Caruana is the top seed the 10-player round robin

# e4stat

The Chess & Stats Blog

## Tuesday, November 21, 2023

## Monday, October 2, 2023

### The 2023 National Memory Championship

The host asked all the finalists why they decided to compete.

"Not sure exactly - it's very difficult and stressful."

That was my response. Back in July, I had made it through the qualifier (results). I won the numbers event, memorizing 80 digits perfectly in 31.92 seconds. This compensated for my performance in the other events. Then in September, I joined the other finalists and reigning champion John Graham in Orlando, Florida.

The first event was a Lumosity game, Pinball Recall.

## Saturday, May 6, 2023

## Wednesday, May 3, 2023

## Wednesday, April 5, 2023

### World Chess Championship 2023

Nepo and Liren are only 7 points apart, so there is a decent chance that the 14-game match will end in a draw (100% - 45.685% - 38.685% = 15.63%)

## 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.

*np*). The standard deviation is the square root of the variance, so that will be

*(np(1-p))*. Thus, the upper end of the 95% range is

^{1/2}*np + 1.96(np(1-p))*. Therefore, your MinStrength is the rating such that

^{1/2}*score = np + 1.96(np(1-p))*

^{1/2}