Calculate Your Chance Of Winning With MATH (Based on Elo)
Have you ever wondered what the chance of you beating a 2000 elo is? What about Magnus? What about your 300 elo friend? What about Hans Niemann??? Using the wizardly powers of MATHEMATICS, I, SupremeChessFailure, have developed the Golden Formula to calculate your chances of winning (comparable to the space equation or Einstein's E=mc^2).
The Equation:
Now, if you cannot do this simple calculation mentally, just copy and paste this:
((50*0.875^(0.02 x))/(1+(0.0025 x)^(2)))
into https://www.geogebra.org/calculator
To find the percentage of you winning, add another equation of X=(how much more elo your opponent has than you)
So if you are 1000 elo and want to find your winning chance against a 1400, copy and paste the equation and then add another equation X=400 (because a 1400 is 400 elo higher than you, if you are a 1000). Then just find the intersect, and that is the percent chance of you winning
IMPORTANT: This equation only finds the chance of beating a higher rated opponent; if you are 1000 and want to find your chance of beating a 750 elo, you can't input use X=-250 (your opponent is 250 less elo than you), because negative numbers don't work. Instead, you can still find out the chance of beating a lower rated opponent by calculating 100-(the difference in elo)
So, to find the chance of beating a 750 as a 1000, the difference in elo is 250. Using the equation, you find that your opponent's chance of winning is 18.44%. So to find your chance of winning, it will be 100-18.44, or 81.56%
But what about draws?
Technically this equation calculates your expected score against an opponent in a tournament where you play infinite games. A draw is worth 0.5 (50%), a win is 1 (100%), and a loss is 0 (0%). So if your chance of winning is 25%, it can also mean that if you play 2 games, 1 will be a loss and the other a draw (0.5/2=0.25). However, there is an equal chance that you will win 1 out of 4 to get the 0.25% as there is the chance to draw 2 out of 4 and lose the rest.
Note that this is merely a decently accurate approximation
