for
Mark Berg

USATT#: 6250

##### Introduction
This page explains how Mark E. Berg (USATT# 6250)'s rating went from 1761 to 1787 at the Westchester 2014 May Open held on - 25 May 2014. These ratings are calculated by the ratings processor which goes through 4 passes over the match results data for a tournament. The following values are produced at the end of each of the 4 passes of the ratings processor for Mark E. Berg for this tournament.

Initial Rating Pass 1 Pass 2 Pass 3 Final Rating (Pass 4)
1761 1775 1761 1761 1787

You can click here to view a table of all the resultant values from each of the 4 passes (and the initial rating) of the ratings processor for all of the 120 players in this tournament. Sections below for further details on the initial rating and the 4 passes of the ratings processor.

Note: We use mathematical notation to express the exact operations carried out in each pass of the ratings processor below. Whenever you see a variable/symbol such as for example ${X}_{i}^{3}$, we are following the convention that the superscript part of the variable (in this case "3") indicates an index (such as in a series), and it should not be misconstrued to be an exponent (which is how it is used by default).

##### Initial Rating
The initial rating of a player for a tournament is the rating the player received at the end of the most recent tournament prior to the current tournament. If this is the first tournament the player has ever participated in (based on our records), then the player has no initial rating.

The initial rating for Westchester 2014 May Open held on - 25 May 2014 for Mark E. Berg, and its source tournament are as follows:
Initial Rating From Tournament Start Day End Day
1761 LYTTC May Open n/a 18 May 2014

Click here to view the details of the initial ratings for all the players in this tournament.

##### Pass 1 Rating
In Pass 1, we only consider all the players that come into this tournament with an initial rating while ignoring all the unrated players. If a rated player has a match against an unrated player, then that match result is ignored from the pass 1 calculations as well. We apply the point exchange table shown below to all the matches participated in by the rated players:

Point Spread Expected Result Upset Result
0 - 12 8 8
13 - 37 7 10
38 - 62 6 13
63 - 87 5 16
88 - 112 4 20
113 - 137 3 25
138 - 162 2 30
163 - 187 2 35
188 - 212 1 40
213 - 237 1 45
238 and up 0 50

Suppose player A has an initial rating of 2000 and player B has an initial rating of 2064, and they played a match against each other. When computing the impact of this match on their rating, the "Point Spread" (as it is referred to in the table above) between these two players is the absolute value of the difference their initial ratings. When the player with the higher rating wins, presumably the better player won, which is the expected outcome of a match, and therefore the "Expected Result" column applies. If the player with the lower rating wins the match, then presumably this is not expected, and therfore it is deemed as an "Upset Result" and the value from that column in the table above is used. So, in our example of player A vs player B, if player B wins the watch, then the expected outcome happens, and 5 points are added to player B's rating and 5 points are deducted from player A's rating. Looking at Mark E. Berg's match results and applying the point exchange table, gives us the following result:

##### Mark E. Berg's Wins
Winner Loser
Point Spread Outcome Gain Player USATT # Rating Player USATT # Rating
533 EXPECTED 0 Mark E. Berg 6250 1761 Hank Roy 83208 1228
92 EXPECTED 4 Mark E. Berg 6250 1761 MICHAEL REFF 24318 1669
116 EXPECTED 3 Mark E. Berg 6250 1761 David (Pavlovich) Nelyubin 76519 1645
330 EXPECTED 0 Mark E. Berg 6250 1761 John F. Mcfadden 29798 1431
143 EXPECTED 2 Mark E. Berg 6250 1761 Lily Xiong 88715 1618
151 EXPECTED 2 Mark E. Berg 6250 1761 David Maisel 76188 1610
60 UPSET 13 Mark E. Berg 6250 1761 Alok Kumar 29925 1821
58 EXPECTED 6 Mark E. Berg 6250 1761 Steven Chen 90786 1703

##### Mark E. Berg's Losses
Winner Loser
Point Spread Outcome Loss Player USATT # Rating Player USATT # Rating
73 UPSET -16 Trevor Salmon 88519 1688 Mark E. Berg 6250 1761
272 EXPECTED -0 Dmitry Tokmakov 80594 2033 Mark E. Berg 6250 1761
267 EXPECTED -0 Adnan Medunjanin 83321 2028 Mark E. Berg 6250 1761

You can click here to view a table of outcomes and points gained/lost from all the matches with all the players in this tournament.

