Does the Elo system work with different expected value functions?

Question

In many Elo rating systems, the expected value for player A with rating $R_A$ against an opponent with rating $R_B$ is

$$E_A = \frac1{1+10^{(R_B - R_A)/c}}$$

The original Elo system is based on a normal distribution instead of a logistic distribution, so it used the error function instead of the logistic function in the equation for $E_A$.

Would the Elo system still work if we chose a simpler function like this?

$$E_A = \frac{R_A}{R_A + R_B} = \frac1{1 + \frac{R_B}{R_A}}$$

For example, if c=400, and 1000 rating players in the logistic Elo system still have a 1000 rating in this system, then 1400 rating players in the logistic Elo system would have a rating of 10000 in this system.

If this still works, then why would they have switched to the logistic function? Is it that they wanted player ratings to follow a normal distribution, and ratings in this simpler version would follow a log-normal distribution?

Answer 1

Let me answer your second question, why they switched to the logistic function. The Elo rating system comes from the Chess world. The initial Elo's premise was a normal distribution, but since more chess statistics became available, FIDE (The World Chess Federation) realized that it was better to consider the logistic function. Have a look here for a description of Elo's rating system.

Regarding your first question, if your formula will work, well, c is the logistic parameter used to fine-tune your rating system. The value c = 400 is for the chess world. But in reality, you can pick any which fits your rating requirements. The same happens with the Distribution you pick. First, maybe you will have to draw your players' initial ratings and see which distribution they follow.

Answer 2

This subject is discussed in Prof. Mark E. Glickman, A Comprehensive Guide To Chess Ratings.

It appears as though there is very little distinction between the shape of the logistic distribution in Figure 2 and the normal distribution in Figure 4. Figure 5 shows both curves superimposed, with the logistic distribution drawn as a solid line. In fact, statistics professor Hal Stern in a 1992 article[11] has shown that when analyzing paired comparison data it makes virtually no difference whether one assumes the logistic distribution or the normal distribution for differences in players' strengths. So, empirically, the choice between the Bradley-Terry model and the Thurstone-Mosteller model is a moot issue. Mathematically, however, the Bradley-Terry model tends to be more tractable to work with. This is the most likely reason that most organizations administering a probabilistic rating system (e.g., FIDE, USCF) use the Bradley-Terry model, which uses the logistic distribution assumption, rather than the Thurstone-Mosteller model, which uses the normal distribution assumption.

It is easy to see from the "FIDE Rating Regulations 2022" that the original Elo tables are still in use by FIDE.

The following distributions are discussed by Elo in The Rating of Chess Players.

• 8.4 Logistic Probability As a Rating Basis
• 8.5 Rectangular Distribution as a Ratings Basis
• 8.8 Binomial Distribution and Small Examples
• 9.1 The Maxwell-Bolzmann Distribution and Chess Ratings

Since the Elo system is self-correcting using the K-factor formula, any reasonable form of the probability distribution function can be used as a starting point for a rating system (ch. 8.75).

Answer 3

The original Elo system is based on a normal distribution instead of a logistic distribution

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If this still works, then why would they have switched to the logistic function?

There hasn't been a switch. The FIDE uses Elo's original system for which a table is computed. See the Fide handbook.

The win-probability (or score,since it includes draws), is computed as

$$p(\text{A win from B})= \Phi\left( \frac{\text{Elo}_A - \text{Elo}_B}{200 \cdot \sqrt{2}} \right)$$

For example, a $100$ Elo difference will relate to a $0.6381632$ win probability for the higher valued player. Or $0.64$ as it is in the tables of the FIDE.

The use of the logistic function/distribution is a useful trick to make an estimate of the values in the table. Also, several other competitions may be using the logistic function. There is not really a large difference between the two and it is just a practical trick to make computations easy.

Would the Elo system still work if we chose a simpler function like this?

$$E_A = \frac{R_A}{R_A + R_B} = \frac1{1 + \frac{R_B}{R_A}}$$

The system is arbitrary, so yes this would work as well. However, the computations with a linear Elo scale might be easier.

• For instance, one could compute an average Elo rating for a pool of chess players and approximated the win probability of a player based on the difference with the average Elo.
• Another example is that differences in Elo rating are more easier to evaluate. If some top player has rating 13102 and another has a rating 4996, than it is a bit awkward to compute the difference in the level between the players because this involves a division instead of a subtraction.
• Also, the updating works as a zero sum game. Whenever a player increases by some value $x$, then another player decreases by a value $x$. If you would do this for an exponentially increasing score, then the people at the top will be changing their scores only very slowly (in some sense the system does have a way to reduce the speed of changes in scores for high level players, but it does not reduce the speed that much as what would be the case with an exponentially increasing score)