Wow. You can't do math. Whatever you think of the election or these students, attacking their argument with such blithe ignorance does more to hurt your apparent 'cause' than convince your readers. The summary you critique is a remarkably simplified report of accurate statistical analysis of the election. I haven't read the math, nor even know if it's been published, but I can reproduce their results with the same numbers. Your ignorance of their calculations is a pitiful attempt to discredit logical argument while remaining blissfully unaware of your own shortcomings. And no, the first two probabilities are not multiplied. They are not independent.
alchemytoday
· 5 months ago
What 'cause' do you think I'm taking up? I can reproduce their numbers as well (except for one error which they've recognized).
The two probabilities are entirely independent in the case where the last two digits are random. In the definition used here, any last digit has 7 out of 10 unique, non-adjacent digits (everything that's not a repeated digit or an adjacent digit, with 0 being adjacent to both 9 and 1). No matter what the last digit, a random second-to-last digit will have a 70% chance of being non-adjacent. Beber and Scacco agree with this in the annotated version of their article.
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· 3 months ago
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· 2 months ago
Acne Treatment As a point of comparison, we can analyze the state-by-state vote counts for John McCain and Barack Obama in last year’s U.S. presidential election. The frequencies of last digits in these election returns never rise above 14 percent or fall below 6 percent, a pattern we would expect to see in seventy out of a hundred fair elections.
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The two probabilities are entirely independent in the case where the last two digits are random. In the definition used here, any last digit has 7 out of 10 unique, non-adjacent digits (everything that's not a repeated digit or an adjacent digit, with 0 being adjacent to both 9 and 1). No matter what the last digit, a random second-to-last digit will have a 70% chance of being non-adjacent. Beber and Scacco agree with this in the annotated version of their article.
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It’s trivial to continue and think of dozens of equivalent events all with a 3.5% probability. In fact, there is a 100% chance that a string of 116 random digits will feature such a pattern (update: I suspect this, but I’m not remotely capable of proving it).
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