Chess ELO as a martingale

I just learned about martingales and the martingale gambling strategy, so I'm no expert on the topic, but I'll summarize it the best I can (with the help of Wikipedia, of course!).

https://en.wikipedia.org/wiki/Martingale_(probability_theory)

A martingale is a situation where the conditional expectancy (i.e. average expected value) is equal to the current / most recent value. For example, in a random walk (e.g. taking n steps in random directions starting at the origin of Cartesian xy graph), the conditioned expectancy is the starting position.

I am wondering: does chess ELO counts as a martingale? My gut tells me that it does, but I don't understand martingales or ELO well enough to say for sure.

Chess is not a game of chance. Yes, computers can evaluate the likelihood that you will do so well in a tournament etc, but it's not the same.

Also, say you're improving. You are rated 1000, but you kick trash in a tournament and play at a 2000 level and win the tournament. Maybe you're a 2000 player. But your rating goes up to 1100. Next tournament, is your expected play going to be 1100? No, it's not. You will likely do better than that. In ELO it seems that the direction someone is traveling has some effect on the future outcomes, i.e. your expected value depends not only on where you're at but also on which direction you're going.

Alas, expected performance at a tournament also depends on other factors like whether or not, and how well, you studied/prepared, how well or sick you're feeling that day, etc.

Anyway, does this answer your question?

As is the case always with mathematical modeling, the answer could be yes under a few idealistic assumptions.  One assumption you would need is that you do not improve over time.  However, you can bypass this assumption by introducing a drift.  This is done, for example, when modeling stock price movements.  If we introduce an upward drift, then an adjusted ELO score would be a Martingale process: e.g. e^{- c t} E(t) would be Martingale where E is your ELO score and c is a drift parameter.  A discrete model would be a different drift factor.  To make things simpler, I would assume that there is no improvement over time so that the expected value of change over any time increment is zero.

Another assumption is that the past values do no influence on the future ELO changes.  This is a natural assumption with sufficiently random assignment of opponents.  Under idealistic assumptions, one can model ELO with a continuous-state discrete-time Markov chain (if games are infrequent), or a Brownian motion (if games are very frequent, like Blitz rating at chess.com).

I've done something similar when I was teaching this course (Stochastic Processes).  My students know that I like chess, and I word a lot of problems in terms of chess.  I tell my students that my chess player doesn't learn from past mistakes, which is a comical but an idealistic assumption.  Here is one homework problem from my course:

Sarah is playing chess at her computer (the fact that she never gets better is to the benefit of
our course). There are 4 levels, level 4 being the hardest. If she wins, she advances a level (unless she cannot go higher); if she losses, she drops a level (unless she cannot go lower); if she draws, then she remains at the current level. Of course, each level carries different probabilities for all three outcomes (win, loss, draw).  She begins with level 1.

(a) Construct a reasonable transition probability matrix for this Markov process and explain in words
why your choice of transition probabilities make sense.

(b) What is the probability that she will play at level 4 at the fifth game? 

(c) What is the expected level at the fifth game?

(d) What is the probability that she will ever have played at level 4 within 10 games? 

(e) (extra credit) What is the expected number of games until she reaches level 4?

Interesting. Thanks guys, I understand martingales better now!

If I understand correctly, Chess ELO would be a martingale if:

no improvements of skill happened

your true rating = your initial rating (taking into account short term factors)

opponent's true rating = opponent's initial rating (taking into account short term factors)

characteristics of past games (move sequences, style, etc.) are forgotten/ irrelevant

Now to try the problem (I am no mathematician by any means, and I just learned what Markov chains are too)...

a.    1    2     3     4

1    .2    .8    0     0

2    .3    .4    .3    0

3    0     .6    .3    .1

4    0     0     .9    .1

She has a skill level near 2, and draws a lot with evenly matched players.

b. you would sum the probabilities for each path that involves 3 advances and 1 stagnation

(.2)(.8)(.3)(.1) + (.8)(.4)(.3)(.1) + (.8)(.3)(.3)(.1) + (.8)(.3)(.1)(.1) = .024 = 2.4%

c. Now it's just guesses on my part without much calculative proof.

Level 2 looks correct as the probability strongly leads toward it.

d. I have some ideas of the pathways, but I forget the stats techniques needed for them, and even then I think there are complication involving the discrete states of the Markov chain that I haven't learned yet.

a) and b) are correct.  If you just learned Markov chains, you will soon learn an easier way to solve (b) without generating all possible paths.  Also this is needed for the rest of the parts.

c) is correct to a degree.  Expected value comes to about 2.0944.

For d), a special trick is required for this one, and perhaps you will encounter it later in your course. 

The ELO chess system works such. If you  win against a lower player, you gain few points. If you draw, you lose 10-15, and if you lose you will drop 30 points. If you beat a higher rated player, even if you play like a 1700 but your rating is 1200, you will go up to 1350. In my opinion, this is unfair considering that if you play an underrated player, you will lose many points, and overrated player may play worse than you yet be 200 points above you. That is why there is provisional rating.