How to get previous ELO score from current and match result












3












$begingroup$


I want to get a previous ELO value knowing the opponent's value and the result of the match. There are two equations, one for a loss and one for a win:



Win



$$ELO_{Current} = ELO_{Previous} + K cdot left(1 - frac{1}{1 + 10^{{(ELO_{Opponent} - ELO_{Previous})}/{400}}}right)$$



Loss



$$ELO_{Current} = ELO_{Previous} + K cdot left(0 - frac{1}{1 + 10^{{(ELO_{Opponent} - ELO_{Previous})}/{400}}}right)$$



$K, ELO_{Opponent}, ELO_{Current}$ and whether it was a win or not are known, which is why I figured that it should be possible to get $ELO_{Previous}$ by rearranging the equation, but I have no clue how to do it.






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I find it a bit puzzling that you know the opponent's score before the match but not the player's score, and you know the player's score after the match but not the opponent's. Is that really the information you have?
    $endgroup$
    – David K
    Dec 22 '18 at 21:53












  • $begingroup$
    @DavidK Yes, that is strange, but also the information I need to work with. This will never be the case with sufficient information, but what I can work with is really limited.
    $endgroup$
    – creativecreatorormaybenot
    Dec 22 '18 at 22:14






  • 1




    $begingroup$
    This is sufficient information in the mathematical sense (in that it completely determines the remaining data you want); it's just strange and somewhat difficult to work with. I hope you're able to work out a method that you find suitable.
    $endgroup$
    – David K
    Dec 22 '18 at 22:22


















3












$begingroup$


I want to get a previous ELO value knowing the opponent's value and the result of the match. There are two equations, one for a loss and one for a win:



Win



$$ELO_{Current} = ELO_{Previous} + K cdot left(1 - frac{1}{1 + 10^{{(ELO_{Opponent} - ELO_{Previous})}/{400}}}right)$$



Loss



$$ELO_{Current} = ELO_{Previous} + K cdot left(0 - frac{1}{1 + 10^{{(ELO_{Opponent} - ELO_{Previous})}/{400}}}right)$$



$K, ELO_{Opponent}, ELO_{Current}$ and whether it was a win or not are known, which is why I figured that it should be possible to get $ELO_{Previous}$ by rearranging the equation, but I have no clue how to do it.






algebra-precalculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I find it a bit puzzling that you know the opponent's score before the match but not the player's score, and you know the player's score after the match but not the opponent's. Is that really the information you have?
    $endgroup$
    – David K
    Dec 22 '18 at 21:53












  • $begingroup$
    @DavidK Yes, that is strange, but also the information I need to work with. This will never be the case with sufficient information, but what I can work with is really limited.
    $endgroup$
    – creativecreatorormaybenot
    Dec 22 '18 at 22:14






  • 1




    $begingroup$
    This is sufficient information in the mathematical sense (in that it completely determines the remaining data you want); it's just strange and somewhat difficult to work with. I hope you're able to work out a method that you find suitable.
    $endgroup$
    – David K
    Dec 22 '18 at 22:22
















3












3








3


1



$begingroup$


I want to get a previous ELO value knowing the opponent's value and the result of the match. There are two equations, one for a loss and one for a win:



Win



$$ELO_{Current} = ELO_{Previous} + K cdot left(1 - frac{1}{1 + 10^{{(ELO_{Opponent} - ELO_{Previous})}/{400}}}right)$$



Loss



$$ELO_{Current} = ELO_{Previous} + K cdot left(0 - frac{1}{1 + 10^{{(ELO_{Opponent} - ELO_{Previous})}/{400}}}right)$$



$K, ELO_{Opponent}, ELO_{Current}$ and whether it was a win or not are known, which is why I figured that it should be possible to get $ELO_{Previous}$ by rearranging the equation, but I have no clue how to do it.






