Requesting Feedback - A Potential Alternate Rule Set

[-ACE-]

Only two stages were banned for this specific reason. Hyrule and Great Bay. Those two stages will be banned in singles in the actual ruleset for this, whenever I get around to writing it up, but probably not doubles. I actually wouldn't mind seeing an intense game of last player last stock on each team playing circle tag as the time runs out, desperately trying to land a hit and take back the lead. Sounds awesome.

Yoshi's island?

 

[Kal]

Pipes was banned because it was broken for a variety of reasons (mostly present with Fox). Not because of the potential for laser camping.

Or are we talking about the one with the clouds? In which case it wasn't circle camping, just that there was a spot any character could camp once in the lead that was more or less unapproachable.

 

[Cactuar]

Yoshi's Island was not banned because of Fox camping.

 

[Bl@ckChris]

are there any characters that can't make it back from the second cloud? i think in my scrub days i used to think that mario couldn't, but i feel like he can, and we were just bad at managing jumps or something.

that would be annoying to live, but still be dead lol.

 

[-ACE-]

My bad, I don't read your entire posts anymore. Yoshi's island would be banned though, correct?

 

[da K.I.D.]

i tested this actually. like... 5 years ago lol. Every character in the game can make it to either the main stage or the high right cloud from teh low right cloud without interference.

but what cact said is right, if you have a lead and go to that cloud its game breaking.

Since we can hack things now, somebody should remake that stage with the right side blastzone about halfway between the 2 clouds on the right side, that would make that stage effing amazing.

 

[Dan Salvato]

Since we can hack things now, somebody should remake that stage with the right side blastzone about halfway between the 2 clouds on the right side, that would make that stage effing amazing.

If only booting a Melee mod was as simple as inserting an SD card, we could have made so many great changes for tournament play by now.

 

[sanchaz]

My bad, I don't read your entire posts anymore. Yoshi's island would be banned though, correct?

http://www.smashboards.com/showthread.php?t=317562&page=2&highlight=sanchaz

not at my tournament, :D. matches will be recorded.

possibly livestreaming with commentary from people that think kirby's down b is the greatest move in the game

 

[BigD!!!]

that cloud stage would actually still be really really big, the top platform is like twice as high as the top platform on any other stage

 

[Warhawk]

Since we can hack things now, somebody should remake that stage with the right side blastzone about halfway between the 2 clouds on the right side, that would make that stage effing amazing.

Its already been done actually, changed blast zones and removed the clouds and all, I just can't remember where I found it... Probably on Megaupload, so it may very well be gone now.

 

[KrIsP!]

If only booting a Melee mod was as simple as inserting an SD card, we could have made so many great changes for tournament play by now.

If only, if only. Then screw the stage, we could make 10 better ones, add 2 more to the neutral list and MANY more to the CP list and we'd be set.

 

[Kal]

Math is hard when you struggle with reading comprehension.

I'm a little late on this response. Sorry, it took me a day to download MikTeX and get TeXniccenter working.

This is false. First of all, it's not even well-defined. If we define the better player as the player with a probability of winning a given match that is greater than 50%, the "better player" in a given pair of players might change depending on the number of stocks per game. I'll elaborate later in this post. (Note: I am bringing up this definition because it is the one Kal mentioned before.)

How isn't this well-defined? The "game" is what is being played. Being better at the game with four stock doesn't mean being better at the game with two. As I've said before, and as you would have observed if you took the time to read the thread instead of awkwardly stalking me via AIM and private messages, a different ruleset means a different game.

lol, you are a hypocrite, also a jerk (no offense)

Super constructive addition to your post. Why are you acting like a creepy, passive-aggressive ex-girlfriend? We haven't even slept together yet. Take a chill pill.

In order to allow me to explain why you are wrong, perhaps you should define "better player" formally, so there's no confusion and/or possibility of you dodging my points.

I've explained this several times. You're getting stuck on nonsense because you stubbornly refuse to admit that an incorrect application of basic first-order logic does not mean I'm wrong.

You are wrong, as I have explained / will explained. Notably, I should point out that the probability for an individual match can change from > 0.5 to < 0.5 by adjusting the number of stocks.

I've already explained, over and over and over in this thread, that the probability will change depending on the number of stock.