The "Outcome" column above, shows whether the match had an expected (player with the higher rating wins the match) or an upset (player with the higher rating loses the match) outcome. Based on this outcome, and using both the player's initial rating, we apply the point exchange table from above and show the ratings points earned and lost by Mark E. Berg in the "Gain" column. Matches are separated out into two tables for wins and losses where points are gained and lost respectively. We get the following math to calculate the Pass 1 Rating for Mark E. Berg:

Initial Rating Gains/Losses Pass 1 Rating
1761 + 0 + 4 + 3 + 0 + 2 + 2 + 13 + 6 - 16 + 0 + 0 $=\mathrm{1775}$

You can click here to view a table of pass1 calculations for all the rated players in this tournament.

##### Pass 2 Rating
The purpose of this pass is solely to determine ratings for unrated players. To do this, we first look at the ratings for rated players that came out of Pass 1 to determine an “Pass 2 Adjustment”. The logic for this is as follows:

1. We calculate the points gained in Pass 1. Points gained is simply the difference between the Pass 1 Rating and the Initial Rating of a player:

${\rho }_{i}^{2}={P}_{i}^{1}-{P}_{i}^{0}$
where,

 Symbol Universe Description ${P}_{i}^{0}$ ${P}_{i}^{0}\in \mathrm{{ℤ}^{+}}$ the initial rating for the $i$-th player. We use the symbol $P$ and the superscript $0$ to represent the idea that we sometimes refer to the process of identifying the initial rating of the given player as Pass 0 of the ratings processor. ${P}_{i}^{1}$ ${P}_{i}^{1}\in \mathrm{{ℤ}^{+}}$ the Pass 1 rating for the $i$-th player. ${\rho }_{i}^{2}$ ${\rho }_{i}^{2}\in ℤ$ the points gained by the $i$-th player in this tournament. Note here that we use the superscript $2$ to denote that this value is calculated and used in Pass 2 of the ratings processor. Further, ${\rho }_{i}^{2}$ only exists for players who have a well defined Pass 1 Rating. For Players with an undefined Pass 1 Rating (unrated players), will have an undefined ${\rho }_{i}^{2}$. $i$ $i\in \left[1,\mathrm{120}\right]\cap ℤ$ the index of the player under consideration. $i$ can be as small as $1$ or as large as $\mathrm{120}$ for this tournament and the i-th player must be a rated player.

2. For rated players, Pass 1 points gained, ${\rho }_{i}^{2}$, is used to calculate the Pass 2 Adjustment in the following way:
1. If a player gained less than 50 points (exclusive) in pass 1, then we set that player's Pass 2 Adjustment to his/her Initial Rating.
2. If a player gained between 50 and 74 (inclusive) points in pass 1, then we set the player's Pass 2 Adjustment to his/her Final Pass1 Rating.
3. If a player gains 75 or more points (inclusive) in pass 1, then the following formula applies:
• If the player has won at least one match, and lost at least 1 match in the tournament, then the player's Pass 2 Adjustment is the average of his/her Final Pass 1 Rating and the average of his/her opponents rating from the best win and the worst loss, represented using the formula below:

$\mathrm{{\alpha }_{i}^{2}}=⌊\mathrm{\frac{\mathrm{{P}_{i}^{1}}+\mathrm{\frac{\mathrm{{B}_{i}}+\mathrm{{W}_{i}}}{2}}}{2}}⌋$

where ${\alpha }_{i}^{2}$ is the Pass 2 Adjustment for the current player, ${P}_{i}^{1}$ is the Pass 1 Rating, ${B}_{i}$ is the rating of the highest rated opponent against which the current player won a match, and ${W}_{i}$ is the rating of the lowest rated opponent against which the current player lost a match.
• If a player has not lost any of his/her matches in the current tournament, the mathematical median (rounded down to the nearest integer) of all of the player's opponents initial rating is used as his/her Pass 2 Adjustment:

