ELO Scoring

[whatshisbucket]

There is also a multiplier applied to the rating change based on the number of players. The multiplier is 1 for two player games and \smaller for larger players. The reason is as follows: If you are playing a 3 player game and you beat player A, you probably also beat player B. Thus you should receive not as much rating for beating B as you might in a two player game with B.

The constants actually used are not the best and also appear to differ from game to game (for the same number of players), but the idea is right.

[Jest Phulin]

I think the correction factor is based on the BGG voting of what the game is best at. So something like Can't Stop is considered best at 3, whereas 7 Wonders is best at significantly more players.

[whatshisbucket]

I think the correction factor is based on the BGG voting of what the game is best at. So something like Can't Stop is considered best at 3, whereas 7 Wonders is best at significantly more players.

In what way? As far as I can see, there is an objectively best constant for each player count. Also, I think at the very least the constants for 3 and 4 players are the same across all of BGA.

[Jest Phulin]

First, let me say that this is being pulled from the back of my mind, and I may be way off.

I think the scaling factor is based on ( (best at count) / (actual count) ) if actual count is larger than best at count. So, for a 3-player game best at 2, each ELO variance will be multiplied by 2/3rds. A 3-player game best at 4 would get no multiplication, but a 5-player game would be multiplied by 4/5ths.

Again, this is based on old memories of forum posts probably by other users (not BGA staff), so there's a good probability that it is somewhat incorrect.

I checked my own stats at Can't Stop, and if I am correct it is considered best at 2, so that doesn't seem to add up. Me being lazy, I don't feel like looking at results of larger games such as 7 Wonders and 6 Nimt! at various player counts to determine the actual values.

[Orsizuma]

Can anybody elaborate on the scores when there are 3 players ?
Does it not related to how close you were for winning ?

ELO does not consider how big the win was. It was designed for chess, where the score is always 1-0, 0-1, or 0.5-0.5. As such, it doesn't matter if the win in Can't Stop is 3-0, 3-1, or 3-2; it treats all of these as 1-0.

In a multiplayer game, ELO calculation is done as if several 2-player games were done. So if the final scores are A:3, B:2, C:2, D:1, the final ELO will be treated as if:
A defeats B, A defeats C, A defeats D
B loses to A, B ties C, B defeats D
C loses to A, C ties B, C defeats D
D loses to A, D loses to B, D loses to C.

The amount of points shifted in each matchup depend on what the initial ELO difference was. It is more likely a higher-ranked player will defeat a lower-ranked player, so there will be less of a points shift if it happens that way or if the lower-ranked player wins. Then, too, there is a factor based on how many games the player has played (to more quickly get a newer player's ranking closer to their accurate value).

Hovering over the ELO on the end screen will bring up a pop-up showing more detail on these calculations.

Thanks for taking the time and answering me . It is much clearer now .

[Orsizuma]

ELO does not consider how big the win was. It was designed for chess, where the score is always 1-0, 0-1, or 0.5-0.5. As such, it doesn't matter if the win in Can't Stop is 3-0, 3-1, or 3-2; it treats all of these as 1-0.

In a multiplayer game, ELO calculation is done as if several 2-player games were done. So if the final scores are A:3, B:2, C:2, D:1, the final ELO will be treated as if:
A defeats B, A defeats C, A defeats D
B loses to A, B ties C, B defeats D
C loses to A, C ties B, C defeats D
D loses to A, D loses to B, D loses to C.

The amount of points shifted in each matchup depend on what the initial ELO difference was. It is more likely a higher-ranked player will defeat a lower-ranked player, so there will be less of a points shift if it happens that way or if the lower-ranked player wins. Then, too, there is a factor based on how many games the player has played (to more quickly get a newer player's ranking closer to their accurate value).

Hovering over the ELO on the end screen will bring up a pop-up showing more detail on these calculations.

I just played a game that contradicted this and now I am confused again.

There was 3 player started with player A:ELO 282, player B:ELO 230, player C: ELO 229.
Player A won and player B and C both had 2 columns .
Player A got +13 for winning over B and C, which I understand .
Player B got an ELO -6 and player C got an ELO -7. Why?
Player B and C both got defeated by player A and they had a tie with each other. Played C started with the lower (even if just by 1) ELO score, but still got a bigger score deduction for the the same result.

[Tulivuori]

Rounding.

Look at the details. B lost 6.41 and C 6.33.

[Orsizuma]

Rounding.

Look at the details. B lost 6.41 and C 6.33.

That still doesn’t explains it . A slightly higher penalty like 6.51( 6.41 would also be rounded down to 6) would be rounded to 7 for player B and 6.33 should be rounded to 6 for player C . So player B should get the -7 and player C should get the -6, but the opposite happened .

[Tulivuori]

230.4 -> 224.0 = 230 -> 224 = -6
228.6 -> 222.3 = 229 -> 222 = -7

Just an example. How hard can it be?

[gordoz]

yall notice there are no 500 elo Can't Stop players? or almost none...

Even though its a very popular game.

I think that's because even if you are very good at the game you only will win slightly more than 50% because there is so much luck.

Whereas if you are very good at carcassone or other type of game its almost impossible to lose to a noob.

just intresting