About fraud in general, we know it exists, we know our rulers admit it exists, or used to. We need to be careful about claims of fraud now, putting forth only our best evidence (example, example). Time is short and we don’t need distractions.
It needs to be us, not the enemy, discovering our flaws.
See code update at end: 952 PM, Thursday night.
Here’s the gist of the method in this picture (or, see updates, one method, all of which are similar):
People in Michigan, the greatest state, could vote straight R or D ballots for all offices, or individual ballots, selecting a mix of candidates for offices.
One plot: the x-axis is the total number who voted a straight Republican ballot, divided by all votes, straight or individual, inside a precinct. In other words, the percent of all votes that are straight R ballots. Update: it appears IOE does total number straight Republican divided by total of all straight ballots.
One y-axis is the number who voted for R for president (Trump) out of those voting individual ballots minus the x-axis value. Update: Another is percent R among all individual ballots minus the Republican straight percentage. Yet a third possibility, suggested in the comments, is that y-axis is R votes divided by all votes of any kind minus x.
In any of these techniques that minus is the problem.
Here is a brief explanation of the first interpretation of the technique. For others, which are similar, see the Updates.
A friend of mine gave me this first example. Suppose x = 100%. Then there are no votes left for individual ballots, because everybody in the precinct voted straight R. Then necessarily y = 0%, it’s maximum.
Assume no fraud. When x = 99%, then 1% of the total votes are individual ballots (assuming no straight D ballots). If Trump wins 80% of these individual ballots—a huge amount!—he’s in the hole, because y = 80% – 99% = -19%.
Thus even though Trump got 99% + 0.8* 1% = 99.8% of all votes in the precinct, his “score” is -19%. This kind of thing repeats for all values of x. The decrease in y as x increases is exactly what we’d expect to see under normal voting conditions.
That is really it for my analysis. The rest of this article is fleshing this counter-argument out. It will not be easy going.
Some are saying IOE meant x = R straight ballots/(R straight ballots+D straight ballots).
And y = (R individual / R individual + D individual) – x.
Grant this. But it doesn’t change anything, really.
If x = 100%, then any number of individual ballots can remain, as commenters are saying. But then Trump would have to win 100% of them for y to equal it’s maximum of 0.
If x = 99% then again it could be any number of individual ballots remains, and if Trump wins 80% of them, still a huge amount, then y = 80-99% = -19%, just as before.
If x = 50%, then again there could be any number of individual ballots, and Trump has to win more than 50%, a heavy task, to keep y > 0.
I don’t see how this saves his method.
Consider x is not a as good measure of R strength in a district as what I thought he originally meant. I mean this interpretation is worse for this theory.
It could be there are 100 votes in the district, say. If only 1 is straight R, then x = 100%. Then 99 are individual. If Trump wins 90, a huge take, then y = 90/99 – 100% ~ -9%. More realistically, if x = 100 in this scenario, Trump may take as many as half of the remaining, then y ~= 50% – 100% = -50%.
Or if x = 90/100, i.e. no straight D, leaving 10 individual, and if Trump takes 80%, then y = -10%.
Later this evening I’ll add comments on the examples.
x = # straight R ballots / # total ballots,
y = # individual R ballots / # individual ballots – x,
# total ballots = # straight R + # straight D + # individual
# total ballots = # straight R + # individual,
where # individual includes # straight D. The x and y are converted to percentages by multiplying by 100.
Inventor of Email never makes it clear which # total ballots he’s using. I’m guessing (as you’ll see) the first.
Suppose x = 0%, i.e. no Rs voted a straight ballot. It could be all the ballots are D straight ballots, or there was a mix of straight D and individual. Surely there were some straight Ds.
Since the max x (for Rs) in all Inventor of Email’s Oakland example was ~70% (this maximum changed by county), we might suppose the same is true for max D straight tickets. Which is to say, there were not any 100% D straight tickets in any precinct, but maybe there were some 70%s in Oakland. That would leave about 30% independent votes.
Any number of individual votes in this 30% (or whatever number it was) for Trump would give a y > 0 when x = 0. We gather from the 5-6% average y when x was near zero (see the figure) that Trump got the few Rs who didn’t vote straight tickets, and perhaps a D or two who also didn’t vote straight ticket. R support is low in these precincts, because x is small. And we know Oakland is a predominately D county.
As x increases, it’s true the precinct becomes more R, and hence more votes go to Trump. But then there’s fewer individual ballots out there, so Trump would have to capture higher and higher percentages of the individual tickets to keep the y positive. The larger x is, probably the smaller the precinct is in voting size, meaning it’s likely a good deal of the remaining individual ballots will go to Ds.
In other words, the higher x is, the more likely y is negative. And the smaller x is, the more likely y is positive. The exact “0 crossing” for y depends, of course, on the D/R ratio in the county. For example, here is Inventor of Email’s Macomb county example:
The max R straight ticket precinct is ~90%, which suggests this county has more base R support, and then we see y doesn’t dip below 0 until around x = 70%, which confirms it.
This county has the same shape, the x-y curve, for the same reason, though.
Let’s go through some concrete examples, using Oakland county as a base.
