer the 5 man shaman group myself.

BTW...w00t w00t...I took out 16 levels above me yesterday in Un'goro crater. The guy kept coming by and killing my group. So I dropped Fire Nova(5) then hit him with Fire Shock(5) and melee. I think he was more surprised than I was when he dropped faster than a turd on a hot summer morning. The count was 5 deaths to his 1 but it well worth the dying just to see him drop that one time. I think if it was me I would not tell my friends that a level 55/54 shaman group just tagged you.

oooo Tet...you have to give him credit for at least knowing how to type. Math is really an option. :thumbup:

[keyclone]

i believe these 2 are in conflict:
B: 1 mboxer = 1 player - correct
F: 1 player = 1 character - correct

the implication would be that 1 mboxer == 1 character

i see where you are trying to go with the logical proof. let me try:
(note: mb== multi-boxer sb == single boxer)

A: 1 mb == 5 characters
B: 1 sb == 1 character
C: 5 characters == 5 characters
D: 1 mb != 1 sb

but this is obvious since the power wielded by a mb is vastly superior to a sb

but, as far as the game world is concerned...

E: 1 mb == 5 characters
F: 5 sb == 5 characters
G: 1 mb == 5 sb

i see this conclusion to be inaccurate as it does not factor in the response times and strategy differences.
so i modified it:
(note: rt == response times + strategy difference)

E: 1 mb = 5 characters + 1 rt
F: 5 sb == 5 characters + 5 rt
G: 1 mb != 5 sb

so... you end up with a multi-boxer is not the same as 5 individual players.
but, since 1 mb == 5 characters + 1 rt and 1 sb == 1 character + 1 rt ... you end up with:

H: 1 mb > 1 sb

which is obvious as well. now we have proof.

Multi-boxers Rock!

[Ðeceased]

Erm, don't mean to be a hardass, but that's not really math...

details details :P

Erm, don't mean to be a hardass, but that's not really math...

i believe these 2 are in conflict:
B: 1 mboxer = 1 player - correct
F: 1 player = 1 character - correct

the implication would be that 1 mboxer == 1 character

hmm I guess it would be a case or redefining as you did

i should probably distinguish betweeen player and person or so..

can't remember now tho, because i was a little bit of a rush, but what i wanted to show that nr of players did not have anything do with balance within this subject. as a boxer neither equals 5 nor 1 sb. the only way to accurately state it is with character numbers..

i get the feeling that's why i separated the two blocks of statments.. can't remember now

thanks for pointing it out I'll work on it a little bit taking urs into account too, and repost it on the wow forum

[kayb]

As far as Blizzard are concerned: 1 character = 1 character, end of story.

[Ðeceased]

Somebody hit the bong pretty hard before writing this one..

eeek, I've been rumbled!

[crebble]

As far as Blizzard are concerned: 1 character = 1 character, end of story.

nah they just count the $ inflow!

[geoffdavison]

Its certainly not a mathematical proof, but looks more like an attempt for a logical theorem.

[Sanctume]

Fact: 1 account = 10 character per server.
Fact: 1 account may log in 1 character at any given time when servers are up.

Let A = player A
Let B = player B
Let C = player C
Let D = player D
Let E = player E
Let F = player F
Let X = skill to play Single box
Let Y = skill to play Multi-box
Let n = number of accounts owned by player
Let G = 5 players in a group

Assume players A, B, C, D and E owns 1 account each.
Assume player F owns 5 account each.

Account Player functions
(n=1): A(n) = A = 1
(n=1): B(n) = B = 1
(n=1): C(n) = C = 1
(n=1): D(n) = D = 1
(n=1): E(n) = E = 1
(n=5): F(n) = F = 5

Group functions
G(A) = 1/5 G
G(A+B+C+D+E) = 5/5 G = 1
G(F) = 5/5 G = 1

Since players A, B, C, D and E player single box, they each have X skills.
Since player F multi box, he has Y skills.

G(A+B+C+D+E) * X = 5/5 G * X = 1 * X = X
G(F) = 5/5 G * Y = 1 * Y = Y

Question: Is X = Y ?
Answer: Yes.

Given in the Group function:
G(A) = 1/5 G
G(F) = 5/5 G = 1

We can say that one-fifth of G(F) function is the same as the G(A) function:
1/5 G(F) = G(A)


Assume the skill of A is equal of B, C, D and E.
If A = 5, them G(A) = G(F)

G(A+A+A+A+A) * X = 5/5 G * X = 1 * X = X
This is the same as
G(A) * X = 5/5 G * X = 1 * X = X
G(F) = 5/5 G * Y = 1 * Y = Y

Player A happens to be 5-boxing in Server 1, and it's the same person as player F in Server 2.
Therefore, X = Y.

So, G(G) = X skill to Multi Box