I have a new idea based on 12 games all at once. Who wins most is the Dominator. I'm also going to suggest how to break the ties. I'm suggesting games with 5 players and every player would play 5 games.
For example in this way:
Game 1:
1 lifeinpixels
2 Spartakus
3 Matty
4 thaithai
5 Antonis_xania
Game 2:
6 nikeboix69
7 Dferguson
8 suldam
9 copperbird
10 Fendi
Game 3:
11 Joca
12 oliver
1 lifeinpixels
2 Spartakus
3 Matty
Game 4:
4 thaithai
5 Antonis_xania
6 nikeboix69
7 Dferguson
8 suldam
Game 5:
9 copperbird
10 Fendi
11 Joca
12 oliver
1 lifeinpixels
Game 6:
2 Spartakus
3 Matty
4 thaithai
5 Antonis_xania
6 nikeboix69
Game 7:
7 Dferguson
8 suldam
9 copperbird
10 Fendi
11 Joca
Game 8:
12 oliver
1 lifeinpixels
2 Spartakus
3 Matty
4 thaithai
Game 9:
5 Antonis_xania
6 nikeboix69
7 Dferguson
8 suldam
9 copperbird
Game 10:
10 Fendi
11 Joca
12 oliver
1 lifeinpixels
2 Spartakus
Game 11:
3 Matty
4 thaithai
5 Antonis_xania
6 nikeboix69
7 Dferguson
Game 12:
8 suldam
9 copperbird
10 Fendi
11 Joca
12 oliver
This is just using the classification order but also a random matching can be fine.
8, 9 or 10 starting territories are fair in my opinion, having 5 players it means to chose maps with 40, 45 or 50 territories. There are 14 maps that match this: Basket, Caribbean Double, East Asia, Eld World, Europe, Falkland, House, Korea-Japan, South Africa, Tor, United States, USA, World Classic, World Modified. Maybe World Classic (and its Modified version) can be excluded giving an advantage of the knowledge of the specific maps of the site. So we can set 12 games in 12 different maps, randomly chosen.
12 players, 12 games, 12 maps. Do you like the idea?All at once since it's like 1 round tournament: I mean they would take long like playing 1 game only.
Being 5 games with 5 players the average result is to win 1.
For example a common outcome can be like this (just an example):
2 games won:
-lifeinpixels
-Spartakus
-Matty
-thaithai
1 games won:
-Antonis_xania
-nikeboix69
-Dferguson
-suldam
0 games won:
-copperbird
-Fendi
-Joca
-oliver
In a context like this who wins 3 (or more) games will be probably the winner. And a player winning 3 games out of 5 can occur much more often than a player winning 2 games out of 2. But still there are several possibilities of a tie.
I'm studying how to break the ties without playing a final game, which it would be weird in case of 2 players and stalemating in case of 3 players (but even 4...). And in any case it would take long.
First I thought about the 'direct match comparison' (I don't know how you call it in English) that consists to compare the played games (of the tournament) where both the tied players were in. In case of 2 players is simple, in case of 3 or more still it can be done but it's a bit more complex. Btw I'm not a fan of this method.
A second method that can be added or completely substitutes the previous is based on who (among the players tied with most games won) has those games faster, counting the rounds delayed to win. For example, let say, lifeinpixels wins 3 games and Spartakus too. Lifeinpixel have won a game in 16 rounds, another in 12 and another in 7, ok, totally are 35 rounds; Spartakus 12 rounds, 23 and 15, totally 50 rounds: lifeinpixels has been the player who won most games in less rounds, then he is the winner.
A third method that can be used in case the previous fails or also always matched with that. It consists of counting how many rounds a players survived in the games he lost ('minus' how many rounds he delayed to win in the games he won).
In few words. Who win most games is the winner. In case of a tie it would be awarded the fastest winner (the least rounds to win) or the longest loser (the most rounds to die).
Can be good?
I have a new idea based on 12 games all at once. Who wins most is the Dominator. I'm also going to suggest how to break the ties. I'm suggesting games with 5 players and every player would play 5 games.
