# The ating problem

dukemper
Posers and Puzzles 18 Feb '03 23:14
1. 18 Feb '03 23:142 edits
(oh, π that should read &quot;rating problem&quot;..)
My problem is simple:

Say you have a rating of 1400. And there are two games pending, just one move away from your certain defeat (so you could resign).
one opponent has 1200 the other one 1300 - given that your rating at RHP (which is a bit unlucky) is calculated with the point basis at the time you end the game (!), not with the values when you started it.
What game to end first? (in order to choose the least deteriorating order..)

The same problem when you are about to defeat two different opponents.
What and how much changes the difference of their ratings?

Can you guess it? Can you appoximise it? Can you solve the general case analytically? What to do if there are games waiting for the two cases?
(so thats why people tend to let you wait when the game is practically over π)

From the FAQ-section:
Players are rated using the following formula:
New Rating = Old Rating + K * (1 - Win Expectancy)
K is a constant (32 for 0-2099, 24 for 2100-2399, 16 for 2400 and above)
The Win Expectancy is calculated using the following formula:
Win Expectancy = 1 / (10^((OpponentRating-YourRating)/400)+1)

Have fun...

2. 10 Mar '04 21:18
Beat the 1300 playerπ
3. 11 Mar '04 02:22
From the FAQ-section:
Players are rated using the following formula:
New Rating = Old Rating + K * (1 - Win Expectancy)
K is a constant (32 for 0-2099, 24 for 2100-2399, 16 for 2400 and above)
The Win Expectancy is calculated using the following formula:
Win Expectancy = 1 / (10^((OpponentRating-YourRating)/400)+1)
There is a problem with the formula you have presented here: it refers only to a winning situation, but your problem deals with two losing situations. It is not possible to figure out what decrease will result in either loss with this formula. I'm too lazy to go looking for the losing situation's fomula, so if you can find it, I'll then try to figure out what change will result in your rating.
4. richjohnson
TANSTAAFL
11 Mar '04 02:41
Originally posted by econundrum
There is a problem with the formula you have presented here: it refers only to a winning situation, but your problem deals with two losing situations. It is not possible to figure out what decrease will result in either loss with this formula. I'm too lazy to go looking for the losing situation's fomula, so if you can find it, I'll then try to figure out what change will result in your rating.
I believe the formula is the same except for replacing the + with a - after &quot;Old rating&quot;.
5. 11 Mar '04 02:52
Originally posted by richjohnson
I believe the formula is the same except for replacing the + with a - after "Old rating".
I might be wrong in assuming this, but if that were true then the absolute value of the change in your rating, whether you won or lost would be the same. I doubt that because if a person with a 2400 rating lost against a person with a 1000 rating, I expect that the decrease in the first player's rating would be much greater than the increase would be for a win. Again, I'm too lazy to look it up, so I'm not sure, but it makes sense for what I just said to be true.
6. richjohnson
TANSTAAFL
11 Mar '04 17:55
Originally posted by econundrum
I might be wrong in assuming this, but if that were true then the absolute value of the change in your rating, whether you won or lost would be the same.
I think that's right - points are conserved with the winner receiving the same amount lost by the loser. So if a 1200 player lost to a 2000 player, the 2000 player would win only one point (if that), while if a 1200 beat a 2000 player, the 2000 player would lose a lot of points. I'm too lazy to look it up too, but we could always play a game to find out first hand - send me a challenge if you're interested.
7. 11 Mar '04 18:06
Originally posted by econundrum
There is a problem with the formula you have presented here: it refers only to a winning situation, but your problem deals with two losing situations. It is not possible to figure out what decrease will result in either loss with this formula. I'm too lazy to go looking for the losing situation's fomula, so if you can find it, I'll then try to figure out what change will result in your rating.
The definition of &quot;score&quot; sorts this problem out. Basically your rating change by a lot if either you were expected to win and lost, or is you were expected to lose but won. The full text from the help is:

New Rating = Old Rating + K * (Score - Win Expectancy)

K is a constant (32 for 0-2099, 24 for 2100-2399, 16 for 2400 and above)

Score is 1 for a win, 0.5 for a draw and 0 for a loss.