(Also, notably, the probability of winning an individual match isn't necessarily independent of the outcomes of other matches, etc. To take a simple (but extreme) counterexample, let's say we have Player A vs Player B, where Player A excels on FD+PS and Player B excels on Brinstar+DL64, to such a large degree that it's impossible for either one to win on the other one's counterpick. Then the outcome of the set will be determined solely by the outcome of the first match. Of course, you might argue that this counterexample wouldn't happen in real life, but regardless, it almost certainly breaks whatever assumptions you use in your supposed proof. (If not, feel free to explain your proof.))

A simple modification of "better player" takes this into account. You're right, however, that if we are defining being better solely by the probability of winning a single match on a single stage, this becomes an issue. In that case, it suffices to change the definition to something along the lines of "player A is better than player B if the probability that player A wins is greater than 50% on at least half of the stages."

We haven't been ignoring counterpicks because they prove us wrong. We've been ignoring them because they obfuscate the issue. It's clear that a reasonable definition of "better" exists to accommodate for a being able to lose on a counterpick.

Spoiler

Now, here we go: explanation of how the probability of winning a given match can shift from Player A's favor to Player B's favor, simply by changing the number of stocks.

Here is a simple, yet realistic, counterexample to thinking the "better player" can be well-defined as the player with the greater chance of winning a given game, regardless of stock count. Suppose Player A is an excellent aggressive player, with good technical skill, that sometimes SDs. Let's say that the rate of his SDs is related to the rate of his kills on Player B as follows: over the course of him taking one stock from Player B, there is a 20% chance of him suiciding (and effectively throwing away, on average, half a stock, since at any given time Player A could have any fraction of his current stock remaining; this is a little fuzzy but acceptable). This is reasonable because the amount of damage/KOs dealt to the opponent is proportional to how much you play, as is the number of SDs suffered. Now, let's say Player B's style is heavily into trading hits and more defensive, and VERY consistent, and he never SDs. For every 10 stocks Player A takes from him, Player B manages to take 9 stocks in return.

Now, let's consider a 2-stock game. As long as Player A doesn't suicide before eliminating Player B's two stocks, he will win, since Player B's damage output is a little lower. The probability of this happening is (1 - 0.2)^2 = 0.64, so Player A has a 64% chance of winning the game. (Technically, I didn't prove that it was exactly 64%, only that it was at least 64%, since I didn't formally show Player A only wins if he doesn't suicide, but bounding it below at 64% is good enough for our purposes, as we'll see.)

Now, let's consider a 4-stock game. As long as Player A has at least one suicide, Player B will win, because Player A loses the equivalent of 0.5 stock per suicide, and Player B will take the other 3.5 of Player A's stocks while only losing (10/9)*3.5 < 4 of his own stocks. The probability of this happening is 1 - (1 - 0.2)^4 > 0.59, so Player B has a greater than 59% chance of winning the game. (A similar parenthetical note applies here as in the 2-stock game analysis.)

So, it is clear that in this simplified model, which still seems fairly realistic, the "better player" actually depends on the number of stocks per game.

Spoiler

This model is a little silly because it sort of pretends Melee is a traditional fighter with health bars, with its notion of continuous damage until a KO point, but it seems like a reasonable model for certain matchups + playstyles, where combos and gimps or other early kills are rare. Like, maybe bair-only Puff dittos, haha. Anyway, that isn't really relevant. If a simplified model can disprove a general statement, saying that Melee is more complex than my example assumes it is, without actually doing any analysis, doesn't really prove the general statement... just wanted to cut that off in case Kal tried; it would basically be saying "oh yeah you can't disprove what I said so I'm right."

Spoiler

I was going to use A = Mango and B = Armada, but that is unfair to the two players. Also, Armada would tap regardless of stock count.

None of this is relevant. Yes, given two players, A and B, the chance of winning an individual game will depend on stock count. I have said this before. However, your chance of winning an individual game is not equal to your chance of winning. Your goal isn't to win matches. It's to win sets.

Your "counterexample" suggests:

1) That you have a very poor understanding of basic math
2) That you are unable to comprehend basic English

As I've said, a billion times already, the probability of winning a single match will change according to the number of stock. This is almost guaranteed.