For example in this way:
[spoiler]Game 1:
1 lifeinpixels
2 Spartakus
3 Matty
4 thaithai
5 Antonis_xania
Game 2:
6 nikeboix69
7 Dferguson
8 suldam
9 copperbird
10 Fendi
Game 3:
11 Joca
12 oliver
1 lifeinpixels
2 Spartakus
3 Matty
Game 4:
4 thaithai
5 Antonis_xania
6 nikeboix69
7 Dferguson
8 suldam
Game 5:
9 copperbird
10 Fendi
11 Joca
12 oliver
1 lifeinpixels
Game 6:
2 Spartakus
3 Matty
4 thaithai
5 Antonis_xania
6 nikeboix69
Game 7:
7 Dferguson
8 suldam
9 copperbird
10 Fendi
11 Joca
Game 8:
12 oliver
1 lifeinpixels
2 Spartakus
3 Matty
4 thaithai
Game 9:
5 Antonis_xania
6 nikeboix69
7 Dferguson
8 suldam
9 copperbird
Game 10:
10 Fendi
11 Joca
12 oliver
1 lifeinpixels
2 Spartakus
Game 11:
3 Matty
4 thaithai
5 Antonis_xania
6 nikeboix69
7 Dferguson
Game 12:
8 suldam
9 copperbird
10 Fendi
11 Joca
12 oliver
This is just using the classification order but also a random matching can be fine.[/spoiler]
8, 9 or 10 starting territories are fair in my opinion, having 5 players it means to chose maps with 40, 45 or 50 territories. There are 14 maps that match this: Basket, Caribbean Double, East Asia, Eld World, Europe, Falkland, House, Korea-Japan, South Africa, Tor, United States, USA, World Classic, World Modified. Maybe World Classic (and its Modified version) can be excluded giving an advantage of the knowledge of the specific maps of the site. So we can set 12 games in 12 different maps, randomly chosen.
[u]12 players, 12 games, 12 maps. Do you like the idea?[/u]
All at once since it's like 1 round tournament: I mean they would take long like playing 1 game only.
Being 5 games with 5 players the average result is to win 1.
For example a common outcome can be like this (just an example):
[spoiler]2 games won:
-lifeinpixels
-Spartakus
-Matty
-thaithai
1 games won:
-Antonis_xania
-nikeboix69
-Dferguson
-suldam
0 games won:
-copperbird
-Fendi
-Joca
-oliver[/spoiler]
In a context like this who wins 3 (or more) games will be probably the winner. And a player winning 3 games out of 5 can occur much more often than a player winning 2 games out of 2. But still there are several possibilities of a tie.
I'm studying how to break the ties without playing a final game, which it would be weird in case of 2 players and stalemating in case of 3 players (but even 4...). And in any case it would take long.
[spoiler]First I thought about the 'direct match comparison' (I don't know how you call it in English) that consists to compare the played games (of the tournament) where both the tied players were in. In case of 2 players is simple, in case of 3 or more still it can be done but it's a bit more complex. Btw I'm not a fan of this method.
A second method that can be added or completely substitutes the previous is based on who (among the players tied with most games won) has those games faster, counting the rounds delayed to win. For example, let say, lifeinpixels wins 3 games and Spartakus too. Lifeinpixel have won a game in 16 rounds, another in 12 and another in 7, ok, totally are 35 rounds; Spartakus 12 rounds, 23 and 15, totally 50 rounds: lifeinpixels has been the player who won most games in less rounds, then he is the winner.
A third method that can be used in case the previous fails or also always matched with that. It consists of counting how many rounds a players survived in the games he lost ('minus' how many rounds he delayed to win in the games he won).[/spoiler]
In few words. Who win most games is the winner. In case of a tie it would be awarded the fastest winner (the least rounds to win) or the longest loser (the most rounds to die).
Can be good?
«God doesn't play dice with the World» ~ Albert Einstein