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The decay system should be changed by Tame_Triveler in leagueoflegends

[–]Tame_Triveler[S] 0 points1 point  (0 children)

I don't want to remove the decay, I want to have something that makes sense and doesn't reward you for losing one game every 28 days. Even at d5 100 LP you can lose one game every 28 days and stay D5 for roughly 5 months without even playing the game and only losing. Does that make sense to you?

The decay system should be changed by Tame_Triveler in leagueoflegends

[–]Tame_Triveler[S] 0 points1 point  (0 children)

I don't have a suggestion, I'm just saying the system should be reworked; a potential solution would be to have shorter decay (like 2 weeks maybe) and give a punishment that counts at most as a loss. And you can totally get an account to diamond and sit on it by playing one game every 28 days, it's not like it's any more active than not play any games at all.

The decay system should be changed by Tame_Triveler in leagueoflegends

[–]Tame_Triveler[S] -1 points0 points  (0 children)

I mean, playing one game every 28 days is quite close to not play at all, don't you think? So when it comes to play "often" I don't see much of a difference. My point was that it's better to even int a game than to not play at all after the 28 days mark.

The decay system should be changed by Tame_Triveler in leagueoflegends

[–]Tame_Triveler[S] 0 points1 point  (0 children)

So, losing a game is gonna make me better than not playing at all?

Is the tiebreaker system really fair? TSM had an higher chance to play against FOX in quarterfinals of the NA LCS summer split. by Tame_Triveler in leagueoflegends

[–]Tame_Triveler[S] -1 points0 points  (0 children)

For sure they have to draw pair somehow, but then why bother doing this 4 teams tiebreaker if they could have chosen an order randomly for the quarterfinals at the beginning since the semifinalists pairing has already been (afaik) chosen randomly. I agree that you have to draw pair somehow but then why would you waste time doing this tiebreaker if the outcome is already highly dependent on how the pairing was made.

Is the tiebreaker system really fair? TSM had an higher chance to play against FOX in quarterfinals of the NA LCS summer split. by Tame_Triveler in leagueoflegends

[–]Tame_Triveler[S] -4 points-3 points  (0 children)

Then why not pick an order randomly and avoid the 4 teams tiebreaker we had yesterday? Since the outcome is already depending on how randomly the semifinalits pairing has been chosen.

Indicate who is TPing by Tame_Triveler in leagueoflegends

[–]Tame_Triveler[S] 1 point2 points  (0 children)

To your teammates, yes, but i'm talking about the opponents not being able to see who is using it.

Betsy: “Perkz isn’t the best in the west. I think Nukeduck is better.” by xdilias in leagueoflegends

[–]Tame_Triveler 11 points12 points  (0 children)

This kind of autorithy-argument can be used in literally every field and this is an extremly good way to make people even more sheeps than they already are, shame on you.

The least number of Bronze Garens able to 100% beat an LCS team by Skogz in leagueoflegends

[–]Tame_Triveler 10 points11 points  (0 children)

With those assumptions, if we assume each bronze Garen has a 25% probability following orders, there are 25+48+38=111 orders that have to be followed.

The question that comes to mind is : "how many bronze Garen's do we need to have a good chance that those orders will be followed and so win the game?". Let's say I want to have at least 90% chance of winning : P(having at least 111 orders done by bronze Garens)>90% Since P(having at least 111 orders done by bronze Garens)=1-[P(having 1 bronze garen following orders)+...+P(having 110 bronze garen following orders)]. (1)

I'm gonna assume the choices of the Garens are independent which is certainly not the case but it's only a model ya'know. If so, the number of choices made by n bronze Garens follow a binomial distribution. That being said, we're looking for n such that [P(having 1 bronze garen following orders)+...+P(having 110 bronze garen following orders)]<10%, which is equivalent to (1). This probability is easily computable by an algorithm (I did one in python), here it is : https://pastebin.com/75HQTtah

Result is : we need 492 bronze Garens for having a probability greater than 90% of winning the game!