Chess Referee Course – ARPO Tiebreaker System (Average recursive performance of opponents)

Curso de árbitro de ajedrez – sistema de desempate ARPO (Average recursive performance of opponents)

Hello everyone in this video I will explain how to calculate the tiebreaker called ARPO which is based on calculating the recursive average of the performance of the opponents It is not commonly used because people it is complicated to calculate and understand but I think more precisely it is a long calculation, not complicated To understand how it works I will first explain some concepts some basic concepts that will help in calculations The first concept is very simple

It is the average rating of the opponents that is usually called in the FIDE manual Average rating (Ra) Nothing more than explaining Takes the ELOs, the rating of the opponents and calculating the average Sum them and dividing them by the number of opponents Suppose the player we will use in the example has 1756 I repeat Calculate this by adding the ratings of the opponents and dividing by the number of opponents

The second concept In the FIDE manual is usually called "dp" or qualifying difference and let's say that this player has played five games and won four of them in the FIDE manual (I wrote the link in the description) there is a table where you can find each percentage and a corresponding qualifying difference

Looking up at the table For a player who scored four points in five games, the assigned number would be 240 The next concept is a little more complicated but not so much, it is a weighted average let's call it "c" this weighted average can be calculated by the addition for all opponents who played against a player the amount of games that that opponent played and multiply the difference of points of this opponent For example , the first of the opponents in our example has played four games and the table we have a 190 because he scored three points in four games the second opponent has played four games and the same, also 190 dp the next opponent has also played four games but only scored a point and a half of four, so from the table we get -89 points because he scored badly in the tournament

The next opponent Also played four games and the table assigns 88 points difference because it scored two and a half points of four AND the last opponent has played five games and the table assigns -241 points because he scored a point of five In the denominator we put only the number of games of each opponent A simple calculation of the weighted average of the points differences The result for "c" would be 14,8095 This is probably the most complicated concept

In these calculations because the next variable is "c" with a hat or whatever it is called and is calculated subtracting from the "dp" The average we just calculated For this player the result is 225.19 The following concept is the performance in the FIDE rules consists in adding the average rating of the opponents and the difference in qualification (dp) So simple Then the performance of this player is 1756 + 240 that is to say , 1996 Once these concepts are introduced, we can begin the calculation of our tiebreaker This calculation is done by repeated additions

By hand there are many additions That's why you usually use a computer As you can see, are very simple calculations But demanding time The first iteration of these additions consists in adding all the actions of the opponents In this case it is 1852.5 + 1802.5 + 1743.5 + 1813 + 1365 and in the denominator the number of opponents the player had and the result of this addition is added to the "c" with hat we previously calculated which is 225.19 then , the first iteration for this player is 1940,49 These are the basic concepts that must be understood for the calculation of this tiebreaker Now we will see on a table how to calculate it for a simulated tournament There is the table I told you about This is a simulated tournament in which there are eight players each have their opponents and with this results

"w" represents a win "l" represents a loss and "d" means a tie The first column has nothing to explain It's the rating of each one The second column is also very simple The average rating Take the ratings of the opponents, make a sum and divide by the number of them The third is the big difference

I explained earlier how to get it You have to open the FIDE manual in fidecom in section B02 about the rating standards And this is the table I was talking about At this table For each percentage of score is assigned a "dp" Next to the player score you will see which is the "dp" that we have to use

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For example, a player who scored all points the dp is 800 As you can see For a player who got 60%, the number would be 72 For a player who got 50%, the dp would be 0 And the same for any score Once this column is full We can calculate what is the weighted average of which we speak letter "c" The next column is the "c" with hat that is calculated the number obtained for "c" The next column is not used Only for reference, it is the player's score The following is the performance It is very simple Sum the average of the opponents and the "dp" and we get this number Y at this point we are ready to calculate the tiebreaker This is the first iteration that is calculated To add the performances of the opponents of each player dividing by the number of opponents and adding the "c" with hat is the way to calculate for each player

With this column finished, we do exactly the same in the next column For each player We calculate the average of each opponent of this column divided by the number of opponents and add the "c" with hat And so we repeat the same I did so many times I stopped counting It does not matter, you can see that they are many I did quite and so I did it until It is very clear that I wrote the differences of a column to the previous subtraction each of the calculations in the previous column And I did it until all the subtractions are 0 They are very close to zero much earlier In these results you see here, E-11 means zero point eleven zeros and then one were already very close to zero But here it is even clearer I don't know the accuracy of the cell Then , we continue the iterations until we reach a difference very close to zero

Each software could assign a threshold It will never be an absolute zero The last I will explain how to test the calculations are correct You do this way The first is that the addition of all the "c" with a hat has to be zero This is the first test And then, in each iteration The addition of all iterations has to be the same for each column

As you can see, it is correct in the example It is not that I wrote the same number You can see the formulas up here You can see that the formulas are different But the results are all the same And if we take the results of the last iteration We will have the ARPO tiebreaker for each player That's all Thank you for your attention