Probability and Vote Rigging

A fellow in one of my astro clubs is a vehement Trump supporter — mostly because of 2nd amendment issues, he told me. But he also greatly dislikes Hillary Clinton, believing that she rigs everything. I believe he said that the fact that Hillary beat Bernie in 5 out of 6 coin tosses used to settle dead heats in some small mostly-white northern caucus state (I forget which), proved that she cheated.
I’d like to go into that here.
He also wrote: “Read the leaked DNC emails. Look at the videos of voter fraud. Tell me they are not true. The emails alone are dam[n]ing…”
“Lessee, the leaked DNC emails are, what, a million or so pages? It’d take me HOURS to read all that (sike – decades!) Mind giving me a clue as to what to look at first? Help me out here?
“And which voter fraud videos? The ones where this one person, or another person (gosh, possibly as many as a full DOZEN?) voted twice? Or the ones where racist politicians wipe tens of thousands of their political enemies off the voter list, and enact policies that they know good and well will further reduce the voter turnout of their enemies’ supporters by tens of thousands more? Which one do you think is more serious []?” — I continued…
I haven’t seen his response yet.
And by the say, to get 5 heads out of six flips is not all that unlikely – I think it will happen about 10% of the time if you reproduce the experiment a few million times on a computer. Here’s how I calculate that: we only really need to figure out what are the odds of getting exactly one tails in the experiment, which is much easier to calculate. For each of the six coin tosses, I am going to assume that the probability of getting heads = p(tails) = 1/2 or 50% or 0.5. [Obvviously if the game is rigged, then the probabiolity is gonna be different, and wer’ll look into that in a bit.
So the probability of getting tails on the first toss and all the others heads is 1/2 * 1/2 * 1/2 * 1/2 * 1/1 * 1/2 because the coin tosses in our ideal experiment are completely independent – nobody’s cheating, no magnets or tiny weights or two-headed coins. Or (1/2)^ 6 , or 1/64. Now if you think about it a bit, the probability of getting tails on throw #2 and heads everywhere else is exactly the same: 0.5 ^ 6, or 1/64. And in fact, there are six places that your solitary heads can come up – first, second, 3rd, 4th, 5th, or 6th, and the probabilities are all the same, so we can just add them or all together, or else multiply 1/64 by 6, and we get 6/64, of 3/32, 9.375% of the time. So I was off a bit, it’s closer to 9% than 10% . Not a big deal.
In fact, if you use something called the binomial theorem, or better yet, Pascal’s triangle which you can write on a piece of scratch paper in a minute or less, you can calculate what is P(0 tails), P(1 tail), and P(2 tails) all the way up to P(6 tails.)
P0 = 1/64 = P6 = 1.5625%
P1 = 6/64 = P(5) = 9.375%
P2 = 15/64  = P(4) = 23.4375%
P(3) = 20/64 = 31.25%
I hope you underrstand my shortcuts. if not, please tell me and I’ll explain more clearly.
In any case, the chance of getting exactly 1 head or exactly 1 tail adds up to about 19% of the time — not impossible.
If, howebver, the coin (or whatever it was they were using) was rigged, then things are different, and I’ll look into that later. Gotta run now.
Published in: on October 20, 2016 at 6:12 pm  Comments (1)
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