If you had 4 players, left of dealer goes first, and play steps by 2, then only the left of dealer and right of dealer would ever get to play.
If you step by 1, everyone plays. If you step by 5, it's the same as by 1. If you step by 3, everyone plays and it's like you're going in reverse order.
But in general, if you had N players, and stepped by n (where n and N are coprime), will everyone get to play an equal number of times?
Here's an example not coprime but not divisible either. If N is 15, and you step by 6. Let the first player be 0.
0, 6, 12, 3, 9, 0, 6, ...
It's different players for the first couple rounds but then it repeats, and those players not divisible by 3 never get to play.
Here's an example where n is coprime to N. 10 players, step by 3.
0, 3, 6, 9, 2, 5, 8, 1, 4, 7, 0, ...
Here everyone got to play once before it cycled.
Is this always true?
What I tried:
I noticed we start at $0$ and keep adding $n$. Once we get above $N$, we drop down again, to a value less than $n$.
We can't add to exactly $N$, because it's coprime. And we couldn't add from below $N$ to a value above $N+n$, so we'd get $N+k$, where $k<n$, and that'd be the $k$th player.
Then next would go the $(k+n)$th player, then the $(k+2n)$th player and so on. At some point it'll pass above $N$. Or will it? Could it end up at exactly $N$ again, preventing players other than multiples of $n$ and $k+$ multiples of $n$ from getting to play?
What I know:
$k$ was the difference between some multiple of $n$ and $N$. Let the number of times $n$ goes into $N$ be $a$, so $k = (a+1)n - N$.
Then we keep adding $n$ to $k$, let's say until we get to player $k+bn<N$.
$b$ is $a$ if $k<N-an$ or $a-1$ otherwise.
Then we add $n$ to go back through the start and we're on player $l$, who is again less than $n$. $l$ also couldn't be $k$, because we can't have gotten $N+k$ by adding multiples of $n$ to $k$ (since $n$ and $N$ are coprime). But could $l$ be $0$?
Well, it could be at least if $k$ also is $N - an$. That would mean $N$ is exactly between two multiples of $n$.
So, let's solve this equation: $k = (a+1)n - N = N - an$. We find $2N = (2a+1)n$. Since $2$ is even and $(2a+1)$ is odd, this equation could only be true if $n$ has a factor of $2$ and $N$ has an odd factor $(2a+1)$. In addition, if $n$ had any other factors, they would be in $N$, according to the equation. For them to be coprime, $n$ has to be $2$, with no other factors, and $N$ has to be odd. If $n$ is $2$ and $N$ is odd, we'd do all the even players the first round, then all the odd players the second round, then restart at $0$ and every player would have gotten to play. For example, with $N=7$:
0, 2, 4, 6, 1, 3, 5, 0, ...
Is that the only way $l$ can be $0$? What about after a few iterations? Is it possible to skip players? This question's probably really easy if one thinks about it in the right way. I'm probably overcomplicating it.