algebra-precalculus







share|cite|improve this question











$endgroup$




I want to get a previous ELO value knowing the opponent's value and the result of the match. There are two equations, one for a loss and one for a win:



Win



$$ELO_{Current} = ELO_{Previous} + K cdot left(1 - frac{1}{1 + 10^{{(ELO_{Opponent} - ELO_{Previous})}/{400}}}right)$$



Loss



$$ELO_{Current} = ELO_{Previous} + K cdot left(0 - frac{1}{1 + 10^{{(ELO_{Opponent} - ELO_{Previous})}/{400}}}right)$$



$K, ELO_{Opponent}, ELO_{Current}$ and whether it was a win or not are known, which is why I figured that it should be possible to get $ELO_{Previous}$ by rearranging the equation, but I have no clue how to do it.






algebra-precalculus




algebra-precalculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 22 '18 at 20:39









David K

55.8k345121




55.8k345121










asked Dec 22 '18 at 14:56









creativecreatorormaybenotcreativecreatorormaybenot

1234




1234








  • 1




    $begingroup$
    I find it a bit puzzling that you know the opponent's score before the match but not the player's score, and you know the player's score after the match but not the opponent's. Is that really the information you have?
    $endgroup$
    – David K
    Dec 22 '18 at 21:53












  • $begingroup$
    @DavidK Yes, that is strange, but also the information I need to work with. This will never be the case with sufficient information, but what I can work with is really limited.
    $endgroup$
    – creativecreatorormaybenot
    Dec 22 '18 at 22:14






  • 1




    $begingroup$
    This is sufficient information in the mathematical sense (in that it completely determines the remaining data you want); it's just strange and somewhat difficult to work with. I hope you're able to work out a method that you find suitable.
    $endgroup$
    – David K
    Dec 22 '18 at 22:22
















  • 1




    $begingroup$
    I find it a bit puzzling that you know the opponent's score before the match but not the player's score, and you know the player's score after the match but not the opponent's. Is that really the information you have?
    $endgroup$
    – David K
    Dec 22 '18 at 21:53












  • $begingroup$
    @DavidK Yes, that is strange, but also the information I need to work with. This will never be the case with sufficient information, but what I can work with is really limited.
    $endgroup$
    – creativecreatorormaybenot
    Dec 22 '18 at 22:14






  • 1




    $begingroup$
    This is sufficient information in the mathematical sense (in that it completely determines the remaining data you want); it's just strange and somewhat difficult to work with. I hope you're able to work out a method that you find suitable.
    $endgroup$
    – David K
    Dec 22 '18 at 22:22










1




1




$begingroup$
I find it a bit puzzling that you know the opponent's score before the match but not the player's score, and you know the player's score after the match but not the opponent's. Is that really the information you have?
$endgroup$
– David K
Dec 22 '18 at 21:53






$begingroup$
I find it a bit puzzling that you know the opponent's score before the match but not the player's score, and you know the player's score after the match but not the opponent's. Is that really the information you have?
$endgroup$
– David K
Dec 22 '18 at 21:53














$begingroup$
@DavidK Yes, that is strange, but also the information I need to work with. This will never be the case with sufficient information, but what I can work with is really limited.
$endgroup$
– creativecreatorormaybenot
Dec 22 '18 at 22:14




$begingroup$
@DavidK Yes, that is strange, but also the information I need to work with. This will never be the case with sufficient information, but what I can work with is really limited.
$endgroup$
– creativecreatorormaybenot
Dec 22 '18 at 22:14




1




1




$begingroup$
This is sufficient information in the mathematical sense (in that it completely determines the remaining data you want); it's just strange and somewhat difficult to work with. I hope you're able to work out a method that you find suitable.
$endgroup$
– David K
Dec 22 '18 at 22:22






$begingroup$
This is sufficient information in the mathematical sense (in that it completely determines the remaining data you want); it's just strange and somewhat difficult to work with. I hope you're able to work out a method that you find suitable.
$endgroup$
– David K
Dec 22 '18 at 22:22












