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- Oct 19, 2009
No, I don't!(But you knew all that.)
Gonna keep working on it.
Actually, I have no idea how I need to fix this and it's bothering me. Is there something I'm missing in our conception of this problem?
Is it easier to cheat in 6.0 or CoP?
- Thread starter Mathman
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Gonna keep working on it.
Actually, I have no idea how I need to fix this and it's bothering me. Is there something I'm missing in our conception of this problem?
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I dunno, things like this should be monitored by someone and the judge should be asked if that is really what they meant by a person overseeing the scoring.
I don't like the idea of any scores being thrown out without an actual reason behind it.
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- Jun 21, 2003
Change 12 and 11 to 9 and 8 and you'll get the right answer (There are 9 judges in all..)
The random draw happens first, and the assumption is that all of the conspirators' scores are higher than the scores of all of the other six.
Your method, however, works only in the cases x = 0 and x = n, where n is the total number of judges eliminated in the random draw. For the intermediate cases -- in our example, the case where exactly one conspirator and exactly one honest judge is eliminated -- this method of multiplying conditional probabilities does not work.
Try it out on this example and you will see why. In a five-card poker hand, what is the probability of getting exactly 3 red cards and exactly 2 black cards.
Wrong answer: 26/52 times 25/51 times 24/50 times 26/49 times 25/48 = .032. :no:
Right answer: (26x25x24)/(3x2x1) times (26x25)/(2x1) divided by (52x51x50x49x48)/(5x4x3x2x1) = .325. :yes:
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- Oct 19, 2009
Yes, I didn't want to get into the intermediate examples; I wanted to start with the simpler cases first. 
I didn't remember what the denominator factorials conceptually stood for.
And we have 9 total judges instead of the previous 12?
Also, am I correct in believing it's worse that the random dropping of judges happens first as far as allowing for more cheating influence?
ETA: Yes, it would be worse.
I didn't remember what the denominator factorials conceptually stood for.And we have 9 total judges instead of the previous 12?
Also, am I correct in believing it's worse that the random dropping of judges happens first as far as allowing for more cheating influence?
ETA: Yes, it would be worse.
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Gosh, I've been away from serious math for awhile. *goes back to working on the script*
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- Jun 21, 2003
Here is the reason behind it.
Here is a sample of numbers.
{ 8 10 10 10 12}
What is the best "measure of central tendency" for these numbers?
Well, any way you go about, 10's the guy. You could take the mean (add them all up and divide by 5), you could take the median (just grab the mddle number), or you could do something half-way in between. The in-between idea is the trimmed mean. Throw out the highest and lowest and take the average of the three that remain. (In this case, the three10's remain.)
So this is no problem. Any way of doing it gives the same answer.
Now consider this problem. I am thinking about applying for a job at a certain company. What salary can I expect? So I do some research and find that the salaries of all the employees are
{$ 25,000, $30,000, $30,000, $30,000, and (the president) $1,000,000}
What is the average? Well, add them all up and divide by five and you get $223,000 as the average salary of everyone who works at this company. That sounds pretty good to me. I am an average guy, so I should get the average salary, right?
This is an example where the median, $30,000, or the trimmed mean, also $30,000, gives me a more useful estimate than the mean using all five numbers.
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Yeah, medians are more resistant than means to extreme values.
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Do you think it actually turns out advantageous in certain conditions to have a smaller panel of judges, in some paradoxical way? Because as my botched calculation using a 12-judge panel shows, it's a lot more difficult to get rid of the all the cheaters when the judging panel is big and you randomly toss out 2, rather than a smaller panel (of 9 judges) randomly tossing out 2.
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- Feb 9, 2009
They are? The hypergeometric distribution. For x = 0, 1, 2, the probability that x of the three conspirators are discarded in the random draw, together with 2-x of the honest judges -- to me, it seems like that should be (binomial coefficient 3 choose x) times (binomial coefficient 6 choose 2-x) divided by (binomial coefficient 9 choose 2).
I was more worried that I didn't figure out the points right -- how the scores of one judge affect the averages and totals.
