ELO change, placing second in a 4 player?

[asd123321]

Suppose I was ranked 1750 and all 3 others are 1550, what would my new rating be?

[sprockitz]

Suppose I was ranked 1750 and all 3 others are 1550, what would my new rating be?

I'd expect you to lose about 8 points in this situation for a new rating of 1742 or 1743.

Each matchup is considered head to head. A 200 points difference means you should win each head to head matchup ~75% of the time. In a 2 player game this means you'd gain 10 for a win and lose 30 for a loss. With 4 players a little less than 40 points is exchanged per matchup, I checked (see post below), and the multiplier for 3 (and presumably 4+) players is 30. So this means you lose 3/4*30 = 22.5 points to the person who beat you and gain 1/4*30 = 7.5 points off both people you beat. So the net is -22.5+7.5+7.5 = -7.5 points.

So new rating is 1742 or 1743.

Edit: See post below for more details, but it seems the 4 player multiplier is 20, meaning you would gain 5 points against each person you placed ahead of and lose 15 to the person you finished behind, for a net loss of 5 points.

[Oso]

200 ELO difference is: (If you're the higher player)

[+30 matches] WIN: +10 / LOSS: -30 / TIE: -10
[0-30 matches] WIN: +14 / LOSS: -46 / TIE: -16

[sprockitz]

200 ELO difference is: (If you're the higher player)

[+30 matches] WIN: +10 / LOSS: -30 / TIE: -10
[0-30 matches] WIN: +14 / LOSS: -46 / TIE: -16

This is only true for 2 player games. (and yes my example above assumes 30+ games, for < 30 multiply by 1.5).

for more than 2 players the 40 points per matchup number goes down.

Here is a 3 player example pulled from the most recent completed games list:
http://en.boardgamearena.com/#!table?table=20653105

Use jvanhoest since he is over the 30 game threshold.
He lost 38 points in this matchup.
His rating: 1692.
Opponent's rating: 1552 and 1647.

If it was 40 points per matchup he would have lost: 40/(1 + 10^((1552-1692)/400)) + 40/(1 + 10^((1647-1692)/400)) = 50 points

Since he only lost 38 points, this means the 40 multiplier is incorrect. The correct multiplier is 30.
30/(1 + 10^((1552-1692)/400)) + 30/(1 + 10^((1647-1692)/400)) = 38 points

Here is a 4 player example:
http://en.boardgamearena.com/#!table?table=11655144

Use sprockitz here.
My rating: 1594. Opponent's ratings: 1516, 1524, 1478
I lost 37 points, meaning the multiplier for this case is 20.

So assuming the multiplier is based on number of players (and not different for different games), the multipliers are as follows:
2 players: 40 points
3 players: 30 points
4 players: 20 points
and again, if # games < 30, multiply this number by 1.5

[Silene]

Only loosely connected to the subject but a good occasion to bring it up:

I think the head-to-head-comparision in a 3+ player matchup leads to too big ELO-change for one game. Especially because the outcome is not only dependent of your own skills but also of decision your opponents make that can favour you or not.

A careful approach to this (very-strong-oscillation-of-ELO-) problem would be to see it like this: You cannot play 3 games at once. So when you play a 4-player game only 1/3 of the game is the fight vs. player A etc. So for each head-to-head-matchup the ELO given/lost should be treated like it was only 1/3 of a game and therefore leading to 1/3 of the ELO-change it's resulting in by now.

I think this solution might be easy to implement (only deviding by the number of opponents has to be calculated) and will reduce oscillation very much while still keeps the same (significant) ELO-change in 2-player-matchups.

[Oso]

This is only true for 2 player games. (and yes my example above assumes 30+ games, for < 30 multiply by 1.5).

In 3 or more players matches is the same case, if are a triple tie, one player tie with both players, so you should calc the ELO with each player.

Only loosely connected to the subject but a good occasion to bring it up:

I think the head-to-head-comparision in a 3+ player matchup leads to too big ELO-change for one game. Especially because the outcome is not only dependent of your own skills but also of decision your opponents make that can favour you or not.

A careful approach to this (very-strong-oscillation-of-ELO-) problem would be to see it like this: You cannot play 3 games at once. So when you play a 4-player game only 1/3 of the game is the fight vs. player A etc. So for each head-to-head-matchup the ELO given/lost should be treated like it was only 1/3 of a game and therefore leading to 1/3 of the ELO-change it's resulting in by now.

I think this solution might be easy to implement (only deviding by the number of opponents has to be calculated) and will reduce oscillation very much while still keeps the same (significant) ELO-change in 2-player-matchups.

Not really...

When you're playing a free-for-all in 4 players, you've only 25% chances of win -it depends with each player skill too, but it's just theory-, so is nice that you get 3x ELO if you win, and -3x ELO if you get last place... -really, in 4p games you've only 25% chances to be in last place-

Your idea is good in "team play", because ELOs are converted in a "team average ELO"...

[Silene]

Fair point.

If i compare 2p and 4p game:

Game time doubles
Chance to win halves

Sounds like a factor of 1/2 would be appropriate then (not 1/3 as i suggested before). I'll try to make a formula for n players to suggest here when I have more time.

[Oso]

Fair point.

If i compare 2p and 4p game:

Game time doubles
Chance to win halves

Sounds like a factor of 1/2 would be appropriate then (not 1/3 as i suggested before). I'll try to make a formula for n players to suggest here when I have more time.

BGA has the system like in some Chess tours. Just an example:

"You should play a round robin with 7 players, so you will play 7 games. After one game, the ELO doesn't change; so if you started the tournament with 2200, all matches you will starts with 2200.
When the tournament ends and the winner is revealed, each player change their ELO basing in their victories, defeats and ties"

In BGA is the same case, so it's just in one game: the first beat all players, the second beat all except first player, etc.

[sprockitz]

Fair point.

If i compare 2p and 4p game:

Game time doubles
Chance to win halves

Sounds like a factor of 1/2 would be appropriate then (not 1/3 as i suggested before). I'll try to make a formula for n players to suggest here when I have more time.

look at my above example. 4 players is already 1/2. (3 players is 3/4)