 for  Amgalan Bayanbat

USATT#: 1172377

##### Introduction
This page explains how Amgalan Bayanbat (USATT# 1172377)'s rating went from n/a to 949 at the 2021 Washington D.C. Table Tennis Giant Round Robin held on 28 Aug 2021 - 29 Aug 2021. 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 Amgalan Bayanbat for this tournament.

Initial Rating Pass 1 Pass 2 Pass 3 Final Rating (Pass 4)
n/a 766 1000 949

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 27 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 2021 Washington D.C. Table Tennis Giant Round Robin held on 28 Aug 2021 - 29 Aug 2021 for Amgalan Bayanbat, and its source tournament are as follows:
Initial Rating From Tournament Start Day End Day
n/a n/a

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 Amgalan Bayanbat's match results and applying the point exchange table, gives us the following result:

##### Amgalan Bayanbat's Wins
Winner Loser
Point Spread Outcome Gain Player USATT # Rating Player USATT # Rating
0 0 Amgalan Bayanbat 1172377 0 James Nix 1171942 0
0 0 Amgalan Bayanbat 1172377 0 James Nix 1171942 0

##### Amgalan Bayanbat's Losses
Winner Loser
Point Spread Outcome Loss Player USATT # Rating Player USATT # Rating
0 -0 Lennox Douglas 3212 0 Amgalan Bayanbat 1172377 0
0 -0 Aaron Zhang 218476 0 Amgalan Bayanbat 1172377 0
0 -0 Benicio Nix 1171941 0 Amgalan Bayanbat 1172377 0
0 -0 David Sakai 9668 0 Amgalan Bayanbat 1172377 0
0 -0 Benicio Nix 1171941 0 Amgalan Bayanbat 1172377 0
0 -0 Ryan"TheNightmare" Gibson 89377 0 Amgalan Bayanbat 1172377 0
0 -0 Sameer Wadkar 216784 0 Amgalan Bayanbat 1172377 0
0 -0 David Sakai 9668 0 Amgalan Bayanbat 1172377 0
0 -0 Liam Draper 1170916 0 Amgalan Bayanbat 1172377 0
0 -0 Liam Draper 1170916 0 Amgalan Bayanbat 1172377 0
0 -0 Jasmine Lennert 94083 0 Amgalan Bayanbat 1172377 0

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 Amgalan Bayanbat 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 Amgalan Bayanbat:

Initial Rating Gains/Losses Pass 1 Rating
$=0$

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 ${\rho }_{i}^{2}$. $i$ $i\in \left[1,\mathrm{27}\right]\cap ℤ$ the index of the player under consideration. $i$ can be as small as $1$ or as large as $\mathrm{27}$ 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
$i$ $i\in \left[1,\mathrm{27}\right]\cap ℤ$ the index of the player under consideration. $i$ can be as small as $1$ or as large as $\mathrm{27}$ for this tournament and the i-th player must be a rated player.
$q$ $q\in \left[1,\mathrm{133}\right]\cap ℤ$ the index of the match result under consideration. $q$ can be as small as $1$ or as large as $\mathrm{133}$ 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.

• Since Amgalan Bayanbat did not come to this tournament with an initial rating, she does not get a Pass 2 Adjustment. 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 Amgalan Bayanbat's matches were against unrated players. So this rule does not apply to her.
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, \begin{equation} 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. \end{equation}
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.

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. Amgalan Bayanbat will be skipped from Pass 3 since he has all wins and no losses. 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 Amgalan Bayanbat's wins and losses and applying the point exchange table, gives us the following result:
##### Amgalan Bayanbat's Wins
Winner Loser
Point Spread Outcome Gain Player USATT # Rating Player USATT # Rating
0 0 Amgalan Bayanbat 1172377 0 James Nix 1171942 0
0 0 Amgalan Bayanbat 1172377 0 James Nix 1171942 0