Now, allow me to write a proof that you can choose a sufficiently long set length to guarantee as great a chance of winning as you want. Note, however, that I am assuming a fixed probability between matches to avoid an overly cumbersome computation. As I said earlier, you can account for this by simply requiring the better player to be have a greater than 50% chance of winning on sufficiently many stages (this is already assumed in the current MBR ruleset, we just assume "sufficiently many" is two or three). Note also that I am assuming specific probabilities of particular players. I do not know whether there exists a single set-length that will guarantee the same success across all players. I will see about finding the actual E below for each probability p and taking the supremum across all corresponding N, though I have no guarantee the supremum is finite.

[COLLAPSE="Math Thing That Anand Probably won't Understand Because He Doesn't Know How to Read"]Definition: Given a game and two players, player A, and player B, player A is better than player B if his probability of winning is greater than 50%.

Theorem: Suppose player A is better than player B with probability p of player A winning. Then, for every q in (0,1) there exists an integer N such that the probability of player A winning a best of N set is greater than q.

Proof: In order to choose N, we must first discover the probability of player A winning with arbitrary odd n = 2k - 1. We do this via a binomial distribution: the probability of getting exactly m successes in n trials (i.e., winning exactly m times in a best of n-set) is:

nCm p^m(1 - p)^(n - m)

We see that player A wins k matches exactly nCk p^k(1 - p)^(n - k), times, k + 1 matches exactly nC(k + 1) p^(k + 1)(1 - p)^(n - (k+1)), and so on, so our probability of player A winning a best of n set is the sum, taken from i = k to i = n, of

P_n = nCi p^i(1 - p)^(n - i)

The limit of the above sequence (the sum of P_n from i = k to i = n) is 1 (the proof of this is rather technical, but is in PDF I link to at the end of this post). Equivalently, for every E > 0 there exists a natural number N such that for all n > N, |P_n - 1|< E. In other words, for every E > 0, there exists a natural number N such that for all n > N, 1 - P_n < E. Basic algebra gives:

P_n > 1 - E

Now, take E = 1 - q. As shown above, there exists a natural number M such that, for all n > M, P_n > 1 - E = 1 - (1 - q) = q. Take N = M + 1. Then P_N > q. QED
[/COLLAPSE]
tl;dr No, you haven't proven anything, except that "being better" will vary according to the number of stock, which is something I already said would happen. This doesn't make the definition less well-defined, because there is nothing wrong with "being better" depending on the number of stock. Different rulesets make different games, and you would expect players to not have a fixed probability of winning across all games. While you're clearly quite capable of stating your point of view with the fervor of a thousand angry feminists, this doesn't make your point of view sound. What makes it bull ****, on the other hand, is that you don't seem to understand what a "counterexample" is. The fact that you're a math major just makes this worse.

And here is the full PDF of the above proof, written in TeX.

 

[xianglongfa]

A simple modification of "better player" takes this into account. You're right, however, that if we are defining being better solely by the probability of winning a single match on a single stage, this becomes an issue. In that case, it suffices to change the definition to something along the lines of "player A is better than player B if the probability that player A wins is greater than 50% on at least half of the stages."

Yet your 'definition' doesn't even guarantee that the 'better' Player A would have a higher chance of winning in a BoX set against player B. (as player B could just have much stronger CP's than player A's) Learn to write something that makes sense before criticizing others' on their ability to read. Both of you should probably try to calm down a little bit and settle your disputes in later posts by clarifying your statements and assumptions more rigorously.

Your proof may or may not be relevant to your argument with Anand, but it was a good demonstration of your nice LaTeX skills. I'm impressed

 

[Kal]

For sufficiently large X, it does guarantee it, I think. Though the proof is likely much more convoluted than the one posted above. I don't think having a bunch of 99% guarantees wins would, for example, offset having more 51% chance wins, as long as you played a long enough set. I could be mistaken here, in which case you just need to modify the definition further. The point was to avoid addressing the issue because, while it can be accounted for it, it makes an otherwise simple issue rather complicated.

And keep in mind that I wrote "something along the lines of" to indicate that it was just an afterthought. Not necessarily something flawless.

 

[xianglongfa]

Yet for practical purposes, anything more than bo7 is probably not really worth considering.

 

[Kal]

Depending on the number of stock, I agree. I just want to point out to naysayers that a slight decrease in the success rate of better players might be justified by an overall more enjoyable game, and more so when it can be compensated for by an increase in set length. Especially when the increase in set length is more practical as a result of the new ruleset.