$\mathrm{{\alpha }_{i}^{2}}=\mathrm{⌊\stackrel{\sim }{\mathrm{\left\{\mathrm{{P}_{k}^{0}}\right\}}}⌋}$

where ${P}_{k}^{0}$ is the initial rating of the player who was the i-th player's opponent from the k-th match.
Symbol Universe Description
$i$ $i\in \left[1,\mathrm{120}\right]\cap ℤ$ the index of the player under consideration. $i$ can be as small as $1$ or as large as $\mathrm{120}$ for this tournament and the i-th player must be a rated player.
$q$ $q\in \left[1,\mathrm{473}\right]\cap ℤ$ the index of the match result under consideration. $q$ can be as small as $1$ or as large as $\mathrm{473}$ for this tournament and the q-th match must be have both rated players as opponents.
$g$ $g\in \left[1,5\right]\cap ℤ$ the g-th game of the current match result under consideration. $q$ can be as small as $1$ or as large as $5$ for this tournament assuming players play up to 5 games in a match.
${P}_{k}^{0}$ ${P}_{k}^{0}\in \mathrm{{ℤ}^{+}}$ initial rating of the i-th player's opponent from the k-th match.

• Therefore, the Pass 2 Adjustment for Mark E. Berg is calculated as follows:
• Given the initial rating of 1761,
• and the Pass 1 rating of 1775,
• the Pass 1 gain is 1775 - 1761 = 14.
• Since the Pass 1 gain of 14 is less than 50, the Pass 2 Rating (also referred to as Pass 2 Adjustment) is reset back to the initial rating.
• Therefore the Pass 2 Adjustment for Mark E. Berg is 1761.

You can click here to view a table of Pass 2 Adjustments for all the rated players in this tournament.

3. After calculating the Pass 2 Adjustment for all the rated players as described above, we can now calculate the Pass 2 Rating for all the unrated players in this tournament (which is the main purpose of Pass 2). Pass 2 Rating is calculated using the following formula:
1. If all of the matches of an unrated player are against other unrated players, then the Pass 2 Rating for that player is simply set to 1200. You can click here to view these players who received a 1200 Pass 2 Rating. Not all of Mark E. Berg's matches were against unrated players. So this rule does not apply to him.
2. For unrated players with wins and losses, where at least 1 of the opponents has an initial rating, the Pass 2 Rating is the average of the best win and the worst loss (using the Pass 2 Adjustment of all rated players) as defined by this formula here:

$\mathrm{{P}_{i}^{2}}=⌊\mathrm{\frac{\mathrm{{B}_{i}^{2}}+\mathrm{{W}_{i}^{2}}}{2}}⌋$

where ${P}_{i}^{2}$ is the Pass 2 Rating for the i-th player, ${B}_{i}^{2}$ is the largest Pass 2 Adjustment (best win) of the opponenet against whom the i-th player won a match, and ${W}_{i}^{2}$ is the smallest Pass 2 Adjustment (worst loss) of the opponent against whom the i-th player lost a match.
3. For unrated players with all wins and no losses, where at least 1 of the opponents has an initial rating, the Pass 2 Rating is calculated using the following formula:
$Pi2 = Bi2 + ∑k=0Mi-1 I(Bi2-αk2)$
where the function $I\left(x\right)$ is defined as, $$I(x)=\left\{ \begin{array}{ll} 10, & \text{if}\ x >= 1, x <= 50 \\ 5, & \text{if}\ x >= 51, x <=100 \\ 1, & \text{if}\ x >= 101, x <= 150 \\ 0, & \text{otherwise} \end{array}\right.$$
where,
Symbol Universe Description
${P}_{i}^{2}$ ${P}_{i}^{2}\in \mathrm{{ℤ}^{+}}$ the pass 2 rating, of the i-th player in this tournament only applicable to unrated players, where ${P}_{i}^{0}$ is not defined
${B}_{i}^{2}$ ${B}_{i}^{2}\in \mathrm{{ℤ}^{+}}$ the largest of the Pass 2 Adjustments of opponents of the i-th player against whom he/she won a match.
${\alpha }_{k}^{2}$ ${\alpha }_{k}^{2}\in \mathrm{{ℤ}^{+}}$ the Pass 2 Adjustment of the player who was the opponent of the i-th player in the k-th match
$I\left(x\right)$ $I:ℤ↦\mathrm{{ℤ}^{+}}$ a function that maps all integers to one of the values from -- 0, 1, 5, 10.
${M}_{i}$ ${M}_{i}\in \mathrm{{ℤ}^{+}}$ total number of matches played by the i-th player in this tournament
k $k\in \mathrm{\left[0,\mathrm{{M}_{i}}-1\right]\cap {ℤ}^{+}}$ The index of the match of the i-th player ranging from 0 to ${M}_{i}-1$
4. For unrated players with all losses and no wins, where at least 1 of the opponents has an initial rating, the Pass 2 Rating is calculated using the following formula:
$Pi2 = Wi2 + ∑k=0Mi-1 I(Wi2-αk2)$
where $I\left(x\right)$ is defined above and,

