How does chest looting work?

alexo28·7/4/2016, 12:45:10 AM·1 votes·1,896 views

What are the chances to get a chest when getting S(+-) on a champion?

I mean I got a single chest when the hextech crafting came out then I NEVER got a chest again

I am a main support, so I probably am not the type of player that get S on every games, but I still got S at least 5 times since the crafting update was released and never got a chest

I see a lot of players complaining about "too few keys" but WTF I had 5 keys a few days ago before deciding to spend some RP on chests, plese transfer some luck from keys to chests please ._.

So how does this work?

(English is not my primary language, so please dont look at errors in my grammar and structure, but you can still tells them to me so I can improve myself ^^)

7 Comments

GaleWinUnleashed7/4/2016, 12:49:50 AM2 votes

There is no luck or chance involved with getting chests. Each week a new chest becomes available for you to earn, up to a cap of 4 unearned chests at once. If you get an S-rank on a champion while one is available, you get a chest. However, you can only earn one chest per champion per season.

Check your champion list. If you see a golden border around the champion icon, it means you've already earned a chest for that champion.

Example: http://i.imgur.com/MU0GWct.png?noredirect

https://i.imgur.com/1HvNiEO.png?noredirect

Deep Terror Nami7/4/2016, 12:47:35 AM1 votes

You can find all the information you need on the article here; https://support.riotgames.com/hc/en-us/articles/207884233-Hextech-Crafting-Guide

Flaherty7/4/2016, 12:48:00 AM1 votes

Go to profile page, look on the right and it shows you how many chests you have available and how many days until your next one is available. You can only get a chest from earning at least an S- in a normal 5v5 or ranked 5v5 and you can only earn 1 chest per champion per season. So you should start with 4 chests available, that means you need to get an S- or higher with 4 different champions to earn all the chests available.