Again, posting your reservations with the ruleset is fine. Acting as though these reservations prove the ruleset is worthless, on the other hand, is silly. We should playtest it before we come to any conclusions about whether it is worthwhile.

 

[TPoint1BUA]

Super constructive addition to your post. Why are you acting like a creepy, passive-aggressive ex-girlfriend? We haven't even slept together yet. Take a chill pill.

I've explained this several times. You're getting stuck on nonsense because you stubbornly refuse to admit that an incorrect application of basic first-order logic does not mean I'm wrong.

Your "counterexample" suggests:

1) That you have a very poor understanding of basic math
2) That you are unable to comprehend basic English

Math Thing That Anand Probably won't Understand Because He Doesn't Know How to Read

What makes it bull ****, on the other hand, is that you don't seem to understand what a "counterexample" is. The fact that you're a math major just makes this worse.

And here is the full PDF of the above proof, written in TeX.

Super constructive addition to your post. Why are you acting like a creepy, passive-aggressive ex-girlfriend? We haven't even slept together yet. Take a chill pill.

Super constructive additions to your post. Why are you acting like a creepy, passive-aggressive ex-girlfriend? You haven't even slept together yet. Take a chill pill.

While it doesn't mean you're wrong, using ad hominem "arguments" also doesn't mean you're right.


Also, you still haven't addressed the issue of counter-picks.

 

[KrIsP!]

ad hominem.

No need to wiki search ad hominem for us, if someone doesn't know they can do it themselves. Besides he was just reacting to someone who was doing the same thing to him.

As for the matter of CPs, that's something that could be discussed, but overall can't really be proven until some extensive research(playtesting)is done. None of this can be actually, and while it's ok for people to theorize and state opinions there's no need for the hate as no one can be proven right.

The same points keep coming up and the same arguments are always made by different people with different views. I'm starting to feel like this thread is just running in circles while the Cactus is playtesting and others keep coming back to reexplain what has already been explained.

 

[Kal]

TPoint, I have, to a degree, addressed the counterpick issue: you need a definition of "better" than accounts for it. I think the following definition would work:

A better player is a player whose probability of winning a set played on all legal stages exactly once in sequence is greater than .5.

It's possible this has a flaw I haven't noticed. The definition I just posed ignores counterpicks and that the losing player gets to choose the character after the winning one. Any model can be criticized for being too simple. This doesn't make the model wrong. Mostly, I'm just not addressing the counterpick issue because it needlessly complicates things. This doesn't mean it can't be accounted for.

Also, an ad hominem is insulting your opponent in place of an argument. "That guy is wrong, he's clearly an idiot," is an ad hominem. "Anand, you're wrong, here's why. [insert reasons]. Also, you're an idiot" is not an ad hominem.

 

[-ACE-]

Kal you are truly an expert at word vomiting. I've never seen such a worthless post on these boards and that is saying quite a bit. Everyone here is now dumber for having read that if they were somehow bored enough to do so. I award you no points, and may God have mercy on your soul.

 

[Kal]

Thanks -ACE-. As a good rule of thumb, I disagree with whatever stance you take on any issue. That you think my posts are worthless just reinforces my opinion that I am right. Especially since you seem to have a hard time thinking.

 

[xianglongfa]

I think that while we're on the topic of discussion of BoX's, it's worth clarifying the following important notion as an aside. Just as clarification, this has little to do in particular with the comparison between our current ruleset, and the new proposed ruleset.

Given:
Fixed stock count
Fixed stagelist
Unfluctating player stamina

Definition: The better player is the player with the higher weighted average win percentage across all valid stages. (with non-uniform stage weighting)

This is just to give us something to work with.

It can be seen that stage balance as well as the notion of who is the better player are intimiately related with the X in a BoX construct as well as the other rules (neutrals/CP's/DSR, etc.) which are in theory all free to vary, but for now lets focus on X.

Goal: Choose a ruleset such that the probability of the 'better' player winning the set should be as high as possible within practical limitations.

Given two players who are very close in skill,(within the framework of my 'better' definition along with the same givens), increasing X in BoX will often boost the 'better' player's chances of winning the set drastically. However, given DSR, limited stage count, and varying degrees of strength of different CP stages for any given matchup within the framework of any BoX number (say X<7 or even 11), many simplifying assumptions break down in justifying arguments matchup specifically.