Symbol Universe Description
${P}_{i}^{2}$ ${P}_{i}^{2}\in \mathrm{{ℤ}^{+}}$ the pass 2 rating, of the i-th player in this tournament only applicable to unrated players, where ${P}_{i}^{0}$ is not defined
${W}_{i}^{2}$ ${W}_{i}^{2}\in \mathrm{{ℤ}^{+}}$ the smallest of the Pass 2 Adjustments of opponents of the i-th player against whom he/she lost a match.
${\alpha }_{k}^{2}$ ${\alpha }_{k}^{2}\in \mathrm{{ℤ}^{+}}$ the Pass 2 Adjustment of the player who was the opponent of the i-th player in the k-th match
$I\left(x\right)$ $I:ℤ↦\mathrm{{ℤ}^{+}}$ a function that maps all integers to one of the values from -- 0, 1, 5, 10.
${M}_{i}$ ${M}_{i}\in \mathrm{{ℤ}^{+}}$ total number of matches played by the i-th player in this tournament
k $k\in \mathrm{\left[0,\mathrm{{M}_{i}}-1\right]\cap {ℤ}^{+}}$ The index of the match of the i-th player ranging from 0 to ${M}_{i}-1$
5. For the rated players, all the work done in Pass 1 and Pass 2 to undone and they have their ratings reset back to their initial ratings while the unrated players keep their Pass 2 Adjustment as their final Pass 2 Rating. Since Mark E. Berg is a rated player, his Pass 2 Adjustment of 1761 will be ignored, along with him Pass 1 Rating of 1775 and his Pass 2 Rating will be set to his initial rating of 1761 with which he came into this tournament.

Click here to see detailed information about the Pass 2 Ratings of all the other players in this tournament.

##### Pass 3 Rating
Any of the unrated players who have all wins or all losses are skipped in Pass 3. Since Mark E. Berg has an initial rating of 1761, he is not an unrated player, and therefore this rule does not apply to him. You can click here to view list of all the players that are skipped in this Pass 3.

Pass 3 Rating is calculated using 2 steps described below:
1. In the first part of Pass 3, we apply the point exchange table described in Pass 1 above except this time by using all the players' Pass 2 Ratings. Looking at Mark E. Berg's wins and losses and applying the point exchange table, gives us the following result:
##### Mark E. Berg's Wins
Winner Loser
Point Spread Outcome Gain Player USATT # Rating Player USATT # Rating
533 EXPECTED 0 Mark E. Berg 6250 1761 Hank Roy 83208 1228
92 EXPECTED 4 Mark E. Berg 6250 1761 MICHAEL REFF 24318 1669
116 EXPECTED 3 Mark E. Berg 6250 1761 David (Pavlovich) Nelyubin 76519 1645
330 EXPECTED 0 Mark E. Berg 6250 1761 John F. Mcfadden 29798 1431
143 EXPECTED 2 Mark E. Berg 6250 1761 Lily Xiong 88715 1618
151 EXPECTED 2 Mark E. Berg 6250 1761 David Maisel 76188 1610
60 UPSET 13 Mark E. Berg 6250 1761 Alok Kumar 29925 1821
58 EXPECTED 6 Mark E. Berg 6250 1761 Steven Chen 90786 1703

##### Mark E. Berg's Losses
Winner Loser
Point Spread Outcome Loss Player USATT # Rating Player USATT # Rating
73 UPSET -16 Trevor Salmon 88519 1688 Mark E. Berg 6250 1761
272 EXPECTED -0 Dmitry Tokmakov 80594 2033 Mark E. Berg 6250 1761
267 EXPECTED -0 Adnan Medunjanin 83321 2028 Mark E. Berg 6250 1761

You can click here to view a table of outcomes and points gained/lost from all the matches with all the players in this tournament for Pass 3 Part 1.

