21 Dec 06
2 edits
Originally posted by dmnelson84It's a puzzle .. not a statement of fact my friend .. take a deep breath .. there, now don't you feel better?
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.
Now if you had called me sadistic that would be a different matter 😉
EDIT: low Narcissism i'm afraid but moderate Schizoid and borderline .. go figure ..
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21 Dec 06
Originally posted by idiomsYou'd have to play (1800-1357)/32 games to reach that rating (don't forget to round up to get the real figure).
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?
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.
21 Dec 06
Originally posted by IraxYep 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?
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.
There certainly are lots of combinations but the number is finite as your opponents rating has an upper bound of 2455
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21 Dec 06
Originally posted by idiomsFor 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
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
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!
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21 Dec 06
Originally posted by idiomsThe 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.
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
21 Dec 06
1 edit
Originally posted by IraxExactly!*
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!
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
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21 Dec 06
Originally posted by XanthosNZIs there a way to show this calculation mathematically?
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).
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.
21 Dec 06
1 edit
Originally posted by IraxIt's an iterative computation where:
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.
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.
21 Dec 06
1 edit
Originally posted by XanthosNZFind 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.
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.
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.
22 Dec 06
Originally posted by FabianFnasDoes that work for iterative series that don't have polynomial equations?
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.
22 Dec 06
Originally posted by XanthosNZI'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?
Does that work for iterative series that don't have polynomial equations?
However, if fibonacci serie is not polynomial, then it works for non-polynomials too.
22 Dec 06
Originally posted by FabianFnasFor fibonacci the iterative series is:
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.
t(n+2) = t(n) + t(n+1)
which is linear (and therefore a polynomial).