ELO calculation change

[FrankJones]

True.

I reviewed a 6-player game and the Elo gain was 6.17 compared to 10.3 expected, which is almost exactly 60%.

So:

2 players: 100% of expected.
3 players: 75% of expected
4 players: 66.66666etc% of expexcted.
5 players: ??
6 players: 60% of expected.

It wouldn't surprise me if 5 players is 62.5%, since that would seem to fit the pattern of scaling and also happens to be expressed as a common fraction like all the others.

[FrankJones]

The common denominator in those fractions is 120.

So,

3 players: 90 / 120
4 players: 80 / 120
5 players: 75 / 120
6 players: 72 / 120

If there is a formula, it is eluding me. I wish I could figure this out !

[Meeplelowda]

The common denominator in those fractions is 120.

So,

3 players: 90 / 120
4 players: 80 / 120
5 players: 75 / 120
6 players: 72 / 120

If there is a formula, it is eluding me. I wish I could figure this out !

The formula is 1/2 x [N/(N-1)]

So your prediction for 5 players was correct.

For whatever reason they wanted it to approach 1/2 for large N.

It just occurred to me that the more rational thing would be for it to approach 2 for large N, so that you reward beating more players, but it doesn't reach a ridiculous multiplier. Maybe that's what was intended, but someone programmed the reciprocal by accident. If the formula were 2 x [(N-1)/N] it would trend toward 2.

[FrankJones]

Nice work!

Did you use a technique to figure that out, or did you get it by inspection?
I vaguely remember methods to find the next term in a sequence and/or finding the closed form expression of the terms, but I've forgotten the specifics.

I was on the wrong track by focusing on 120 as the denominator, since that breaks down for 8 players, yielding a fraction of 8/14. Of course, things break pattern once 7 gets involved, as usual

And, as you noted, the sequence very slowly converges to 1/2.

[Meeplelowda]

Nice work!

Did you use a technique to figure that out, or did you get it by inspection?

I noticed that the end result seemed to be 1 - [1/2 x [(n-2)/(n-1)]]. From there it was just algebra.

[Meeplelowda]

It just occurred to me that the more rational thing would be for it to approach 2 for large N, so that you reward beating more players.

I guess I'll push back on myself. This factor also applies to the losses, so would we want the last place player to lose more and more at higher player counts? Maybe the trend toward 1/2 is more about a concern for the losing players.

[FrankJones]

It just occurred to me that the more rational thing would be for it to approach 2 for large N, so that you reward beating more players.

I guess I'll push back on myself. This factor also applies to the losses, so would we want the last place player to lose more and more at higher player counts? Maybe the trend toward 1/2 is more about a concern for the losing players.

That seems like a reasonable explanation.

[dschingis27]

AFAIK, the factors depending on player number are different between different games. The general Elo formula is always the same, but as you found out, on top there is a hidden factor based on player number, it is documented nowhere.

For 7Wonders, the factors can be found here:
https://boardgamearena.com/forum/viewtopic.php?p=56770

I once saw a thread where we had all the factors for 6Nimmt!, but I can't find it right now. I think only 2p games work the same across all games.

[Meeplelowda]

AFAIK, the factors depending on player number are different between different games. The general Elo formula is always the same, but as you found out, on top there is a hidden factor based on player number, it is documented nowhere.

For 7Wonders, the factors can be found here:
https://boardgamearena.com/forum/viewtopic.php?p=56770

I once saw a thread where we had all the factors for 6Nimmt!, but I can't find it right now. I think only 2p games work the same across all games.

It would seem to be a lot of effort to try to tune the factor differently for different games, which is contrary to human nature. And the 3 and 4 player factors in that thread are exactly as above, but the 5 player factor deviates dramatically. They don't link to the game that they used for those numbers so we can't double check. But I'll inspect some 7 Wonders and 6 Nimmt games.

[dschingis27]

Ahhh, I think sprockitz got it all right already (2nd post in the link I shared).

This is for 7Wonders, I verified it for 5p and 6p:
players -> factor * opponents = "2p equivalent"
(2p -> 1.0 * 1 = 1)
3p -> .75*2 = 1.5
4p -> .66*3 = 2.0
5p -> .50*4 = 2.0
6p -> .40*5 = 2.0
7p -> .33*6 = 2.0
Since 7Wonders is recommended with 4p, the net multiplier does not go up when you increase players above 4.
Example: Winning in a 3p game is equivalent to winning a 2p game 1.5 times. Winning a 4p game is equivalent to winning a 2p game 2 times. Winning a 5p game is also equivalent to winning a 2p game 2 times. It seems that the Elo change calculation wants to keep this parameter at 2 when the player number is above the recommended number for a game. That means no one has to finetune these factors but they depend on the recommended player number.

So, here is my take:
Up to the recommended number of players for a game, it choses the factor based on the formula found by Meeplelowda:
1/2 x [N/(N-1)]
Beyond that, it choses the factor based on this:
2/(N-1)