Alice and Bob are playing a game. They take $M$ turns, starting with one point each. The winner is the player with the highest score at the end of the game. In each round the players can make "increments" up to $N=3$ times. Incrementing means increasing your score to the next number that is coprime to your opponent's score. For example, if you have $7$ points and your opponent has $12$, one increment would result in you having $11$ points.
In a $N=3, M=2$ game, the optimal play is for Alice and Bob to increment $2, 3, 3, 3$ times. The scores are: $1-1 \rightarrow 3-1 \rightarrow 3-5 \rightarrow 7-5 \rightarrow 7-9 $, so Bob wins. Alice chooses not to make the three increments in Round 1, as she would score $4$ points and Bob would be allowed to make three increments: $1\rightarrow 3\rightarrow 5\rightarrow 7$.
Using a computer to compute optimal strategies up to about $M<1000$, it appears that there is always a winning strategy for Bob, regardless of the value of $M$. This is interesting because for most combinations of $N$ and $M$, Alice has a winning strategy. My question is whether it is true that, for $N=3$, Bob always has a winning strategy regardless of the value of $M$. (I am also interested in $N$ values for which Bob has a winning strategy for any $M$. I am considering asking a separate question about this.)
Update
I posted a proof to this question last week, but I have found a flaw in it. I have highlighted the problematic part in bold.
Thanks to helpful comments, I was able to come up with a proof.
Let $A_i$ and $B_i$ denote the scores of Alice and Bob, respectively, at the end of the $i$-th round. Suppose for a particular $i$, $B_i$ is odd and $B_i >= A_i + 2$.
In this case, since $B_i-1$, $B_i+1$, and $B_i+2$ are pairwise coprime with $B_i$, Alice cannot step over these values. Consequently, her score in the next round is limited by $A_{i+1} <= B_i + 2$.
If $A_{i+1} <= B_i$, Bob can increment his score twice or three times to ensure that $B_{i+1} >= A_{i+1} + 2$, and $B_{i+1}$ is odd.
If $A_{i+1} = B_i + 1$, then $A_{i+1}$ is even. In this case, Bob can increment his score twice, ensuring that $B_{i+1} >= A_{i+1} + 3$, and $B_{i+1}$ is odd.
If $A_{i+1} = B_i + 2$, Bob's second increment skips $A_{i+1}$, so $B_{i+1} = B_i + 4 = A_{i+1} + 2$. This also guarantees that $B_{i+1}$ is odd.
Since the initial condition $B_i >= A_i + 2$ and $B_i$ being odd are satisfied at $i=1$, it follows that for all $i$, $B_i >= A_i + 2$.
The bold section would be incorrect, for example, in cases where $A_{i+1}=55$ and $B_i=73$. In such a scenario, $B_{i+1}$ can only be $74$, $76$, or $78$, and it cannot be an odd number.