1 Answer
1






active

oldest

votes


















1












$begingroup$

To make the formulas a little easier to read and write, let
begin{align}
E_C &= ELO_{Current}, \
E_P &= ELO_{Previous}, \
E_O &= ELO_{Opponent}. \
end{align}



If we let $w = 1$ for a win, $w = 0$ for a loss, then we can summarize the two equations as
$$
E_C = E_P + K left(w - frac1{1 + 10^{(E_O - E_P)/400}}right).
$$



The unknown value is $E_P.$
Now let's start rearranging this equation in an attempt to solve it.
Distributing the $K$ over the parentheses and collecting all terms on the left side,
we get
$$
frac K{1 + 10^{(E_O - E_P)/400}} + E_C - wK - E_P = 0.
$$
Clear the fraction and separate the exponents:
$$
K + (E_C - wK - E_P) left(1 + 10^{E_O/400}left(10^{-1/400}right)^{E_P} right) = 0
$$



There are still other ways to try to arrange the terms, but the fact that $E_P$
occurs as a linear term in on of the factors and an exponent in another factor seems to be a clear indication that this equation will not have a solution using ordinary operations and functions.
The best you can do is to use numerical methods, which is a kind of sophisticated guesswork.








share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    For example, we could use the bisection method or Newton's method to solve for $E_P$ numerically. Bisection method would be simple and reliable, I think.
    $endgroup$
    – littleO
    Dec 22 '18 at 21:59












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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









1












$begingroup$

To make the formulas a little easier to read and write, let
begin{align}
E_C &= ELO_{Current}, \
E_P &= ELO_{Previous}, \
E_O &= ELO_{Opponent}. \
end{align}



If we let $w = 1$ for a win, $w = 0$ for a loss, then we can summarize the two equations as
$$
E_C = E_P + K left(w - frac1{1 + 10^{(E_O - E_P)/400}}right).
$$



The unknown value is $E_P.$
Now let's start rearranging this equation in an attempt to solve it.
Distributing the $K$ over the parentheses and collecting all terms on the left side,
we get
$$
frac K{1 + 10^{(E_O - E_P)/400}} + E_C - wK - E_P = 0.
$$
Clear the fraction and separate the exponents:
$$
K + (E_C - wK - E_P) left(1 + 10^{E_O/400}left(10^{-1/400}right)^{E_P} right) = 0
$$



There are still other ways to try to arrange the terms, but the fact that $E_P$
occurs as a linear term in on of the factors and an exponent in another factor seems to be a clear indication that this equation will not have a solution using ordinary operations and functions.
The best you can do is to use numerical methods, which is a kind of sophisticated guesswork.








share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    For example, we could use the bisection method or Newton's method to solve for $E_P$ numerically. Bisection method would be simple and reliable, I think.
    $endgroup$
    – littleO
    Dec 22 '18 at 21:59
















1












$begingroup$

To make the formulas a little easier to read and write, let
begin{align}
E_C &= ELO_{Current}, \
E_P &= ELO_{Previous}, \
E_O &= ELO_{Opponent}. \
end{align}



If we let $w = 1$ for a win, $w = 0$ for a loss, then we can summarize the two equations as
$$
E_C = E_P + K left(w - frac1{1 + 10^{(E_O - E_P)/400}}right).
$$



The unknown value is $E_P.$
Now let's start rearranging this equation in an attempt to solve it.
Distributing the $K$ over the parentheses and collecting all terms on the left side,
we get
$$
frac K{1 + 10^{(E_O - E_P)/400}} + E_C - wK - E_P = 0.
$$
Clear the fraction and separate the exponents:
$$
K + (E_C - wK - E_P) left(1 + 10^{E_O/400}left(10^{-1/400}right)^{E_P} right) = 0
$$



