4 ms·
HOW? I just did a toy trial with two random integers between 0 and 1,000,000. In Excel. I'm getting between 24% and 31% of the products ending in a '0'. Benfor
by function_seven 5y ago
HOW? I just did a toy trial with two random integers between 0 and 1,000,000. In Excel. I'm getting between 24% and 31% of the products ending in a '0'.
Benford's law makes intuitive sense to me, in that as soon as you run through the leading digits 1-9, you then spend just as long as all those combined with a '1' in front of the "next set".
But I always figured even digits would be more probable, but uniform across the [0,2,4,6,8] set.
- lucasmullens 5y agoWell, if either number ends in a 0, the product will end in a 0, so that's almost 20% of the cases right there. And if you get any even number and a number ending in 5 you'll get a product ending with 0 also, so 24-31% total sounds about right.
- function_seven 5y agoSuch a simple answer. Thanks. I just did a 10x10 times table and got the following prevalences: [27, 4, 12, 4, 12, 9, 12, 4, 12, 4] So '0' has 27% probability, '5' has 9%, and the rest of the evens 12%, the rest of the odds 4%
- inimino 5y agoAnother way to explain it is consider the prime factorization of both numbers. The product will contain all of the prime factors from both, so if there is at least one 2 and at least one 5 in then the result will be divisible by 10.
- brilee 5y agoif you draw out the 10x10 multiplication table, anything 0 * x or x * 0 ends in 0 (19 out of 100 cases), and any even number times 5 also ends in 0 (8 out of 100 cases), which lines up with the 27% you're seeing. Since this is on the units digit, Benford's law doesn't apply and you can basically expect the distribution of units digits to be uniform.