This is why I hate COP, all this mathamatical who-ha, at least if you cheat in the 6.0 system, you have a much better chance of getting busted. In my opinion if a judge cheats they should be tossed for life
COP just gives judges so many more ways to cheat, with all the different scoring, that how I feel though.
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- Jun 21, 2003
Yes, that is a new rule this year, and it set off quite a stir. Last year the ISU changed it to 9 for all ISU championships "to save money." This year they decided to extend the change to the Olympics also, "to make it the same as Worlds, Europeans, etc."
There were two reasons why people complained about this. First, statistically, the more judges the better. The random error (statistical noise that obscures the true result) goes down proportionally to the square root of the samle size. So by decreasing the size of the judging panel from12 to 9 (ignore the random draw, etc.) that increases the sampling error by a factor of (sqrt(12)/sqrt(9) = 1.15 = a fifteen percent increase in statistical error.
(Elimination of judges' scores by the random draw makes it even worse than that. The effect on the standard error effect of trimming the mean is not so easy to quantify by a simple formula. There is a big literature in statistical journals about this.)
The second reason that people complained is that the new rule for the Olympics was simply ordained by Cinquanta and his cronies, without going through the proper channels according to the by-laws of the ISU. (Nothing came of these protests, of course.)
I think it is the same in the situation of this example. i will have to think about that question in general.
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- Oct 19, 2009
Regarding a larger or smaller panel (12 judges vs. 9 judges), it seems that which panel is "better" depends on which judges are tossed out.
If the cheaters aren't randomly tossed, then you get a total score that is more "off" from a smaller panel than a larger panel since their influence can have a larger effect.
However, on a smaller panel, you are more likelier to toss out all the cheaters than from a larger panel.
So it depends on the degree of the cheating per cheater. Larger degree of cheating might favour a larger panel. In smaller degrees of cheating, it may actually be better to keep the smaller panel since the possible benefits of getting all the cheaters may outweigh the risks that one of them slips by.
If the cheaters aren't randomly tossed, then you get a total score that is more "off" from a smaller panel than a larger panel since their influence can have a larger effect.
However, on a smaller panel, you are more likelier to toss out all the cheaters than from a larger panel.
So it depends on the degree of the cheating per cheater. Larger degree of cheating might favour a larger panel. In smaller degrees of cheating, it may actually be better to keep the smaller panel since the possible benefits of getting all the cheaters may outweigh the risks that one of them slips by.
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CoP is supposed to determine the average of the judges, though.
It's not like 6.0 where each vote counts the same, regardless of the actual scores they give the skaters. If 5 judges think skater A is slightly better than skater B under CoP, but the other 4 judges overwhelming think skater B is better....skater B wins.
And I believe that's how it should be.
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- Jun 3, 2008
ok please dont laugh but this is addictive. Isnt this case like the random selection of black out of black and white sheeps we did at school?
Question:The probability that neither is chosen from random draw is 0.417. I put total 9, number of bad judges 3, picks 2 and success 0. Why the final number is same before calculating the rest, I mean the probability that furthermore one is thrown out cause of high/ low scores?Or is it sure thing, so probability is 1 for that?
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- Oct 19, 2009
As far as the general question of this thread goes, I don't know how to answer the question.
I'm tempted to conclude that CoP is far more susceptible to cheating judgments than 6.0 was, because there is only a small chance that any score is ever going to be perfectly honest (i.e. it's hard to negate the effects of cheated scores unless somehow you get all the cheaters dropped randomly.)
But bringing up SLC where a big big proportion of the judging panel was split and it supposedly came down to that one French judge giving the Russian pair the "1" placement. I can't see how CoP would have been any better or worse (uhh, the point-system of CoP applied to 6.0's standards, if that makes any sense. Because some people have argued that B&S would have won under CoP, and it seems possible.)
In our hypothetical example, is it not possible that the 6 honest judges' tallies favouring Miki could still overcome the 12-point bonus given to Kiira? I mean, how would a "close" competition actually translate into point differentials? That's the major question.
Also, imagine that a split panel like in SLC happened honestly, and it all came down to the antics of one dishonest judge. There is a lot more power in that one dishonest judge under 6.0 than they would have under CoP.