There are still other ways to try to arrange the terms, but the fact that $E_P$
occurs as a linear term in on of the factors and an exponent in another factor seems to be a clear indication that this equation will not have a solution using ordinary operations and functions.
The best you can do is to use numerical methods, which is a kind of sophisticated guesswork.








share|cite|improve this answer









$endgroup$









  • 2




    $begingroup$
    For example, we could use the bisection method or Newton's method to solve for $E_P$ numerically. Bisection method would be simple and reliable, I think.
    $endgroup$
    – littleO
    Dec 22 '18 at 21:59














1












1








1





$begingroup$

To make the formulas a little easier to read and write, let
begin{align}
E_C &= ELO_{Current}, \
E_P &= ELO_{Previous}, \
E_O &= ELO_{Opponent}. \
end{align}



If we let $w = 1$ for a win, $w = 0$ for a loss, then we can summarize the two equations as
$$
E_C = E_P + K left(w - frac1{1 + 10^{(E_O - E_P)/400}}right).
$$



The unknown value is $E_P.$
Now let's start rearranging this equation in an attempt to solve it.
Distributing the $K$ over the parentheses and collecting all terms on the left side,
we get
$$
frac K{1 + 10^{(E_O - E_P)/400}} + E_C - wK - E_P = 0.
$$
Clear the fraction and separate the exponents:
$$
K + (E_C - wK - E_P) left(1 + 10^{E_O/400}left(10^{-1/400}right)^{E_P} right) = 0
$$



There are still other ways to try to arrange the terms, but the fact that $E_P$
occurs as a linear term in on of the factors and an exponent in another factor seems to be a clear indication that this equation will not have a solution using ordinary operations and functions.
The best you can do is to use numerical methods, which is a kind of sophisticated guesswork.








share|cite|improve this answer









$endgroup$



To make the formulas a little easier to read and write, let
begin{align}
E_C &= ELO_{Current}, \
E_P &= ELO_{Previous}, \
E_O &= ELO_{Opponent}. \
end{align}



If we let $w = 1$ for a win, $w = 0$ for a loss, then we can summarize the two equations as
$$
E_C = E_P + K left(w - frac1{1 + 10^{(E_O - E_P)/400}}right).
$$



The unknown value is $E_P.$
Now let's start rearranging this equation in an attempt to solve it.
Distributing the $K$ over the parentheses and collecting all terms on the left side,
we get
$$
frac K{1 + 10^{(E_O - E_P)/400}} + E_C - wK - E_P = 0.
$$
Clear the fraction and separate the exponents:
$$
K + (E_C - wK - E_P) left(1 + 10^{E_O/400}left(10^{-1/400}right)^{E_P} right) = 0
$$



There are still other ways to try to arrange the terms, but the fact that $E_P$
occurs as a linear term in on of the factors and an exponent in another factor seems to be a clear indication that this equation will not have a solution using ordinary operations and functions.
The best you can do is to use numerical methods, which is a kind of sophisticated guesswork.









share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 22 '18 at 21:50









David KDavid K

55.8k345121




55.8k345121








  • 2




    $begingroup$
    For example, we could use the bisection method or Newton's method to solve for $E_P$ numerically. Bisection method would be simple and reliable, I think.
    $endgroup$
    – littleO
    Dec 22 '18 at 21:59














  • 2




    $begingroup$
    For example, we could use the bisection method or Newton's method to solve for $E_P$ numerically. Bisection method would be simple and reliable, I think.
    $endgroup$
    – littleO
    Dec 22 '18 at 21:59








2




2




$begingroup$
For example, we could use the bisection method or Newton's method to solve for $E_P$ numerically. Bisection method would be simple and reliable, I think.
$endgroup$
– littleO
Dec 22 '18 at 21:59




$begingroup$
For example, we could use the bisection method or Newton's method to solve for $E_P$ numerically. Bisection method would be simple and reliable, I think.
$endgroup$
– littleO
Dec 22 '18 at 21:59


















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