I sent this email to FFG but I would also like opinions off of this board. Any help will be appreciated, thanks all.

Attention FFG Staff:

Please bear with me as I ask this question. We had a 5 person tournament using the season 3 box. Player M sat out and players A, J, B, and D all played against each other.

At the end of round one:

J 1-0

B 1-0

A 0-1

D 0-1

Player A sat out the next game and player M played against player D

At the end of round 2

J 1-1

B 2-0

A 0-1

D 0-2

M 1-0

At this point player D was knocked out of the tournament.

At the end of round 3 (the final round)

J 2-1

B 2-1

M 2-0

A 0-2

D 0-2

Ok so...we all looked at the tournament scores for how wins worked and we did not know how to give points to the players that couldn't play in the first two rounds.

The game that B and M played against each other was super close and M only won by 10 points.

How do you give points to a player that could not play in a round if you have an odd number of players?

We decided that the battle for 1st was to close to call between B and M. With the games that they played B had 276 points and M had 186 points. What we ended up doing was rolling 5 attack dice and whoever got the most hits was the winner. B rolled the most dice and got first. Any help the FFG staff can provide would be appreciated.

I ask all of this because I was the tournament organizer. Thank you for your time

I trust you'll share their answer here as well please?

I'm not a tourney official, but I feel like this was done wrong. When a person sits out of a game, its a bye, and they would receive a match win for sitting out. Bye players should always be the lowest man on the list. So honestly M should of won with the amount of rounds you played due to not being able to play all 3 rounds.

Edit: Meaning no person on this list should equal less then 3 matches on their record.

Yeah - when faced with an odd number of players, usually the way it works is that one player receives a BYE (this being the player with the lowest score, and randomly determined in round 1).

So for example,

Player 1 v Player 2

Player 3 v Player 4

Player 5 BYE

And then the event develops from there.

This is how I would have scored it for what its worth: right or wrong it seems to work and first round Bye is random

WIn=5

Modified Win or Bye)=3

Draw=1

Loss=0

At the end of round one:

• J W=5
• B W=5
• M MW=3 (Bye)
• A L=0
• D L=0

Player A or D could have got the Bye depending on who scored the lowest number of VP's in there game. (You said A so we will go with that)

At the end of round two:

• B W,W=10
• M MW,W=8
• J W,L=5
• A L,MW=3 (Bye)
• D L,L=0

At this point player D would receive the Bye

At the end of round three (the final round)

• M MW,W,W=13
• B W,W,L=10
• J W, L,W=10
• A L,MW,L=3
• D L,L,MW=3 (Bye)

M would be the winner with second place being who had the most VP ( J or B )

Isn't a bye worth 5 points? It may seem like too much, but you get a weaker strength of schedule for tie breakers from byes too.

In the tournament rules it is said that a BYE gets 5 points but ofc. gets a lesser SoS.

If there was a tie at say 3-0, with say... 150 - 200 points at the end. Which one has the harder schedule?

If there was a tie at say 3-0, with say... 150 - 200 points at the end. Which one has the harder schedule?

You have to look at the records of their opponents. In those games, both players beat all their opponents. Player 1's opponents went 2-1, 2-1, and 1-2. Player 2's opponents went 21-, 1-2, and 0-3. Player 1 has the stronger strength of schedule, because his opponents won more games than Player 2's opponents.

If there was a tie at say 3-0, with say... 150 - 200 points at the end. Which one has the harder schedule?

You have to look at the records of their opponents. In those games, both players beat all their opponents. Player 1's opponents went 2-1, 2-1, and 1-2. Player 2's opponents went 21-, 1-2, and 0-3. Player 1 has the stronger strength of schedule, because his opponents won more games than Player 2's opponents.

Yes, but say both players schedules are the same and all players used 100 point lists?

Example:

Player 1: 3-0

Opponents - 2-1; 1-2; 0-3

Player 2: 3-0

Opponents - 2-1; 1-2; 0-3

All lists are 100 points.

If there was a tie at say 3-0, with say... 150 - 200 points at the end. Which one has the harder schedule?

You have to look at the records of their opponents. In those games, both players beat all their opponents. Player 1's opponents went 2-1, 2-1, and 1-2. Player 2's opponents went 21-, 1-2, and 0-3. Player 1 has the stronger strength of schedule, because his opponents won more games than Player 2's opponents.

Yes, but say both players schedules are the same and all players used 100 point lists?

Example:

Player 1: 3-0

Opponents - 2-1; 1-2; 0-3

Player 2: 3-0

Opponents - 2-1; 1-2; 0-3

All lists are 100 points.

If there was a tie at say 3-0, with say... 150 - 200 points at the end. Which one has the harder schedule?

