https://www.thedailybeast.com/arizona-election-officials-find-infinitesimal-rate-of-potential-voter-fraud-in-audit-of-2020-election?ref=home Election officials in Arizona announced Friday that they had identified just 182 cases of potential voter fraud among the 3.4 million ballots cast in the 2020 presidential election. Of the 182 cases referred to investigators, only four have led to charges against those involved, and no one has been convicted. That’s .005 percent of the votes cast. Two of the four cases that resulted in charges were Democratic votes, and two were Republican. … https://apnews.com/article/business-government-and-politics-arizona-election-2020-e6158cd1b0c6442716064e6791b4c6fc No one has been convicted. No person’s vote was counted twice. … The Pima County Recorder’s Office has a practice of referring all cases with even a hint of potential fraud to prosecutors for review, something the state’s 14 other county recorders do not do. Pima County officials forwarded 151 cases to prosecutors. They did not refer 25 others from voters over age 70 because there was a greater chance those errors — typically attempts to vote twice — were the result of memory lapses or confusion, not criminal intent, an election official said. … Didn’t know that before, people older than 70 get a pass because the the county consider it more likely that their mistakes are due to memory lapses. About the few cases where some voters appeared to have attempted to vote twice … ”Franklin noted several factors at play, including worries about U.S. Postal Service delays. In addition, Trump at one point encouraged voters who cast their ballots early by mail to show up at their polling places on Election Day and vote again if poll workers couldn’t confirm their mail ballots had been received.”

@mik said in Going all frou-frou at Chez Mik: Jealousy does not become you. Enjoy your h07 p0ck37s. That was cold. Funny, but cold.

Love dogs...

Totally agree with you, Mik.

It’s the future folks are worried about. One respected predictive analytics firm is represented here: https://covid19.healthdata.org/australia?view=resource-use&tab=trend&resource=all_resources Let’s revisit in October.

I bought a car through Costco in 2012. Doing some research, it appeared that the Costco price is not necessarily the lowest price. However, like CarMax, it is the easiest price. What's the hassle-factor worth? The dealer tried to haggle on the value of my trade-in. I was going to walk until he met my price. He acceded. Would I do Costco again? Perhaps. But I would assuredly go to CarMax for my trade-in. I'm shocked at how much they give for used vehicles.

Damn. I should have listened. Thanks, Kos. [image: 1627167896057-img_0850.jpg]

Thought that was a porn rag.

Three nuns are sitting at a bus stop when a man in a trenchcoat runs up and flashes them.. The first nun has a stroke. The second nun has a stroke. The third nun tries to have a stroke, but her arms weren't long enough to reach it.

All of Israel could march across our southern border right now, it is wide open. If they did, they would probably recognize a lot of the terrorists. They could easily take New Mexico, they are good with deserts.

DOJ needs a serious overhaul.

For the Virtue Signaler in your family.

I think Chaplin also has the first "moonwalk" on film.

[image: 1627126373812-img_0855.jpg]

Official solution: SOLUTION: The key here is to recognize that you start with 100 spaghetti-ends and each tying operation reduces the number of spaghetti-ends by two. The operation results in a new loop only if the first end you pick up is matched to the other end of the same strand. At step k, there are 100 - 2(k-1) ends, thus 100 - 2(k-1) - 1 = 101 - 2k other ends you could tie any given end to. In other words, there are 101 - 2k choices of which only one makes a loop; the others just make two strands into one longer one. It follows that the probability of forming a loop at step k is 1/(101 - 2k). If X(k) is the number of loops (1 or 0) formed at step k, then the expected (that is, average) value of X(k) is 1/(101 - 2k). Since the final number of loops is X(1) + . . . + X(50), we conclude that on the average, the final loop count is 1/99 + 1/97 + 1/95 + . . . + 1/3 + 1/1 which is about 2.93777485; less than three loops!