Practical example: Compare 3 stock bo7 with 4 stock bo5, assuming both are time-viable, as to which grants the 'better' player a higher chance of winning the set.

P.S. I know it may be disagreeable to compare two vastly different games because my "betterness" notion is going to be vary depending on stock format.

The most important notion I want people to get out of my post is this:

I'm going to make a lot of simplifying assumptions to make the following calculation very tractable.

Say we only have 1 stage. Best-of-X matches. Say that player A has a 55% chance of winning on this stage against player B with all factors considered fixed. How often do you think player A would win in a bo3? A bo5? What about a Bo7? How much do you think these probabilities vary? Do you think the disparities between these probabilities of winning the overall set give credence to increasing X more earlier in the bracket? What about for semifinals and onwards? Do you think the increase should lend us to favor Bo7 for semi's and onwards with our current (4 stocks, 5+1 stages)?

 

[-ACE-]

In Kal's opinion, he is right.

Lmao

 

[Kal]

Increasing the set length a reasonable amount really only improves your chances if you are a significant favorite. Again, binomial distribution for simplicity, your chances of winning for the following set-lengths are:

Best of 1||Best of 3||Best of 5||Best of 7||Best of 9
50%||50%||50%||50%||50%
55%||57.5%||59.3%||60.8%||62.1%
60%||64.8%||68.3%||71%||73%
65%||71.8%||76.5%||80%||82.8%
70%||78.4%||83.7%||87.4%||90.1%
75%||84.4%||89.6%||92.9%||95.1%
80%||89.6%||94.2%||96.7%||98%
85%||93.9%||97.3%||98.8%||99.4%
90%||97.2%||99.1%||99.7%||99.9%
95%||99.3%||99.9%||99.9%||99.9%
100%||100%||100%||100%||100%

Interestingly, since we expect matches to be close in finals, it would technically save time and keep roughly the same results to have those sets be shorter. And we would improve results by having earlier sets in bracket be longer. Sadly, nobody wants to watch a best of 1 where a loss resulted from a single mistake, even if results are not far off from what was expected. And there is merit in claiming that this analysis breaks down as the probability of winning between matches is more likely to change as you go farther down the bracket.

Here are graphs of the probabilities as well:

[collapse="Best of One"]

[/collapse]
[collapse="Best of Three"]

[/collapse]
[collapse="Best of Five"]

[/collapse]
[collapse="Best of Seven"]

[/collapse]
[collapse="Best of Nine"]

[/collapse]
[collapse="Simultaneous Graph"]

[/collapse]
And, for the lulz:

[collapse="Best of Nine-Hundred Ninety-Nine"]

[/collapse]

 

[n1000]

I played some more 2 stock matches and they were p. sick, and fun.

 

[xianglongfa]

I did the same calculations myself through summing negative binomials and verified the numbers of one of your rows.(to be the same) The results of these calculations are interesting enough, and certainly do challenge our natural intuition on a basic level. Kal you again impress me on your quick calculation/formatting ability. You would be a great asset to any MCM team

 

[Kal]

In all fairness, Mathematica makes these computations pretty easy.

 

[-ACE-]

Kal, if you want to post something useful, post another chart containing only common sense, except this time, have it involve # of stocks; compare this ruleset to the one we play with today.

 

[n1000]

post another chart containing only common sense

in western economic systems it is considered to be the case that money CAN NOT* be "commonly held" so I'm not sure how Kal could hope to "contain" such a non-concept in chart form.

*one might say credit unions/co-ops are an exception to this claim but in fact they exist w/in the capitalist economic system and current theory says there are simply vehicles for utilizing personal wealth, i.e., the money is neither stagnant, nor privately or communally held but held in a self-perpetuating economic loop.

**also that was a weird way of referring to public funds

 

[xianglongfa]

Kal, if you want to post something useful, post another chart containing only common sense, except this time, have it involve # of stocks; compare this ruleset to the one we play with today.

Taking our base unit to be stock instead of match and let it be an iid Bernoulli by simplifying assumption, we can think of winning a match as the sum of negative binomials, just as with matches in a set. Probability of winning a match would be 1-(Pr(k=1)+...+Pr(k=r-1)) for a NegBin(r,p) r.v. where p is the probability of "winning a stock".