You have to look at the records of their opponents. In those games, both players beat all their opponents. Player 1's opponents went 2-1, 2-1, and 1-2. Player 2's opponents went 21-, 1-2, and 0-3. Player 1 has the stronger strength of schedule, because his opponents won more games than Player 2's opponents.

Yes, but say both players schedules are the same and all players used 100 point lists?

Example:

Player 1: 3-0

Opponents - 2-1; 1-2; 0-3

Player 2: 3-0

Opponents - 2-1; 1-2; 0-3

All lists are 100 points.

That scenario shouldn't happen as both 3-0 players would have to have played each other if you are playing enough rounds ... 9-16 players 4 rounds, 17+ 5 rounds. Only one player will be undefeated after three rounds if 8 or less players. (Also the records shown don't show points for full win or modified win. The likelyhood that all of your opponents scored the exact some points from wins, modified wins, ties, and losses is not very likely. Also you can be 2-1 with 10 points vs. 3-0 with 9 points and the 2-1 player wins ...)

I see what you're saying, but I tied for second in a 10+ tournament at 2-1 with 10 points each.

We ended up sharing the victory by taking only prizes we didn't have yet at the time, so it worked out. Our schedule was the same in the end, with the the 2 wins being against opponents at 100 & 99 points and losing to a 100 point list each.

1st place was a swarm with 3 full victories.

Yes, but say both players schedules are the same and all players used 100 point lists?

Example:

Player 1: 3-0

Opponents - 2-1; 1-2; 0-3

Player 2: 3-0

Opponents - 2-1; 1-2; 0-3

All lists are 100 points.

If you're using swiss pairings, this should never happen. With the logic being that if after two rounds, you have two undefeated opponents, they will play each other in the third round. As such, only 1 of them will be undefeated at the end. In your scenario, they never played each other, which means you did not pair the rounds properly.

You have 5 players, A B C D E.

Round 1

A beats B

C beats D

E has bye

A 5

C 5

E 5

B 0 (use net points to determine who gets bye… I’m saying B destroyed more than D and thus wins that tie breaker over D and D gets bye)

D 0

Round 2

A beats C

E beats B

D has bye

A 10

E 10

C 5

D 5

B 0

Round 3

A beats E

C beats D

B on bye

A 15

E 10 (S of Schedule = B(1-2)+A(3-0)+Bye(0-3) = 4-5)

B 5 (S of Schedule = Bye(0-3)+A(3-0)+E(2-1) = 5-4)

D 5 (S of Schedule = Bye(0-3)+C(2-1)+C(2-1) = 4-5)

So - Swiss pairings give you a clear winner. And 2/3 along with 4/5 are easy enough to figure out. But that's not to say that the system is perfect. C and D played each other twice... But with only 4 opponents, the probability of 2 of them playing a rematch for a non-winning spot is high.

Gah... got beat to it... but I can reply to your following comment too!

I see what you're saying, but I tied for second in a 10+ tournament at 2-1 with 10 points each.
We ended up sharing the victory by taking only prizes we didn't have yet at the time, so it worked out. Our schedule was the same in the end, with the the 2 wins being against opponents at 100 & 99 points and losing to a 100 point list each.
1st place was a swarm with 3 full victories.

In a 10+ tourney (actually, in 9+) you're supposed to use 4 rounds of Swiss in order to determine the winner. As such, one of you (the one with the better SoS) would play the dude that won (assuming he was 3-0 at the time), while the other would be playing someone that was 1-2 (assuming no one else was 2-1). This is another part of the Swiss system that I don't like as it favors the 3rd place player to come in 2nd. Which is why it's very important to say beforehand how many rounds you're playing.

I would also prefer with 10+ players to cut to top 4 after 3. And if players complain that 5 rounds are too much, then still play 3 rounds, cut to top 4, and the winners of those two games can flip a coin over who gets the medal - or no one gets it (note this should not be used for sanctioned tourneys, just game night kits) as the prize support is the same for 1st and 2nd.

Another simple solution is points destroyed instead of strength of schedule. Then if there are ties whoever has the most points destroyed wins the tie breaker. This helps encourage people to continue playing if they are losing, or to not run away for 20min if they are winning.

Note it's a bit tricky with byes, but one way is to have that person's points destroyed as the overall average for that round of play.

While there's nothing wrong with using points destroyed, FFG provided the first two levels of tiebreakers - head to head and then strength of schedule, ignoring strength of victory. The only reason to use points destroyed would be as the next level of TB. And even then, net points would be a more fair idea as some lists are glass cannons and others are tanky. By doing net points, you do nor favor one over the other.