Well couple of days ago I saw my nephews playing a game that I played as a child (back then there was no computers and we had to play outside :)) And it got me thinking. Game revolves around coupple of simple rules and starts by throwing a rock on a $3$X$3$ square drawn on a sidewalk (filled with numbers $1-9$) Once the first player throws a rock and the rock falls on, let say number $2$ square that that player "moves" two steps forward and then decides how it will proceed to reach the "goal" which is $10$ steps from its starting point (I am simplifying the rules and objectives a bit). The decision is, he/she can then move by $2$ steps at the time to reach $10$ or $1$ step at the time or mix his/hers decisions but cannot go higher that $2$. However, once he/she decides to use lower number he/she cannot take the higher one again, so:
2+2+2+2+2+2 = 10 or 2+2+2+1+1+1+1 = 10 or 2+1+1+1+1+1+1+1+1 = 10
2+2+1+2+1+2 = 10 - no,no
Next, once the decision has been made and the player reaches $10$ (let say the $2+2+2+2+2+2$ = $10$ was the choice) all other players cannot make more than $6$ moves to reach the same goal ($10$) and the number of steps they start with has to be higher than $2$, so the next player throws a rock and gets $3$ thus his/hers options are:
3+2+2+2+1 =10 (5 moves) 3+3+3+1=10 (4 moves) 3+3+2+2+1=10 (6 moves) ...
also in each case player cannot move more than the initial number of steps he/she chose (got by throwing a rock if it got $3$ then $3$ steps is highest he/she can do per move) but can decide to change the numbers to $2$ or $1$ obeying the above rules (no going back)
After reflecting on the game I started wondering how many possible strategies there are in any given particular case and how would i compute that? So for numbers $3, 4, 5, 6, 7, 8$ and $9$ how many possible strategies there are in total?
Can anyone help with this ??