3 ms·
I used the first 100 digits of pi (left for even, right for odd) and it guessed correctly 49% of the time. By the way, within those first 100 digits, there are
by jawns 8y ago
I used the first 100 digits of pi (left for even, right for odd) and it guessed correctly 49% of the time.
By the way, within those first 100 digits, there are multiple occurrences of even or odd sequences that go past the 5-gram level.
Within the first 1000 digits of pi, there is one 11-digit sequence of odd numbers, which you will lose money on.
- throwawaymath 8y agoIf you convert Pi to base 2 you won't be penalized for those long sequences of odd numbers. Mapping the decimal digits of Pi to a parity function collapses a lot of the entropy.
- umvi 8y ago> If you convert Pi to base 2 you won't be penalized for those long sequences of odd numbers That doesn't seem correct. How do you fairly map digits 0-9 to base 2? Assuming pi has an even distribution of digits, you are going to get a disproportionate amount of 1's. There are 2 ways of mapping digits to binary: fixed width and minimal. With minimal the digits 0 and 1 require 1 bit, and 8 and 9 require 4 bits. So then 0:9 maps to (0, 1, 10, 11, 100, 101, 110, 111, 1000, 1001). Tally it up: 15 ones, 10 zeros. That means you will be encountering 1 50% as often as 0, which doesn't seem random to me. So then let's try fixed width: 0:9 maps to (0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001). Tally it up: 15 ones, 26 zeros. That means you will be encountering 0 >60% as often as 1, which doesn't seem random to me. Also, with this scheme you can never encounter more than 4 ones in a row ("78"), which is also unrandom
- umvi 8y ago> If you convert Pi to base 2 you won't be penalized for those long sequences of odd numbers There's no way to convert pi to base 2 in your head though...
- pvg 8y agoClearly the solution is to remember pi in hex.