This is a game between two players, A and B. Both players start on $n$ ($n\in\mathbb{Z^+}$) points, and the first player to reach $0$ points wins. A begins and the players alternate taking turns.
A turn consists of choosing a number $p$ ($p\in\mathbb{Z^+}$). Then, that player has a $\frac{1}{p}$ chance of losing $p$ points and a $\frac{p-1}{p}$ chance of gaining $p$ points. Essentially, the player decides between progressing $1$ closer to $0$ with a 100% chance, $2$ closer to $0$ with a 50% chance (and a 50% chance of moving $2$ further away), $3$ closer to $0$ with a 33% chance (and a 66% chance of moving $3$ further away), etc.
The question is: What is the probability of each player winning as a function of $n$, the starting number if both players play to maximize their probability of winning on each turn.
My current understanding:
$n=1 \rightarrow P(A)=1$ because player A takes the 100% chance of losing $1$ point.
$n=2 \rightarrow P(A)=\frac{5}{8}$ because player A takes the 50/50 gamble of losing $2$ points. Half of the time they win automatically. The other half of the time, player B loses $1$ point with 100% certainty and player A has a 25% chance of losing $4$ points and winning on their next turn. This strategy for player B is strictly better than for going for the 50/50 if player A loses the 50/50.
$n=3 \rightarrow P(A)=\frac{5}{9}$. The best strategy for A is to lose $1$. Then player B should go for the $\frac{1}{3}$ and this gives B a $\frac{4}{9}$ chance of winning. I think this is right but I have not been as rigorous with this one due to the number of cases.
$n>3 \rightarrow P(A)=\frac{5}{9}$. Here I am very unsure. It seems counterintuitive that the probability does not approach a half for each player as $n$ increases but it seems to me that the best strategy for both players is to keep taking 1 until they get to the $n=3$ case where the probability is $\frac{5}{9}$ for player A to win.
Thanks for any help.
UPDATE
I have done some more work on the problem and attempted to generalize the solution to any (a,b) starting point. Below is an image of my current conjecture. It seems like going 1 is almost always the best option until you get close to the end. In the image, player A (the horizontal numbers) plays next and their strategy is denoted by the colors. The red square represents the only deviation from the go one unless the opposition is 1 or 2 away strategy.
Thanks for any help verifying the probabilities or correcting what I've done so far.