There is a game I play; it's called $21$, the rules are as follows.
- Each turn a player says either the next $1,2,3$ or $3$ from the player previous.
- The games starts at $0$ and ends at $21$.
I've figured out how to win with $100%$ certainly when $2$ players. It's going second. Most people only find that $20$ and $16$ are the numbers there after because if say those they win why? Because $16$ can always be forced into $20$. $16$ can be played into $17$ $18$ $19$ all of which can go to $20$ and you can go the other way $12$ can force $16$ if you say it and you can keep going down to $4$ which you can force if you go $2$nd.
The formula for this is $(n-5)-4)-4)-4)\ldots\, n $ is $21$ and you all the numbers you take as in the product of this equation are winning numbers so that's $2$ players now this is question for you.
How can I maximize my odds of winning with $3,4,5$ players?