In Pennsylvania, it was reported that there was a vote "spike" where 570,000 went to Biden and only 3,200 went to Trump.
So I crunched some numbers!
In a situation like this, where there is a set number of votes ("trials" in maths-speak), and we're trying to work out the probability of a certain number of voting one direction ("successes" in maths-speak), we can use the Binomial Distribution.
A note on how this works... If we liken it to tossing a coin 100 times, and we want to find the probability of 32 "heads", this can be calculated to be 0.000112817. But this isn't really fair, because this is asking for exactly 32 heads, no more, no less. So it's more appropriate to calculate the probability of getting up to 32 heads, which gives this result: 0.000204389 - still a tiny number, but note that it's almost twice as much as the first calculation.
So let's run the numbers of those votes.
Number of trials, n, is 570,000 + 3,200 = 573,200.
Number of successes is up to 3,200.
But what should we use for the probability? Well, as before, I'll assume a 99% support base for Biden. That means the probability of a success (a vote for Trump) is 1%, or 0.01.
The question can then be written in Maths language as:
Given X~B(573200, 0.01), calculate P(X ≤ 3200).
Which gives 1.65332 × 10^-294
or a chance of 1 in 604 842 061 141 140 377 719 184 915 396 113 341 356 675 744 654 064 379
175 243 351 909 393 292 934 471 553 738 480 228 299 700 589 298 016 113
752 778 743 246 771 405 990 194 608 183 829 027 018 967 783 838 497 507
682 978 450 352 998 589 692 971 200 126 418 069 299 327 043 837 434 678
614 636 455 904 893 217 255 383 372 225 337 155 059 338 158 579 591 565
191 683 605 396 826 505 090 107
That's 1 in 6.05 × 10^293
Make sure you understand Scientific Notation to realise just how tiny this probability is.
Now, as before, let's give a bit of perspective:
I am still more likely to randomly shuffle a deck of cards into perfect numerical order, within their correct suits, just as if the cards had been pulled fresh from their packet... not just once, or twice... but four times in a row than for Trump to have that few votes in such a large pack of ballots.
It's not even close... the chances of shuffling that deck of cards four times in a row is over a sextillion times higher (that's 1,000,000,000,000,000,000,000) than Trump getting that few votes.
And remember: that's assuming Biden has a 99% support base in that area!
As a mathematician, I don't believe it. Something fishy is going on.
