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For numbers in natural form, the probability of the first digit being 1 is about 30%, and the probability of 2 being 17.6%, decreasing in order. The probability of the first digit being 8 and 9 is only 5.1% and 4.6% respectively.
\(d\in\{1, ..., 9\}\)
\[P(d) = \log _{10}(d+1)- \log _{10}(d) = \log _{10}(\frac{d+1}{d}) =\log _{10}(1+ \frac{1}{d}) \]
It also has a limitation, that is, the number is at least more than 3000.
1 30.1%
2 17.6%
3 12.5%
4 9.7%
5 7.9%
6 6.7%
7 5.8%
8 5.1%
9 4.6%
You may be curious, shouldn't this be average? Why is the probability of 1 appearing very high?
Sorry, no one knows why.
This may be a law of this world.
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