Rostigalen
Smash Journeyman
Whenever you want the next tier list, just ask Hax or Pink Reaper...22 of 26 "right" placements. DrPP got 11/11 as well, so yeah...
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The closest to the average list is indeed Hax (and by far). Then it's N64 and Skler.Whenever you want the next tier list, just ask Hax or Pink Reaper...22 of 26 "right" placements. DrPP got 11/11 as well, so yeah...
In the .pdf, Pink Reaper only got Mewtwo, Young Link, Zelda and Roy placed wrong, how can he not be closer than N64 who got 11 characters in the wrong place?The closest to the average list is indeed Hax (and by far). Then it's N64 and Skler.
The farthest is forward, followed by Mike G.
all i heard was falcon players are the smartestThe closest to the average list is indeed Hax (and by far).
I made this ranking using the sum of the differences between the ranking given to each character and their average ranking. Since averages are not integers, it means that someone placing a character at the wrong place can be closer to its average ranking than someone who ranked him at the right place.In the .pdf, Pink Reaper only got Mewtwo, Young Link, Zelda and Roy placed wrong, how can he not be closer than N64 who got 11 characters in the wrong place?
Not necessarily true. g-reg is average.all i heard was falcon players are the smartest
LOL that's the funniest thing I ever ****ing readOkay.
Let's say I have a game with n characters and I have a match-up chart with the characters for that game; since there are some slight variations in what a match-up chart's numbers mean, let's say the number in the row for character A and column for character B represents A's chance of beating B in a bo3 set using current tournament rules, assuming both players of each character are at the top level of play and know the match-up. I want to rank the characters solely based on how they perform against each other; an intuitive idea about how to do this would be to have the match-ups against characters that seem "good" in some sense count more than the match-ups against other characters. To make this more precise, we do the following:
First, label the characters c_1, c_2, ..., c_n and create an nxn matrix P whose i,j-th entry is c_i's chance of beating c_j, taken from the match-up chart; let p_i,j denote the i,j-th entry of P. As a first approximation for how good the characters are, we look at p_i,1 + p_i,2 + ... + p_i,n for each i between 1 and n, i.e. for each character, we find the sum of his/her/its odds of winning against all characters (including itself). Note that if we let X be the nx1 vector whose entries are all 1, then PX is a column vector whose i-th entry is the first approximation of how good c_i is. Let A_1 = PX.
For the second approximation, we do something similar, except that we want the match-ups against the characters that are good from the first approximation to have more weight than the match-ups against the ones that aren't. Let a_i denote the i-th entry of A_1 for each i. Now for each i, we take (a_1)(p_i,1)+(a_2)(p_i,2)+...+(a_n)(p_i,n) as the second approximation of how good c_i is. Note that the i-th entry of the vector P*A_1 is the second approximation for how good c_i is, and let A_2 = P*A_1. Note that A_2 = (P^2)X.
Iterating this procedure, we let A_m = (P^m)X for m>=3. The behavior of the sequence (A_m) varies a lot by P. What actually happens to each individual entry of A_m as m goes to infinity isn't very interesting on its own; what is interesting is what happens to the entries relative to each other. For example, in a game with 2 characters in which c_1 beats c_2 100% of the time, both entries of A_m tend to 0 as m approaches infinity, but c_2's entry approaches 0 much faster than c_1's does. Thus, instead of looking at the long term behavior of (A_m), we do the following: for each positive integer m, let s_m be the maximum entry of A_m. Let Q_m = (1/s_m)A_m. In other words, Q_m tells you how good each character is as of the m-th approximation relative to the best character(s) as of the m-th approximation. It turns out that the sequence (Q_m) does almost always converge, and the vector Q that it converges to provides a nice way of ranking the characters.
The proof that (Q_m) converges unless this one odd condition pops up is kind of ugly and tedious, so I won't go into that here; even the statement of that one condition is messy.
Anyways, in practice, it shouldn't be too hard to find a decent approximation of Q; just use Mathematica or something like that to find A_m for some decently large m and then find Q_m from there. I'd also check Q_(m+1) and maybe a few other later values to see if anything odd happens, but I think the condition for the failure of this procedure is improbable.
It's also worth mentioning that this method can lead to some results which might intuitively seem odd. For example, if you have a game with three characters and c_1 has guaranteed wins against c_2 and c_3, and c_2 has a guaranteed win against c_3, then this procedure gives c_1 the value 1 and both c_2 and c_3 get 0. This might seem strange since c_2 has strictly better match-ups than c_3, but it makes sense since c_2 is worthless in the sense that its only winning match-up is against a character that is clearly worthless.
Thats cuz Hax is the mother ****ing truth.The closest to the average list is indeed Hax (and by far).
My Mewtwo beat Kels.inb4midwestsux
though cosmo beat kels
kels beat taj
oh snap
do we have some inui logic in the making?
Pound 4 ?My Mewtwo beat Kels.
Kels beat my Marth.
The only thing I dont agree with is Mario being SOOOOO ****ing far away from Doc. Theyre almost the same character.
could be because mario is under represented because Doc is seen as superior.it's all bout dat tourney results.
there are a couple docs placing well but there are really zero marios anymore.
ORLY? Care to explain? Cause DogySamich wrote a big *** wall of green telling me otherwise. I could find the thread if you like.
that's like saying that fox and falco are almost the same character.
How? I actually dont know.Uhhh doc's edgeguarding is also significantly better, which is sort of a big part of the game... but w/e
where do i make my threads about loving luigi nowI am not MBR anymore.
sheik boardwhere do i make my threads about loving luigi now
Seems like you're making some pretty general statements here. Every one of their moves (except for uair maybe?) have different properties (and I'll gladly go into detail on each one if you're interested), and some of them are pretty drasticEdit: BTW Doc and Mario differences
Doc has a better kill move, a better projectile game, and his usmash is more of a combo move.
Mario has a better combo game, a better fsmash for spacing, and a slightly better recovery.
Lies.if its any consolation i voted her 7th