The "Outcome" column above, shows whether the match had an expected (player with the higher rating wins the match) or an upset (player with the higher rating loses the match) outcome. Based on this outcome, and using both the player's Pass 2 Rating, we apply the point exchange table from above and show the rating points earned and lost by Mark E. Berg in the "Gain" column. Matches are divided up into two tables for wins and losses where points are "Gain"ed for the wins and "loss"ed for losses. Putting all the gains and losses together, we get the following math to calculate the rating for Mark E. Berg in this first part of Pass 3:

Pass 2 Rating Gains/Losses Pass 3 Part 1 Rating
1761 + 0 + 4 + 3 + 0 + 2 + 2 + 13 + 6 - 16 + 0 + 0 $=\mathrm{1775}$

You can click here to view a table of these calculations for all the players in this tournament.

2. Given the Pass 3 Part 1 rating calculated above, the second part of Pass 3 looks very similar to the part of Pass 2 that deals with rated players where we calculate their Pass 2 Adjustment.
1. First, we calculate the points gained in Pass 3 Part 1. Points gained is simply the difference between the Pass 3 Part 1 Rating and the Pass 2 Rating of a player:

${\rho }_{i}^{3}={p}_{i}^{3}-{P}_{i}^{2}$
where,

 Symbol Universe Description ${P}_{i}^{2}$ ${P}_{i}^{2}\in \mathrm{{ℤ}^{+}}$ the Pass 2 Rating for the $i$-th player. ${p}_{i}^{3}$ ${p}_{i}^{3}\in \mathrm{{ℤ}^{+}}$ the Pass 3 Part 1 rating for the $i$-th player. (Note that since this is an intermediate result, we are using a lower case p instead of the upper case P that we use to indicate final result from each pass of the ratings processor. ${\rho }_{i}^{3}$ ${\rho }_{i}^{3}\in ℤ$ the points gained by the $i$-th player in this tournament in Pass 3. $i$ $i\in \left[1,\mathrm{120}\right]\cap ℤ$ the index of the player under consideration. $i$ can be as small as $1$ or as large as $\mathrm{120}$ for this tournament.

3. Pass 3 points gained, ${\rho }_{i}^{3}$, is then used to calculate the Pass 3 Part 2 Rating in the following way:
1. If a player gained less than 50 points (exclusive) in Pass 3 Part 1, then we set that player's Pass 3 Part 2 Rating to his/her Pass 2 Rating.
2. If a player gained between 50 and 74 (inclusive) points in Pass 3 Part 1, then we set the player's Pass 3 Part 2 Rating to his/her Pass 3 Part 1 Rating.
3. If a player gains 75 or more points (inclusive) in Pass 3 Part 1, then the following formula applies:
• If the player has won at least one match, and lost at least 1 match in the tournament, then the player's Pass 3 Part 2 Rating is the average of his/her Pass 3 Part 1 Rating and the average of his/her opponents rating from the best win and the worst loss, represented using the formula below:

$\mathrm{{\alpha }_{i}^{3}}=⌊\mathrm{\frac{\mathrm{{p}_{i}^{3}}+\mathrm{\frac{\mathrm{{B}_{i}^{3}}+\mathrm{{W}_{i}^{3}}}{2}}}{2}}⌋$

where ${\alpha }_{i}^{3}$ is the Pass 3 Part 2 Rating for the current player, ${p}_{i}^{3}$ is the Pass 3 Part 1 Rating, ${B}_{i}^{3}$ is the rating of the highest rated opponent against which the current player won a match, and ${W}_{i}$ is the rating of the lowest rated opponent against which the current player lost a match.
• If a player has not lost any of his/her matches in the current tournament, the mathematical median (rounded down to the nearest integer) of all of the player's opponents rating is used as his/her Pass 3 Part 2 Rating:
$\mathrm{{\alpha }_{i}^{3}}=\mathrm{⌊\stackrel{\sim }{\mathrm{\left\{\mathrm{{p}_{k}^{3}}\right\}}}⌋}$

where ${p}_{k}^{3}$ is the Pass 3 Part 1 Rating of the i-th player's opponent from the k-th match.

