US Election Mathematical fun!

 Let's do some calculations...

If I were to shuffle a deck of cards, the chance of it ending up in perfect numerical order, within the correct suits, as if the cards had just been pulled from the pack for the first time is:

1 in 80 658 175 170 943 878 571 660 636 856 403 766 975 289 505 440 883 277 824 000 000 000 000

That's 1 in 8 x 10^67.

If I were to do that 4 times in a row, the chances would be 1 in 42 324 668 846 415 660 077 857 202 047 347 746 677 935 967 944 612 046 667 814 114 345 170 207 396 228 703 086 283 421 770 796 366 410 106 374 242 729 407 222 439 001 313 217 490 503 683 356 007 835 033 781 255 324 651 204 466 826 842 852 565 637 467 820 039 406 627 183 944 921 497 480 856 600 576 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000 000

Or 1 in 4 x 10^271

 

Now, let's assume that in one particular US county, 99% of the population supports Biden (which, statistically is never the case, but let's suppose anyway). The chances of a box of 20,000 ballots all for Biden would be 1 in 2 x 10^87.

Already these chances are far smaller than the first shuffle of the deck of cards above.

Let's suppose this 99% Biden-supporting county finds a box of 64,000 ballots. The chances of every ballot in that box being for Biden is 1 in 2 x 10^279.

That's over 10,000,000 times smaller than all four shuffles of the deck of cards above!

The story is that there was a box of 128,000 ballots, all for Biden, in a county that's a fair bit lower than 99% supportive of Biden?!?!?

That would be in the order of 10^560...

So, let's assume it was a "typo". If there were 15,000 ballots all for Biden, with not a single one for Trump, those chances are 1 in 3 x 10^65, and that's still assuming a 99% Biden-supporting populace.

In short, if, there are ballot drops with all the votes going to one candidate, it is pretty much certain that there's something fishy going on.