How Chiles, Barbosu, and Voinea all won Bronze, and how Maduro faked the Venezuelan election results.
Significant figures, or significant digits, represent the number of numbers in a number that can be reliably reported. It’s about precision. And leaving aside the fact that subjective judge scoring defies such precision, it’s worth a look at what precision is even allowed mathematically.
Gymnastics scoring has several factors like degree of difficulty and execution. Yet each of these is scored with a single decimal place precision by any individual judge. Which is to say, a level of difficulty may be 5.1 or 5.2 but not 5.15 or 5.23. Similarly, the deduction of points on execution are –0.1 or –0.2 etc. Voinea, for example, was docked 0.1 for stepping out of bounds (more on that below).
If gymnastics scoring wants to pretend to be objectively mathematical, then they have to play by objectively mathematical rules. You cannot report a score with higher precision like 13.666 (as opposed to 13.7) than the precision of the measurements you actually made.
When averaging the score of 3 of 5 judges, you simply cannot report that score beyond the first decimal place. It’s just wrong. It has no mathematical justification whatsoever.
I defy you to slice a pizza into 6 pieces, take 4, and justify how you can eyeball the difference between 0.666 and 0.7.
So, even if Jordan Chiles is denied her challenge because it was 4 seconds late (talk about over-precision!), her original score can only be reliably reported as 13.7 (not 13.666) and that is identical to the score for Ana Barbosu. Even if there were a hundred judges it would not change the fact that you cannot report these summary or averaged scores beyond the first decimal place on which they are scored in the first place! That would just change your margin of error. And believe me, the margin of error on scores from 3 of 5 judges is huge.
Even without the 4 second late challenge:
Andrade got 14.2 (not 14.166)
Biles got 14.1 (not 14.133)
Chiles and Barbosu both got 13.7.
By the way, the rule is “For the last gymnast or group of a rotation, this limit is one (1) minute after the score is shown on the scoreboard”. It doesn’t say “60 seconds”. It says “one minute”, and mathematical precision speaking, 64/60 seconds is 1.06 minutes; indistinguishable from 1 minute.
And what of Voinea? Were it not for her 0.1 deduction for going out of bounds, she would have scored a 13.8 (like Chiles, but for the failure of a judge to record the time of the challenge). But she did not go out of bounds as is plain from this video:
All of this is a failure of precision in judging. More to the point, it’s gymnastics vainly attempting to be mathematically precise where such precision cannot be had.
There is the failure of the judges on whether Voinea stepped out of bounds. There is the failure of the judges on properly scoring Chiles’ level of difficulty in the first place (which they admitted to). There is the failure of the judges to properly record the time of the Chiles challenge. And I’m not even considering the Romanian conflict of interest issue regarding the appeal panel or the fact that they sent messages to the wrong email account in a manner that prevented timely provision of evidence.
For the love of god, grow-up! Be magnanimous. Admit you royally f*cked up, and just give all three of them the Bronze and be done with this nonsense.
I don’t know how many angels can dance on the head of a pin, but surely there’s room for three!
Statistics is a measurement of surprise.
Consider the over-precision of the Venezuelan vote result.
Those percentages are remarkably precise to the first percentage decimal place! Note the “0000” trailing in each. Almost as though someone sat down and said, “Ok, Maduro gets 51.2% to avoid any run-off. Gonzálaez gets 44.2. How many people voted? Oh, yeah, 1,0058,774, so I guess we just multiply 10058774 by 0.512 – let me punch that into my iPhone calculator, there you go.”
A good example, and one used in the detection of financial fraud, is Benford’s Law. More often than not, someone cooking the books will use random numbers to avoid detection. But real numbers found in faithful financial books do not have a uniform distribution. The number 1 should occur in the first digit much more frequently than a random choice of 1 through 9.
Suppose you repeatedly randomly sampled a large population and suppose some portion (0.512 or 0.442, it does not matter) are in favor of some candidate.