This simple model becomes more truthful the higher level the players, and the more combo/punishment heavy a matchup is. (as close to Armada/Mango as possible) With the aforementioned qualities, this simplified model roughly equates to a model which takes p to be the probability of getting the first comboing hit. (other hits producing neglibible damage strings/changes in stage position not considered)

The problem with reality is that most character matchups have the tendency to depend on invincibility on smaller stages (for faster characters who can more easily punish on reaction), and some combination/balance of punishment ability (which involves and damage and edgeguard setup/execution) and neutral game (the 1st hit game), among other factors. (think Armada's peach v. Mango's fox on Dreamland) This makes stock count dependency of matchups in general a very hard structure to model using any easy simplifying assumptions, since all these additional factors involve residual rounding and strategies may even vary depending on stage, relative stock/percentages, and the timer.

 

[Dan Salvato]

in western economic systems it is considered to be the case that money CAN NOT* be "commonly held" so I'm not sure how Kal could hope to "contain" such a non-concept in chart form.

.....What?

 

[Kal]

ACE is on my ignore list, so if you guys don't quote him, he can't bother us. Well, he can't bother me at least. >_>

I went ahead and added graphs to the earlier post. They won't make things any clearer. I just like graphs. I'm sort of new to plotting in Mathematica so I'm sorry if they're not great.

 

[-ACE-]

Lmao, so kal couldn't do what I asked, but instead provided more useless graphs. I would think someone bragging about having a math major would bring something to the table that was beyond high school level. I guess he expects me to prove I took statistics and algebra too now, lmao. I should be the one blocking him, if I were that childish.

 

[Kal]

Ok, I figured out this Negative Binomial Distribution and churned out a probability table for best of one through best of nine for two, three and four stock Melee. There are two major assumptions:

1) That the probability of taking a stock is independent. This is clearly false, but I currently see no reasonable way to account for it.
2) That the probability of winning a match is independent. Again, false, but it would be overly complicated to try and take this into consideration.

From left to right, the games are two, three and four stock Melee, respectively. We see that a best of seven with the two-stock ruleset very closely resembles the best of three with four stock. On the other hand, to get the same consistency as best of five with four stock, the two stock game would have to go best of eleven. These results are not totally out-of-nowhere, as we expect similar levels of consistency with games whose total stock are close. Two stock with best of seven makes 28 total stock, and four stock with best of three makes 24 total stock. Similarly, two stock with best of eleven makes 44 total stock, and four stock with best of five makes 40 total stock.

The computations were done and spit out using python, so I am reasonably sure there are no errors outside of rounding ones (i.e., that your chances of a winning a best of 7 in two stock Melee are 100% when your chance of winning a stock is 85% is clearly false, but that's what I got for making things legible). If there are any errors discovered, please let me know, and if anyone wants to see the .py file to verify the logic (I suck at probability, so it's not unlikely that I messed up writing the probability distribution function) just ask and I will upload it.

Probability of Taking a Stock||||Best of 1||Best of 3||Best of 5||Best of 7||Best of 9||||Best of 1||Best of 3||Best of 5||Best of 7||Best of 9||||Best of 1||Best of 3||Best of 5||Best of 7||Best of 9

50.0%​

||||50.0%||50.0%||50.0%||50.0%||50.0%||||50.0%||50.0%||50.0%||50.0%||50.0%||||50.0%||50.0%||50.0%||50.0%||50.0%

55.0%​

||||57.5%||61.1%||63.8%||66.0%||67.9%||||59.3%||63.8%||67.1%||69.7%||71.9%||||60.8%||66.0%||69.7%||72.6%||75.1%

60.0%​

||||64.8%||71.6%||76.2%||79.7%||82.5%||||68.3%||76.2%||81.3%||85.0%||87.8%||||71.0%||79.7%||85.0%||88.7%||91.3%

65.0%​

||||71.8%||80.7%||86.0%||89.6%||92.2%||||76.5%||86.0%||91.2%||94.2%||96.2%||||80.0%||89.6%||94.2%||96.7%||98.0%

70.0%​

||||78.4%||88.0%||92.9%||95.7%||97.3%||||83.7%||92.9%||96.7%||98.4%||99.2%||||87.4%||95.6%||98.4%||99.4%||99.7%