##### Amgalan Bayanbat's Losses
Winner Loser
Point Spread Outcome Loss Player USATT # Rating Player USATT # Rating
0 -0 Lennox Douglas 3212 0 Amgalan Bayanbat 1172377 0
0 -0 Aaron Zhang 218476 0 Amgalan Bayanbat 1172377 0
0 -0 Benicio Nix 1171941 0 Amgalan Bayanbat 1172377 0
0 -0 David Sakai 9668 0 Amgalan Bayanbat 1172377 0
0 -0 Benicio Nix 1171941 0 Amgalan Bayanbat 1172377 0
0 -0 Ryan"TheNightmare" Gibson 89377 0 Amgalan Bayanbat 1172377 0
0 -0 Sameer Wadkar 216784 0 Amgalan Bayanbat 1172377 0
0 -0 David Sakai 9668 0 Amgalan Bayanbat 1172377 0
0 -0 Liam Draper 1170916 0 Amgalan Bayanbat 1172377 0
0 -0 Liam Draper 1170916 0 Amgalan Bayanbat 1172377 0
0 -0 Jasmine Lennert 94083 0 Amgalan Bayanbat 1172377 0

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 Amgalan Bayanbat 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 Amgalan Bayanbat in this first part of Pass 3:

Pass 2 Rating Gains/Losses Pass 3 Part 1 Rating
766 $=\mathrm{766}$

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. $i$ $i\in \left[1,\mathrm{27}\right]\cap ℤ$ the index of the player under consideration. $i$ can be as small as $1$ or as large as $\mathrm{27}$ 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 Amgalan Bayanbat is calculated as follows:
• Given the Pass 2 Rating of 766,
• and the Pass 3 Part 1 rating of 766,
• the Pass 3 Part 1 gain is 766 - 766 = 0.
• Since the Pass 3 Gain of 0 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 Amgalan Bayanbat is 766.

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. Amgalan Bayanbat's Pass 3 Part 2 Rating came out to 766. Since this value is greater than Amgalan Bayanbat's initial rating of 0, her Pass 3 Adjustment is set to her Pass 3 Part 2 Rating of 766.

• It is possible for the admin of this tournament to override the Pass 3 Adjustment calculated above with a value they deem appropriate. This tournament's admins have set a manual Pass 3 Rating of 1000, therefore that will become Amgalan Bayanbat's Pass 3 Adjustment.
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 Amgalan Bayanbat's match results and applying the point exchange table, gives us the following result:

##### Amgalan Bayanbat's Wins
Winner Loser
Point Spread Outcome Gain Player USATT # Rating Player USATT # Rating
347 EXPECTED 0 Amgalan Bayanbat 1172377 1000 James Nix 1171942 653
347 EXPECTED 0 Amgalan Bayanbat 1172377 1000 James Nix 1171942 653

##### Amgalan Bayanbat's Losses
Winner Loser
Point Spread Outcome Loss Player USATT # Rating Player USATT # Rating
1003 EXPECTED -0 Lennox Douglas 3212 2003 Amgalan Bayanbat 1172377 1000
441 EXPECTED -0 Aaron Zhang 218476 1441 Amgalan Bayanbat 1172377 1000
633 EXPECTED -0 Benicio Nix 1171941 1633 Amgalan Bayanbat 1172377 1000
1040 EXPECTED -0 David Sakai 9668 2040 Amgalan Bayanbat 1172377 1000
633 EXPECTED -0 Benicio Nix 1171941 1633 Amgalan Bayanbat 1172377 1000
1125 EXPECTED -0 Ryan"TheNightmare" Gibson 89377 2125 Amgalan Bayanbat 1172377 1000
544 EXPECTED -0 Sameer Wadkar 216784 1544 Amgalan Bayanbat 1172377 1000
1040 EXPECTED -0 David Sakai 9668 2040 Amgalan Bayanbat 1172377 1000
121 UPSET -25 Liam Draper 1170916 879 Amgalan Bayanbat 1172377 1000
121 UPSET -25 Liam Draper 1170916 879 Amgalan Bayanbat 1172377 1000
218 EXPECTED -1 Jasmine Lennert 94083 1218 Amgalan Bayanbat 1172377 1000

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 Amgalan Bayanbat 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 Amgalan Bayanbat:

Pass 3 Rating Gains/Losses Pass 4 Rating
1000 + 0 + 0 + 0 + 0 + 0 + 0 + 0 + 0 - 25 + 0 + 0 - 25 - 1 $=\mathrm{949}$

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