Concept: Round 1, there are 10 mystery doors. At the beginning of each round, each and every player must PM me their door choice (from 1-10). When a player enters a door, they will encounter a unknown number of other players (all those choosing the same door). Each round only 1 player can leave the door. If no other players choose the door, then the same rule applies of only 1 player can leave the door, and that player gets a bye. [So for example in round 1 if 4 different players PM me the number "2" and only 1 player PMs me the number "5" , the 4 players will be in a 4 player game, and the 1 player will get a bye]. If more than 8 players choose the same door, then it will be divided in to two games and the two winners will play in another final match.
Round 2, there will be 8 mystery doors. Round 3, 6 mystery doors. And so on and so forth.
Once two players are left, they will automatically chose the same door. If (and only if) the last two players have played less than 3 games, the last round will be a best of 3.
Settings: Game Type: Standard Settings: Automatic, Seq, Chained, Fog and esc spoils.
Map: Random of course.... otherwise it wouldn't be a mystery.
Fuzzy316 wrote:In please, but one question as far as the doors go.
If only one person can go through the doors, what if there are less than 6 players after 2nd round? Say there are only 3 or 4 players.. how many doors would there be?
Doors decrease by 2 each round. So like you said round 2 might only have 6 players. All six players might choose separate doors and then all get byes. The next round the doors would decrease again and thus they will eventually have to play each other.