X = 0%, meaning no straight Rs. Suppose 70% are instead straight D, mirroring the top R percentages. Again, I don’t recall hearing Inventor of Email mentioning straight D ballots. But if they existed, which they surely must have, then the remaining individual ballots are fewer than you’d think. Because that’s a smaller number, there’s a higher chance Trump would score a greater percentage of the ballots (smaller samples have more variability).
A 70% straight D leaves 30% of the remaining vote for individual ballots. We know ys for x near 0 were ~0-10% (see the picture). That means about 90-100% of the remaining 30% went to Ds, the other for 0-10% R (Trump). So y = 0-10%. This makes sense.
If instead Inventor of Email classed straight D ballots as “individual”, then 100% of the ballots when x = 0% would be classed “individual”. That means between 0-10% of people out of all voters would have had to vote for Trump to get y = 0-10%. That could happen, too.
In the video, Inventor of Email thinks small x votes are probably fraud free, and that some algorithm switched on when x hit 20%. But I’d think any precinct with y = 0 when x = 0, or when both are very low, means no or almost no votes were recorded for Trump. I’d find that suspicious.
X = 20%, meaning 20% straight Rs. How many straight Ds we don’t know, except it must be less than 80%. Suppose straight D goes down in proportion; maybe there’s 50%. That again leaves 30% of the remaining vote for individual ballots. In Oakland, with x = 20%, y was around -10-10%. That means Trump got between 10-30% (of the 30%).
In other words, Trump got a little more of the individuals than when R support was higher, than he got when x = 0%, implying R support lower. This fits.
Or, if the 80% were included as “individual” (no straight D), then Trump still had to win 10-30% of these (to get the same y), but that’s much harder, since the number of votes called individual are necessarily larger. This leads me to suspect Inventor of Email did indeed separate out straight D ballots.
X = 50%, meaning 50% straight R. Straight Ds? Had to go down. Maybe 20%? That leaves 30% remaining individual ballots again! It’s almost like there’s a straight line hiding in the average, just like he found.
Trump would have had to win more than 50% of this 30% to keep y positive. Tough job! The ys for Oakland were about -15% (plus or minus) when x = 50%, meaning Trump won on average about 35% of the remaining 30%. This makes sense: Trump scores about twice as many individuals when x = 50% as when X = 20%. It fits.
When ys for x = 50% are about -20% or -25%, it’s likely the straight D ballots were higher than 20%.
Don’t forget, too, that the sizes of this precincts won’t be the same. I’d bet the smaller ones were the ones with the higher xs on average. The smaller ones would have a harder time besting the x percentage.
x = 70%. Ys here are about -25%. How many straight Ds? 10%? 5%? That would leave 20-25% individual ballots. Trump would have had to win more than 70% of these to keep y positive. It’s harder and harder to do.
It looks like Trump won maybe 45% of the individual ballots. So he won more of the individual when x was higher (he won 35% on average when x = 50%). So this fits, too.
This is their picture for Wayne Country (Detroit, my hometown).
Notice the x-axis scale has changed. Most of the x are below 15%, with only a couple of precincts above: the max is maybe 30%.
Inventor of Emails says cheating was likely not used in wayne county. We saw above that when x is small, that it’s easier to get positive y. That’s what we see here. Recall, too, we don’t see presented the straight D ballots (my guess). This means with small x we’re seeing mostly individual ballots.
It appears to me, then, just like some of us suspected, that Trump did very well with non-party blacks, winning on average 10% of the independent black vote. In many precincts, Trump did much better than 10%, too. It still means that Biden got 90% or so on average, though.
I believe Inventor of Email is fooling himself, with the model in general, and with the idea there is some kind of step “fraud” function (my term) that kicks in at 20%. Look at the Oakland picture with fresh eyes and it appears a straight downward sloping line starting at x = 0% fits just as well, or better, than a step function that is flat until 20% and then goes down.
The downward slope is an artifact of the way y is calculated dependent on x. There is no indication of fraud using this method.
I do not say no fraud occurred in Michigan. Indeed, it would be shocking if it didn’t, especially in Wayne county, which has a long and storied history of cheating. It is a Democrat machine town, and everybody knows it, even the people lying and saying they don’t know it.
Everything I said can be checked with the original data. Which I don’t have. Anybody have it?
n = 300 # number of precincts # having no idea what number of votes per precinct are, I assumed these # more individual than straights on average, which seems plausible Rs = round(rnorm(n,1.5e4,4e3)) + 5000 # number of straight R per precinct Ds = round(rnorm(n,1.5e4,4e3)) + 5000 # number of straigh per precinct Ri = round(rnorm(n,2e4,4e3)) + 5000 # number of R individual votes per precinct Di = round(rnorm(n,2e4,4e3)) + 5000 # number of D individual votes per precinct x = Rs/(Rs+Ds) y = Ri/(Ri+Di) - x plot(x,y, xlab='R straight %', ylab='R individual % - R straight %',main="Shiva Simulation") grid(20) abline(h=0) fit = glm(y~x) abline(fit)
One run gives this:
There’s no cheating here, but also no correlation between straight and individual ballots. That correlation would give more spread to the x (closer to 0s and 1s). I leave it as an exercise, or I might do later, to build in correlation, more variable sizes.
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