If we listed the first digits of each of the first 100 powers of 2, we would get:

1, 2, 4, 8, 1, 3, 6, 1, 2, 5,
1, 2, 4, 8, 1, 3, 6, 1, 2, 5,
1, 2, 4, 8, 1, 3, 6, 1, 2, 5,
1, 2, 4, 8, 1, 3, 6, 1, 2, 5,
1, 2, 4, 8, 1, 3, 7, 1, 2, 5,
1, 2, 4, 9, 1, 3, 7, 1, 2, 5,
1, 2, 4, 9, 1, 3, 7, 1, 2, 5,
1, 2, 4, 9, 1, 3, 7, 1, 3, 6,
1, 2, 4, 9, 1, 3, 7, 1, 3, 6,
1, 2, 4, 9, 1, 3, 7, 1, 3, 6

Looking at these digits, not each digit appears with equal frequency. For example, the digit 1 appears 30 times (or 30% of the time), while the digit 9 only appears 5 times.

If you think about it, this makes sense. If you take any number, the first digit of its double will be 1 if the first digit of the original number is anywhere between 5 and 9, while the first digit of its double will be 9 only if the first two digits of the number are between 45 and 49.

If we take a look at the first digits of the first few numbers of the Fibonacci sequence, the results are similar:

1, 1, 2, 3, 5, 8, 1, 2, 3, 5, 8, 1, 2, 3, 6, 9, 1, 2, 4, 6, 1, 1, 2, 4, 7, 1, 1, 3, 5, 8

Looking at the first 30 numbers, 1 appears 9 times, or, again, 30% of the time, while 9 only appears once.

Benford's law, published by Frank Benford in 1938, states that, in various lists of numbers, the digit 1 appears in the leftmost position about 30% of the time, much greater than the 11.1% that would result if each digit occurred with equal probability. In general, in a number list, the probability of a digit appearing in the leftmost position is equal to log(1 + 1/`n`), where `n` is the digit. This gives the following probabilities, approximately:

First Digit | Probability |
---|---|

1 | 30.1% |

2 | 17.6% |

3 | 12.4% |

4 | 9.7% |

5 | 7.9% |

6 | 6.7% |

7 | 5.8% |

8 | 5.1% |

9 | 4.6% |