Maths Problem

idioms
Posers and Puzzles 20 Dec '06 23:12
1. 20 Dec '06 23:12
I am rated 1800 on RHP, I was rated 1357. What is the minimum number of games I could have played to get this rating given that the highest rating of my opponent was 2455?

BONUS: How many different combinations of ratings are possible to achieve this minimum number of games?
2. 20 Dec '06 23:36
http://www.4degreez.com/misc/personality_disorder_test.mv

Take that test. It's from another thread. My guess is that you'll score high in the Narcissistic field.
3. 21 Dec '06 00:072 edits
Originally posted by dmnelson84
http://www.4degreez.com/misc/personality_disorder_test.mv

Take that test. It's from another thread. My guess is that you'll score high in the Narcissistic field.
It's a puzzle .. not a statement of fact my friend .. take a deep breath .. there, now don't you feel better?

Now if you had called me sadistic that would be a different matter ðŸ˜‰

EDIT: low Narcissism i'm afraid but moderate Schizoid and borderline .. go figure ..
4. 21 Dec '06 00:38
Originally posted by idioms
I am rated 1800 on RHP, I was rated 1357. What is the minimum number of games I could have played to get this rating given that the highest rating of my opponent was 2455?

BONUS: How many different combinations of ratings are possible to achieve this minimum number of games?
You'd have to play (1800-1357)/32 games to reach that rating (don't forget to round up to get the real figure).

There are lots of different combinations possible to reach your goal, so long as whenever you beat a new opponent they're 400 points higher than you.
5. 21 Dec '06 01:28
Originally posted by Irax
You'd have to play (1800-1357)/32 games to reach that rating (don't forget to round up to get the real figure).

There are lots of different combinations possible to reach your goal, so long as whenever you beat a new opponent they're 400 points higher than you.
Yep that's right, the formula you gave is the lower bound of this problem. What if the highest rating of my opponents was also 1800? would the formula still be valid?

There certainly are lots of combinations but the number is finite as your opponents rating has an upper bound of 2455
6. 21 Dec '06 02:00
Originally posted by idioms
Yep that's right, the formula you gave is the lower bound of this problem. What if the highest rating of my opponents was also 1800? would the formula still be valid?

There certainly are lots of combinations but the number is finite as your opponents rating has an upper bound of 2455
For the different combinations, I guess you'd have 2455-1357+400 different ratings possible for your first game, then 2455-1357+400-32 different ratings for your second, so that's

1498+32n possibilities for each game where n will run from 0 to 13 (For each of the 14 games).

So multiply them together to get the permatations.

(1498-32*0) * (1498-32*1) * (1498-32*2) * .... * (1498-32*13)

I could calculate a figure at this point, but there must be a better way to express this mathematically, it's been too long since I've done this kind of stuff, my memory fails me right now!
7. 21 Dec '06 02:04
Originally posted by idioms
Yep that's right, the formula you gave is the lower bound of this problem. What if the highest rating of my opponents was also 1800? would the formula still be valid?

There certainly are lots of combinations but the number is finite as your opponents rating has an upper bound of 2455
The formula wouldn't hold if the max rating of your opponent was 1800, and since the formula to calculate rating change isn't linear, I wouldn't know how to rearrange everything to create a new formula, but what I can say is that it'll be no more than ((1800-1357)/32) * 2 games, because you'll be gaining at least 16 rating points each game, until you exceeded 1800.
8. 21 Dec '06 04:361 edit
Originally posted by Irax
For the different combinations, I guess you'd have 2455-1357+400 different ratings possible for your first game, then 2455-1357+400-32 different ratings for your second, so that's

1498+32n possibilities for each game where n will run from 0 to 13 (For each of the 14 games).

So multiply them together to get the permatations.

(1498-32*0) * (1498-32*1) * ...[text shortened]... ically, it's been too long since I've done this kind of stuff, my memory fails me right now!
Exactly!*

Unfortunately there isn't an easy way of expressing this without continued fractions since 32 isn't a divisor of 1498

*Technically this gives a rating of 1805 but that's close enough .. I think it may even be 1803 because the "expected win" factor kicks in sometime in the 1700's but then the calculation becomes horrid
9. XanthosNZ
Cancerous Bus Crash
21 Dec '06 07:19
After 19 wins against an 1800 player your rating would have gone from 1357 to 1807 (assuming rating changes are rounded to the nearest integer).
10. 21 Dec '06 12:45
Originally posted by XanthosNZ
After 19 wins against an 1800 player your rating would have gone from 1357 to 1807 (assuming rating changes are rounded to the nearest integer).
Is there a way to show this calculation mathematically?

It'd be fairly straight forward to implement this as a computer program to work out the final answer, but I'm curious to know if maths has the language to describe it.
11. XanthosNZ
Cancerous Bus Crash
21 Dec '06 17:141 edit
Originally posted by Irax
Is there a way to show this calculation mathematically?

It'd be fairly straight forward to implement this as a computer program to work out the final answer, but I'm curious to know if maths has the language to describe it.
It's an iterative computation where:

t(n+1) = round(n + 32 * (1 - 1/(10^((X-n)/400)+1)))

Where X is the opponent's rating and the function round rounds to the nearest integer.

It is (I think) theoretically possible to turn this from an iterative calculation into a non-iterative one (for a given start value) but the math is beyond me.
12. 21 Dec '06 17:551 edit
Originally posted by XanthosNZ
It's an iterative computation where:

t(n+1) = round(n + 32 * (1 - 1/(10^((X-n)/400)+1)))

Where X is the opponent's rating and the function round rounds to the nearest integer.

It is (I think) theoretically possible to turn this from an iterative calculation into a non-iterative one (for a given start value) but the math is beyond me.
Find an appropriate matrice M in where the iterative formula is coded, and multiply this to itself a number of times, M^n, and find wich n that satisfy the criteria. To find the M, however, is beyond my level of math. I flunked that exam, I'm afraid.

For XanthosNZ's iterative formula it suffice with an 2x2-matrice.

A matrice where the fibonacci series is coded was demonstrated in the class so you could actually see which n satisfied the equation f(n) is greater than a given number. M is a 3x3-matrice.
13. XanthosNZ
Cancerous Bus Crash
22 Dec '06 10:04
Originally posted by FabianFnas
Find an appropriate matrice M in where the iterative formula is coded, and multiply this to itself a number of times, M^n, and find wich n that satisfy the criteria. To find the M, however, is beyond my level of math. I flunked that exam, I'm afraid.

For XanthosNZ's iterative formula it suffice with an 2x2-matrice.

A matrice where the fibonacci seri ...[text shortened]... ly see which n satisfied the equation f(n) is greater than a given number. M is a 3x3-matrice.
Does that work for iterative series that don't have polynomial equations?
14. 22 Dec '06 10:21
Originally posted by XanthosNZ
Does that work for iterative series that don't have polynomial equations?
I'm not sure, I flunked that exam I'm afraid, but if you want me to I could try to find my notes from the class?
However, if fibonacci serie is not polynomial, then it works for non-polynomials too.
15. XanthosNZ
Cancerous Bus Crash
22 Dec '06 10:47
Originally posted by FabianFnas
I'm not sure, I flunked that exam I'm afraid, but if you want me to I could try to find my notes from the class?
However, if fibonacci serie is not polynomial, then it works for non-polynomials too.
For fibonacci the iterative series is:

t(n+2) = t(n) + t(n+1)

which is linear (and therefore a polynomial).