I'm tempted to conclude that CoP is far more susceptible to cheating judgments than 6.0 was, because there is only a small chance that any score is ever going to be perfectly honest (i.e. it's hard to negate the effects of cheated scores unless somehow you get all the cheaters dropped randomly.)
But bringing up SLC where a big big proportion of the judging panel was split and it supposedly came down to that one French judge giving the Russian pair the "1" placement. I can't see how CoP would have been any better or worse (uhh, the point-system of CoP applied to 6.0's standards, if that makes any sense. Because some people have argued that B&S would have won under CoP, and it seems possible.)
In our hypothetical example, is it not possible that the 6 honest judges' tallies favouring Miki could still overcome the 12-point bonus given to Kiira? I mean, how would a "close" competition actually translate into point differentials? That's the major question.
Also, imagine that a split panel like in SLC happened honestly, and it all came down to the antics of one dishonest judge. There is a lot more power in that one dishonest judge under 6.0 than they would have under CoP.
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- Jun 21, 2003
Yes, I didn't want to get into the intermediate examples; I wanted to start with the simpler cases first.
The denominators ...
Time out for etymology lesson.
The numerator of a fraction is the numberer, or counter. The denominator is the "namer" (Latin, "nom' = name.) so in a fraction like 3/7, the namer names the thing you are talking about -- sevenths -- and the counter counts how many of these thing you have -- three.(Somethimes I can't even stand myself. They wanted me to play the part of Sheldon on The Big Bang Theory, but I was too busy with my favorite figure skating board.
. )Anyway...the denomonators take away the effect of ordering. How many ways can you draw a five card poker hand? 52x51x50x49x48.
However, you have now counted each distinct poker hand 120 times. This is because these two hand are the same hand: {Ace of spade, king of diamonds, three of hearts, etc.} and {three of hearts, Ace of spades, king of diamonds, etc.} It is th4 same hand, just listed in a different order.
The number of orderings of the five cards in my hand is 5x4x3x2x1 = 120. So in my first computation I have to divide by 120 to eliminate all the extra 119 times that I counted the same hand again.
So the number of five-card hands that can be drawn from a 52-card deck is (52x51x50x49x48)/(5x4x3x2x1) = 2598960 -- a very famous number to gamblers.
I'll think about that. I doubt is there is any situation where is would be advantageous to have a smaller panel.
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- Oct 19, 2009
I SURRENDER!!! No math etymology please, next time I'll say "the ! thing under the division sign." 
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- Oct 19, 2009
Yes, I remember. I just didn't remember how to use it. It's kind of sad, I used to be very good at Finite math. Now I'm useless at it.
All I know is that when I ever play the lottery, the chances of me winning by choosing the numbers 1, 2, 3, 4, 5, 6 in a 49-choose-6 lottery is the same as any other mix of numbers.
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- Jun 21, 2003
Yes, exactly right. (you can check it out with Mir.
)Yes, we are assuming that all of the three conspirators score Kiira higer and Miki lower than all of the other six judges, so the probabilty is 1 in every case.
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- Jun 3, 2008
oh I feel stupid, it took me an hour to calculate the probability that the scores of the remained bad judges wont be thrown out and I was puzzled why you didn't consider this, I was like how is he sure the bad judge score will be off, then I decided it will be cause you say so
Finally i have no idea why i looked at this, do these results mean we should get back to 6.0?
Oh trust me, I would but it is too late in greece to upset her for such a problem and she would probably banned me from sister for good.Yes, exactly right. (you can check it out with Mir.
Finally i have no idea why i looked at this, do these results mean we should get back to 6.0?
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- Jun 21, 2003
Yes, it does. The question is, is it easier or harder for dishonest judges to manipulate the system if we use averages (CoP) or ordinals (6.0).
That is indeed the test case. "Should" in your last sentence means something like "fairest," or "most likely to produce the 'right' result most of the time," or something like that. To me, it is hard to evaluate that claim.
By the way, even under OBO, if there were more than two contestants, a skater who was judged much better by 4 judges might well finish ahead of a skater who was judged just a little bit better by five judges. This could happen if a thrid skater was placed between the two by the four-judge minority.
