>>10985043yes i have posted it here. actually the problem that i described about your algorithm in

>>10982984 that makes it so the distribution isn't uniform is the same problem every function will have. its in a very terse form

>>10979594 by someone else, and

>>10979615 by me. i will do it slightly more formally here.

proof:

i will use some more specialized language quickly here so anyone still lurking this thread can know that taking a probability measure on one set and "pushing it forward" to get a probability measure on a different one is actually very rigorously defined, its not just some fly by night bs. we have a probability triple ({0,1}^n,power set B, uniform probability measure), now we we will use a function f to push forward this probability measure onto another measurable space, of course this is ({0,1,2,3,4,5,6}, and its power set A). this is called a push forward measure. and its defined like this: probability(a in A)=probability of the preimage of a under f.

easy to read again: of course for each element in {0,1,2,3,4,5,6}, say for {1}, this just means its probability is the same as probability that any one of the binary strings that map to 1 happens. because each binary string has probability 1/(2^n) the probability of {1} is (the number of binary strings that are mapped to it by the function we choose)/(2^n). so if we want probabilty(1)=probability(2)=... then we need (a_1)/(2^n)=(a_2)/(2^n)=(a_3)/(2^n)=..., where a_x is the number of binary strings that map to x. now, there are 2^n binary strings and 7 values we are mapping binary strings to, but 7 does not divide 2^n so every a_i can not be equal to each other. DONE, sorry for not using latex lol.