Saturday, March 18, 2023

MinStrength Methodology

In a game between Players A and B, there is a normal distribution centered at A’s rating and another centered at B’s rating. The standard deviation is 200. In the Elo system, the expected score for Player A is the probability that a random number from A’s distribution is higher than a random number from B’s. This seems to ignore the possibility of draws – there is a 0% chance that both random numbers are equal – but that will be addressed later. The expected score can be approximated with the logistic function:


Next, I model a tournament as n games against your average opponent. This is an approximation (the expected score isn’t a linear function, so a game against an 1800 followed by a game against a 2000 is slightly different from playing two games against a 1900). With this assumption, your score follows a binomial distribution. The mean is np and the variance is np(1-p), where p is your expected score against the average opponent. The issue with this binomial distribution is that there is no accounting for draws. However, the binomial distribution converges to a normal distribution, so I use that as an approximation. The normal distribution is continuous, so scores such as 8.5 are possible. This means that we aren’t ignoring draws.

 

If you pull a random number from a normal distribution, there is a 95% chance that it will be within 1.96 standard deviations from the mean (np). The standard deviation is the square root of the variance, so that will be (np(1-p))1/2. Thus, the upper end of the 95% range is np + 1.96(np(1-p))1/2. Therefore, your MinStrength is the rating such that score = np + 1.96(np(1-p))1/2