2 rating questions: beating up on 10k players, is rank loss proportionate to opponent strength

[shinuito]

The only thing I don’t like so far, small things hang me up, is

If we now assume the log-odds can be written as a linear combination of the inputs (or more probabilistically, if we assume the conditional densities are Gaussian with a shared covariance matrix and equally likely classes), then…

I’ve absolutely no idea if that’s a reasonable assumption to make or not, is it a strong assumption a “normal”/natural assumption

That assumption probably contains a lot of the meat of the “why” I imagine, and it’s over and done with in a sentence.

[Vsotvep]

@yebellz

And again the pattern repeats with her continuously winning and her rank increasing by increments of 1/8, 1/16, 1/32, … and so on.

Surely, this infinite sum of a positive rank increases must diverge to infinity, right?

How much rank would Alice gain if she beats an opponent who is 3k weaker?

[yebellz]

@shinuito

It’s also an intriguing rating system in that it doesn’t seem to care what my rank is, only the rank of the person I beat, and how many times I’ve beaten them

@Vsotvep

How much rank would Alice gain if she beats an opponent who is 3k weaker?

Everything about that hypothetical rating system is completely contrived just to be pedantic.

[shinuito]

I mean while it might be contrived it’s not really too far from a ladder type rating system at say fox.

Replaced with

where a series of winning X out of Y games against fixed rank opponents offers the following increments to one’s rank:

1. +1 rank if X > a
2. +2 rank if X> b
3. -1 rank if X< c
4. -2 ranks if X < d

But of course the decreasing fractional ranks was something about having a converging series even though your rank was updating.

[gennan]

@yebellz

Surely, this infinite sum of a positive rank increases must diverge to infinity, right?

I’m no mathematician, but that infinite sum converges (as I’m sure you knew).

I’m not going to attempt proving by calculation that an infinite number of wins and zero losses against an infinite pool of opponents with rating R will result in a divergent Elo rating (in theory, with infinite precision rating calculation and updates).

But maybe it is enough to consider this:
When a player (A) has a finite rating, there has to be a non-zero probability for them to lose against an anchor player (B) having some finite Elo rating. This non-zero probability determines the (finite) Elo difference between player A and the anchor player B.
But when a different player (C) wins an infinite number of games without ever losing to anchor player B, the probability of player B winning is evidently zero, so player C can’t have a finite rating. No matter how large (but finite) the Elo difference gets, the probability of losing never vanishes completely.

[yebellz]

Don’t ask me how, but I’ve managed to play an infinite sequence of games, where my overall rating change should be given by

Can someone please help me figure out what my new rating will be?

If you think that’s bad, my friend has somehow wound up with this rating adjustment after his infinite series:

Note: s = 0.75

[jlt]

Now someone will create a rating system for uncountably many games.

[yebellz]

It seems tricky to systematically work through so many games in any practical rating system.

For the construction of such rating systems, are we allowing the possibility of a function that can select exactly one game out of each and every tournament that a player has participated in?

[ArsenLapin1]

@jlt

Player B will be tired of losing all the time at some point.

In case a professional player is reading this topic and looking for a player B who wouldn’t tire of losing against them, I volunteer. You can challenge me here on OGS any time!!

[qnpnpmqppnp]

You might as well play Katago?