The Win Expectancy is calculated using the following formula :

Win Expectancy = 1 / (10^((OpponentRating-YourRating)/400)+1)
8. 12 Mar '04 03:19
Originally posted by iamatiger
The definition of "score" sorts this problem out. Basically your rating change by a lot if either you were expected to win and lost, or is you were expected to lose but won. The full text from the help is:

New Rating = Old Rating + K * (Score - Win Expectancy)

K is a constant (32 for 0-2099, 24 for 2100-2399, 16 for 2400 and above)

Score is 1 fo ...[text shortened]... sing the following formula :

Win Expectancy = 1 / (10^((OpponentRating-YourRating)/400)+1)
Thank you, Iamatiger. The &quot;score&quot; makes much more sense, and would definitely make a diference to the score. Now I'll try to figure out the best outcome.
9. 12 Mar '04 03:56
The answer to this is rather simple:
Should you consider the loss to the 1200 rated player first, followed by the 1300 player, the calculation of your new score would be this:

(1400+32(0-1/(10^((1200-1400)/400)+1)))+32(0-1/(10^((1300-1400)/400)+1)) = 1400+32(-1/(10^(-.5)+1))+32(-1/(10^(-.25)+1))

Conversely, losing to the 1300 player first, then the 1200 player would result in:

1400+32(-1/(10^(-.25)+1))+32(-1/(10^(-.5)+1))

Since each case results in the same value (each adding the same terms but in a different order), it would not matter who you lost to first because your final rating would be the same. This is also true for winning two people of different ratings because the only difference between the calculation of the wins would be that 1 would be added to the terms inside of the parenthesis. The only exception to this would be in a situation where the player's rating crossed between one of the three boundries that determin the value of 'k'. In other words, if a player crosses between having a score less than 2100 to more than that with one game than the change in the new score for the next game could be different than if they were finished in a different order.
10. richjohnson
TANSTAAFL
12 Mar '04 06:46
Originally posted by econundrum
The answer to this is rather simple:
Should you consider the loss to the 1200 rated player first, followed by the 1300 player, the calculation of your new score would be this:

(1400+32(0-1/(10^((1200-1400)/400)+1)))+32(0-1/(10^((1300-1400)/400)+1)) =
But after the first loss your rating isn't 1400
11. 13 Mar '04 07:49
Originally posted by richjohnson
But after the first loss your rating isn't 1400
That is true, but that doesn't matter. The formula for figuring out the new rating is &quot;the old rating&quot; + k times all that other stuff. &quot;The old rating&quot; is your rating before that loss, or that win (depending on the outcome of the game). When you lose your first game, your &quot;old rating&quot; was 1400, so you would get your new rating through that formula. Then, for your next loss, your &quot;old rating&quot; would become what you recieved after losing the first game, and that you would plug back into the formula to get your final rating (for the two games). Therefore, the score you get for losing the frist game is then changed again by losing the next game. I simply left the score after losing the first game in the form of the original formula and then plugged that back into the formula to find the new rating.

Everything I've just said may sound confusing, but it's all simple algibra. I did not have to find the value of the rating after the first loss in order to plug that value back in the formula to get the value after a second loss. I left the value in the formula's notation when I plugged it back into the fomula a second time. It's the same thing as finding the actual value, but using algibra allows for a more irrifutable proof than just plugging in numbers would have.

Go back over my proof, and work it out yourself. I think you'll find that it is accurate.
12. 13 Mar '04 08:21
There is a very simple rule (no algebra here): End that game first, with which you gain the least points (or lose the most, but that is negative gaining π). That will give you the highest rating. If two opponents have nearly the same rating, it won't matter because of rounding. If the ratings weren't rounded, it would matter.
13. 14 Mar '04 02:243 edits
Originally posted by econundrum
Go back over my proof, and work it out yourself. I think you'll find that it is accurate.
I worked out the eventual rating both ways by doing the sums. It looks like your rating ends up higher if you lose to the player with the furthest rating to you first. When I tried to expand out the algebra to see why, it got quite horrible. The closest I can come is the following logic:

The eventual change in your rating is proportional to a.b where a and b are your win probabilities computed using the fomula given.

Assuming a and b will sum to about the same whichever way you play the games, their product will be greatest when they are most equal. By taking the big hit first you reduce the size of the smaller hit - this makes the two changes more unequal and reduces the size of their product.
14. 15 Mar '04 21:24
Dang... I just checked over my work and realized that I forgot one very important thing. I'll not bother bringing up the math again because it involves too much typing, but I will say this: iamatiger seems to be right--the first loss does make an impact on the second. I should have gone over my work more carefully. I normally don't make such simple mistakes in math.