I noticed the quantity of numbers from 1-100 with remainder zero modulo nine = quantity of numbers from 1-100 with remainder one modulo nine > quantity of numbers from 1-100 with remainder 2 modulo nine=...= the quantity of numbers of remainder eight modulo nine.
You can generalize this two the remainders modulo three.
I came up with 67/101 including zero up to 100. If I omit zero, do I need to omit something else or should I remove zero? That's why monte-carlo on spreadsheets is confusing me.
but how do you calculate this exactly? Does using given probability help? I'm guessing 2/3 is the probability.
I realize the remainder modulo n form a equivalence class, but the remainder 1 modulo 9 is not comparable with the remainder 1 modulo 3.
Each remainder $\pmod 9$ is equally probable, and the ones that have the property you seek are $\{3,4,5,6,7,8\}\pmod 9$. Hence the answer is $\frac 69=\frac 23$, in line with your simulation and conjecture.
A note on the probability: I am speaking of the limit, of course. For any particular cutoff $N$ there may be some slight bias. Indeed, fix the cutoff $N$ and let $r_N$ be the remainder of $N$ divided by $9$ Then the number of natural numbers $≤N$ with remainder $r$ is given by $$ \begin{cases} \big \lfloor \frac N9\big \rfloor, & \text{if $r>r_N$ } \\ \\ \big \lfloor \frac N9\big \rfloor+1, & \text{if $1≤r≤r_N$}\\ \\\big \lfloor \frac N9\big \rfloor, & \text{if $r=0$ } \end{cases}\\$$
If $N$ is large, the additive factor of $1$ has no significant impact on the probability. Of course, if $N$ is small it does matter.