• Therefore, the Pass 3 Part 2 Rating for Mark E. Berg is calculated as follows:
• Given the Pass 2 Rating of 1761,
• and the Pass 3 Part 1 rating of 1775,
• the Pass 3 Part 1 gain is 1775 - 1761 = 14.
• Since the Pass 3 Gain of 14 is less than 50, the Pass 3 Part 2 Rating is reset back to the Pass 2 Rating.
• Therefore the Pass 3 Part 2 Rating for Mark E. Berg is 1761.

The Pass 3 Part 2 rating ends up becoming the final Pass 3 rating (also referred to as the Pass 3 Adjustment) except as follows:
• In the cases where the Pass 3 Part 2 rating is less than the players' initial rating ${P}_{i}^{0}$, the Pass 3 rating is reset back to that players initial rating. Mark E. Berg's Pass 3 Part 2 Rating came out to 1761. Since this value is greater than Mark E. Berg's initial rating of 1761, his Pass 3 Adjustment is set to his Pass 3 Part 2 Rating of 1761.

• It is possible for the admin of this tournament to override the Pass 3 Adjustment calculated above with a value they deem appropriate. Mark E. Berg does not have a manually overridden value for his Pass 3 Adjustment, therefore the value remains at 1761.
You can click here to view a table of Pass 3 Part 2 Ratings for all the players in this tournament along with any manually overridden values.

##### Pass 4 Rating
Pass 4 is the final pass of the ratings processor. In this pass, we take the adjusted ratings (Pass 3 Adjustment) of all the rated players, and the assigned rating of unrated players (Pass 2 Rating), and apply the point exchange table to the match results based on these ratings to arrive at a final rating. Looking at Mark E. Berg's match results and applying the point exchange table, gives us the following result:

##### Mark E. Berg's Wins
Winner Loser
Point Spread Outcome Gain Player USATT # Rating Player USATT # Rating
469 EXPECTED 0 Mark E. Berg 6250 1761 Hank Roy 83208 1292
92 EXPECTED 4 Mark E. Berg 6250 1761 MICHAEL REFF 24318 1669
116 EXPECTED 3 Mark E. Berg 6250 1761 David (Pavlovich) Nelyubin 76519 1645
330 EXPECTED 0 Mark E. Berg 6250 1761 John F. Mcfadden 29798 1431
93 EXPECTED 4 Mark E. Berg 6250 1761 Lily Xiong 88715 1668
151 EXPECTED 2 Mark E. Berg 6250 1761 David Maisel 76188 1610
60 UPSET 13 Mark E. Berg 6250 1761 Alok Kumar 29925 1821
58 EXPECTED 6 Mark E. Berg 6250 1761 Steven Chen 90786 1703

##### Mark E. Berg's Losses
Winner Loser
Point Spread Outcome Loss Player USATT # Rating Player USATT # Rating
57 EXPECTED -6 Trevor Salmon 88519 1818 Mark E. Berg 6250 1761
272 EXPECTED -0 Dmitry Tokmakov 80594 2033 Mark E. Berg 6250 1761
267 EXPECTED -0 Adnan Medunjanin 83321 2028 Mark E. Berg 6250 1761

You can click here to view a table of outcomes and points gained/lost from all the matches with all the players in this tournament.

The "Outcome" column above, shows whether the match had an expected (player with the higher rating wins the match) or an upset (player with the higher rating loses the match) outcome. Based on this outcome, and using both the players' Pass 3 Adjustment, we apply the point exchange table from above and show the ratings points earned and lost by Mark E. Berg in the "Gain" and "Loss" columns. Matches are separated out into two tables for wins and losses where points are gained and lost respectively. We get the following math to calculate the Pass 4 Rating for Mark E. Berg:

Pass 3 Rating Gains/Losses Pass 4 Rating
1761 + 0 + 4 + 3 + 0 + 4 + 2 + 13 + 6 - 6 + 0 + 0 $=\mathrm{1787}$

You can click here to view a table of Pass 4 calculations for all the players in this tournament.