Suppose the number of significant digits is 7 (which it is in this case because no candidate got more than 10 million votes). The number of significant digits in the Maduro count and the Gonzalez count is seven, so you’re allowed seven in the percentage.
The number reported to only 3 significant digits (e.g., 0.512) obviously includes rounding such that it could represent anything as large as 0.5124999 but not anything as large as 0.5125000.
What are the probabilities of numbers in the 4th significant digit position from such a random sampling of 10,058,774 voters?
That is, the position *after* the 0.512 or 0.442? In 0.512wxyz… what is w expected to be?
You might think that the 4th significant digit could be any of 0 through 9 uniformly. In which case the probability of a zero there for all 3 outcomes is 0.1*0.1*0.1 = 0.001. For the 5th through 7th digits it is equiprobable.
But it’s NOT equiprobable in the 4th figure!
Because of the shape of probability distributions, and rounding,
– about 48% of the time it will be a 5 or a 4.
– about 32% of the time it will be a 6 or a 3
etc
Only less than 1/2 of 1% (0.45%) of the time will it be a zero “0” (same for a “9”).
Thus, the probability that all three (Maduro, Gonzalez and “Other”) each had a “0” in that fourth significant digit position is 0.0045^3 or p < 0.0000001.
Ten thousand times less likely than a random choice of 0 through 9 in the 4th position.
Once in 10 million elections. For realz. Have there even been 10 million elections in the history of the world?
Math fraud isn’t hard to find if you just bother to try. But it won’t make you many friends, trust me, or ask me.
What is also remarkable about the “official” Venezuelan result is that 5,150,092 divided by 10,058,774 is very precisely 51.20000%. Not, for example 51.16378% rounded up, nor 51.23194% rounded down.
Similarly, the number reported for Edmundo Gonzalez was 4,445,978 out of 10,058,774. Also, remarkably precisely 44.20000%. Of course, the votes cast for “Others” tallied 462,704 of 10,058,774, which is precisely 4.60000%.
That is, in each of 3 cases ALL of the last 4 of 7 significant digits is ‘0000’. (The justifiable # of significant digits is the least of the 2 operands, e.g. 7 in 5,150,092 not the 8 in 10,058,774).
This would be laughable were it not for all of the already-dead and soon to be dead on the streets of Caracas.
I am far from the first to point out the obvious fraud apparent in the numbers reported in the Venezuelan vote .
What is the measure of “surprise!” looking at the official Venezuelan vote
It’s easy to just brute force it. Here’s some Python code that find the frequency you would expect all of the last 4 significant figures being the-same (not even all-zero, just the same) for the 3 outcomes.
#!/usr/bin/env python
# BRUTE FORCE proof of Maduros fraud
# in which you will never get a positive result for all three having the same last 4 precision numerals
# no matter how long you wait
import random
total = 10058774
anytwo = 0
allthree = 0
k=0
j=-1
# this will go on forever
# you will never ever get a positive result for all three
while k > j: # infinite loop
. maduro=0
. gonzalez=0
. other=0
. for i in range (0, total):
. n = random.random()
. if n < 0.512 + 0.0005:
. maduro += 1
. elif n < (0.512 + 0.442 + 0.0005):
. gonzalez += 1
. else:
. other += 1
.
. # round – with 10 million votes, you get 7 sign digits
. maduro = round(maduro/total, 7)
. gonzalez = round(gonzalez/total, 7)
. other = round(other/total, 7)
. # convert last 4 digits to string
. maduros = str(maduro)[5:]
. gonzalezs = str(gonzalez)[5:]
. others = str(other)[5:]
. # make a set
. unique_values = {maduros, gonzalezs, others}
. # If the set length is less than 3, then at least two are the same
. if len(unique_values) < 3:
. anytwo += 1
. # If the set length is 1, then all three are the same
. if len(unique_values) == 1:
. allthree += 1
. print (anytwo, allthree)
. if k % 1000000 == 0 and k > 0:
. print (k, anytwo, allthree, ‘Current p-values are < ‘, (anytwo+1)/k, (allthree+1)/k)
. k+=1