75.0%​

||||84.4%||93.4%||97.0%||98.6%||99.3%||||89.6%||97.0%||99.1%||99.7%||99.9%||||92.9%||98.6%||99.7%||99.9%||100.0%

80.0%​

||||89.6%||97.0%||99.0%||99.7%||99.9%||||94.2%||99.0%||99.8%||100.0%||100.0%||||96.7%||99.7%||100.0%||100.0%||100.0%

85.0%​

||||93.9%||98.9%||99.8%||100.0%||100.0%||||97.3%||99.8%||100.0%||100.0%||100.0%||||98.8%||100.0%||100.0%||100.0%||100.0%

90.0%​

||||97.2%||99.8%||100.0%||100.0%||100.0%||||99.1%||100.0%||100.0%||100.0%||100.0%||||99.7%||100.0%||100.0%||100.0%||100.0%

95.0%​

||||99.3%||100.0%||100.0%||100.0%||100.0%||||99.9%||100.0%||100.0%||100.0%||100.0%||||100.0%||100.0%||100.0%||100.0%||100.0%

100.0%​

||||100.0%||100.0%||100.0%||100.0%||100.0%||||100.0%||100.0%||100.0%||100.0%||100.0%||||100.0%||100.0%||100.0%||100.0%||100.0%

Hopefully this proves useful or, at the very least, interesting.

One final addition is what the maximum possible time-spent for each game-mode is, assuming we use the standard rule of allocating a minute for one fewer than the total number of stock:

Number of Stock||Best of 1||Best of 3||Best of 5||Best of 7||Best of 9
2 Stock||3 minutes||9 minutes||15 minutes||21 minutes||27 minutes
3 Stock||5 minutes||15 minutes||25 minutes||35 minutes||45 minutes
4 Stock||7 minutes||21 minutes||35 minutes||49 minutes||63 minutes

which gives you a set of upper bounds that allows you to look at what possibilities are logistically feasible, as well as to assess whether the difference in consistency is worth the difference in time.

 

[Dan Salvato]

That's an interesting table. Putting a little more application into it, picking a really strong CP would add maybe 10-15% probability to the losing player. This is, I think, enough of a gap to fairly determine who the better player truly is between the two and have that player win the match. If the better player is 15+% better, then he will likely defeat the losing player even on unfavorable CPs. If the player is less than 10% better, then the other player deserves the possibility of winning, and whoever wins on the neutral will probably win the set.

I think this only applies in Cactuar's ruleset because with the current stage list, CPs aren't strong enough to give the player a significant (10-15%) advantage. Either way, I'd be interested in seeing a chart that somehow takes CPs into consideration for the probability.

 

[xianglongfa]

1) That the probability of taking a stock is independent. This is clearly false, but I currently see no reasonable way to account for it.
2) That the probability of winning a match is independent. Again, false, but it would be overly complicated to try and take this into consideration.

For the above simplifying assumptions to be as reasonable as possible, think high level falco dittoes on FD only. (maybe on point[mango v. pp])

 

[Kal]

That's an interesting table. Putting a little more application into it, picking a really strong CP would add maybe 10-15% probability to the losing player. This is, I think, enough of a gap to fairly determine who the better player truly is between the two and have that player win the match. If the better player is 15+% better, then he will likely defeat the losing player even on unfavorable CPs. If the player is less than 10% better, then the other player deserves the possibility of winning, and whoever wins on the neutral will probably win the set.

I think this only applies in Cactuar's ruleset because with the current stage list, CPs aren't strong enough to give the player a significant (10-15%) advantage. Either way, I'd be interested in seeing a chart that somehow takes CPs into consideration for the probability.

I can see about writing something more general in python that allows you to input what the probability of winning a counterpick is. I could also allow you to input what the probability of winning a stock is, given how many you've taken without losing any. Of course, any choice for these inputs would be pretty hard to justify.

 

[xianglongfa]

Building a robust Bayesian model for this would be a fun project. Modeling the neutral game with Markov matrices would be another.

 

[Kal]

Ugh, I hate probability.

 

[Cactuar]

Just to point this out, the closer the probability of winning is to 50% in non-even matchups, the bigger the skill difference between two players must be to gain higher odds of winning.