VS Data Reaper #303: Shaman Supremacy Edition

Altair you forget about Deathrattles spawning other minions or effects etc. I’ve seen boards of that sort making it worse for the opponent by throwing an AOE; it’s almost always better to have a bigger board; at the very least: as I said initially: you don’t know if they even have clears (people get unlucky).

I don’t currently play the deck but I play a similar deck of sorts; yes sometimes I’m just cleared horribly; but I’ve lost count of the times the opponent just didn’t have the AOE damage high enough to do it or they just got bad draws and couldn’t clear.

If a deck becomes more popular, then the number of mirror matches increases. This pulls winrate towards 50%, meaning a decrease if it’s over and an increase if it’s under.

That said, most people who are aware of this effect exaggerate its effect. Yes, as popularity→100%, winrate→50%. But that’s an extreme case that will never actually happen.

The formula for non-mirror winrate is
w’ = (w-.5p)/(1-p)
where w’ is non-mirror winrate, w is overall winrate, and p is the popularity of the deck in question. 0 < w < 1, 0 < p < 1.

For example, in VS #303, Reno Druid has an overall winrate of 51.44% and 7.62% popularity in T1KL, giving it a non-mirror winrate of 51.56%. A 0.12% difference. Whoop de doo.

So in this case you are pointing to an effect that is, technically, real, but more importantly it is trivial. It can safely be ignored.

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Just noticed this bullying. Ah it’s still the first half the month. I guess the narcissism will drop again at the end of the month when the better players start ranking.

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Brutal.

I’d say something but I wouldn’t want to pressure anyone to maintain a high rank.

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As I said, the more you wait to stack more spell schools, the higher the chance they draw their AoE, and the payoff is worth less, not more

I say to you that if I have more of something, each additional unit of that something is worth LESS to me, and what do you do? You go and say there are potential DEATHRATTLES to take into account.

What part of “every additional unit is worth less to me” do you not understand?

If you want chances of getting a deathrattle specifically in that one or two spots more, count the number of deathrattle minions and divide it with the number of all 5-drops in the pool. That’s what you’re looking for, lol.

Anyway, suit yourself. At this point I’m no longer having fun. I get no “thank you” for teaching you anything, anyway.

I’m not trying to pressure him to rank. I’m saying it’s nonsense to bully people because he supposedly has a rank; it makes him a worse player being that flamboyantly narcissistic; and it’s a waste of time having half of the forum being “do you know my rank boi?”.

The good players balance it; they do better than the bad players at least; maybe they’ll only spawn 3 with Razzle-Dazzler to be fast but the bad players would be able to spawn 2 or 1 or wait too much.

“Brutal” wasn’t a compliment. Simply an observation.

A light feather, compared to jumping up and down around the forum, shouting “do you know my rank boi?”.

It wasn’t a criticism either.

We do not care for this effect. Rainbow shaman doesn’t have that problem. It’s a surprise if anyone has ever played a Rainbow Shaman mirror xD

So in this case you are pointing to an effect that is, technically, real, but

I wasn’t pointing to that effect. How many mirrors are played doesn’t change the total number of games played of that deck.

I mean, it does in this formula, but it’s also corrected in that formula (by calculating mirror and non-mirror winrates separately)

Absolutely. Agreed.

If this is the formula they use, it’s true that the sample size doesn’t matter.

I’m having trouble understanding it, though.

Why /(1-p) instead of /p?

1-p represents the percentage of all the games tracked which didn’t include rainbow shaman. How is that number relevant to the winrate of rainbow shaman?

I’m curious, don’t get me wrong, that denonimator changes everything. I want to understand how that’s a viable formula, and why I was wrong, since it’s simple enough to understand that when you divide with p, the higher the p, the lower the quotient.

It seems…fitting that would be changed to the exact complement of (1-p), one which increases the quotient with increase of p. Ironic, even.

Also, why 0.5p? Where’s the other half?

Please do notice how I’m trying to be polite and unassuming here, but there’s definitely a lot fishy with the formula. It’s just not something which is consistent with logic. Or maths, for that matter.

Gotta say I love rogue but Sonya needs to go it restricted the fun things and gets rogue nerfed

We start with overall winrate = non-mirror winrate × non-mirror popularity + 50% × mirror popularity, then solve for non-mirror winrate

w = w’ × (1-p) + 50% × p
w-.5p = w’ × (1-p)
(w-.5p)/(1-p) = w’

How is that compatible with

This?

According to your 1st quote, p is the popularity of a deck.

According to your 2nd quote, p is the mirror popularity.

So, which one is it?

The chance of a deck facing the mirror is the same as its own popularity

Lol, no, of course it’s not. It’s proportional to it, yes, but it’s not the same.

No, if you’re playing Reno Druid, then the chance you’re playing against Reno Druid is 7.62%, T1KL , timeframe covered by VS #303.

It’s exactly the same.

It’s not exactly the same, because your games are a part of the total population of the deck, which lowers your actual mirror frequency slightly.

Anyway, on to the more important issues.

The formula you posted is always increasing as long as x < 1, which suggests that the higher the deck popularity, the higher the non-mirror winrate of the deck (or vice versa).

I admit that I was wrong about the direction in which sample size impacts the deck winrate. It doesn’t decrease as the size increases.

But, you were wrong every time in your life when you posted that the popularity of the deck isn’t an indication of the deck’s strength, and the formula you posted literally proves it.

Apparently, the higher the non-mirror winrate of a deck, the higher its’ popularity. The higher in the ladder we go, the more that statement is correct. People just (correctly) assume they’ll play against less mirrors than non-mirrors, so playing the most popular deck is the most effective strategy, although not optimal.

If anything, bigger weight needs to be put on frequencies in top 1k than on tier lists.

That’s my honest conclusion of this debate.

EDIT:

If anything, bigger weight needs to be put on frequencies in top 1k than on tier lists.

Actually, belay that.

Frequencies in top 1k = tier lists.

It’s that simple.

Ofc, there will always be decks which can be strong and SHOULD be much more popular than they are, in theory, but in practice it doesn’t work that way for some reason. My sludgelock should be more popular in all ranks until top 1k, but it’s not. Maybe it’s too hard, maybe it’s too boring, maybe it’s both. Doesn’t matter. If 99,9% of the playerbase finds a deck too boring or too hard to be worth playing, then it’s not worth talking about.

This is too stupid of a position for me to believe that you believe it. It doesn’t even align with all that you’ve said you believe beforehand. I think you’re either deliberately trolling or you’re somehow subconsciously frustrated enough with me to troll on a subconscious level.

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by that logic, odyn warrior would be Tier 0 everywhere despite having garbage winrate

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I’m curious why no one has pointed out one of the obvious flaws of the arguments about “low play rate at 1-2%, deck is insignificant” ???

If there are a solid 25 decks in the meta, a 4% play rate would just be normal. In fact, that would be a bit high since other non-solid meta decks are also being played.

So one of those decks in the 25 deck meta at a 1-2% play rate actually isn’t bad. If anything, that’s exactly what you’d expect some play rates to be, even the good decks.

All this talk about 1-4% play rates being low play rates is such a gaslighting lie when this meta is diverse with tons of decks seeing play.

We’re not in a meta where only 10 decks see play.

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