Computing the probability of reaching any leaf of an $n$-ary infinite probability tree Suppose you have two players $X$ and $Y$ fighting, both of which have $n\in \mathbb{N}, n\geq1$ life.
Each player has a probability $p_i$ of doing $i$ damage for all $i\in[0, n]$. Note that $p_0$ is the probability of doing no damage, and all $p_i$ should sum to 1. $X$ and $Y$ both share probabilities, so all $p_i$ for $X$ equals $p_i$ for $Y$. 
Now play the game as follows (each step is one turn):


*

*Player $X$ attacks, and randomly deals $i$ damage to player $Y$ according to the distribution $p_i$. This means that $Y$ decreases their life by $i$.

*Player $Y$ attacks, and randomly deals $i$ damage to player $X$ according to the distribution $p_i$. This means that $X$ decreases their life by $i$.


The game ends when a player's life is $\leq 0$.
Now a person watching this game would like for it to end after about $t\in \mathbb{N}, t\geq 1$ turns. In what ways can they assign the probabilities $p_i$ such that the expected number of turns is $t$?
Edit: To be clear I am looking for all possible $p_i$ such that the expected number of turns is $t$.
 A: Let $A(x,y)$ be the expected length of the game (counting each player's turn as $1$) if you start with the players having life points $x$ and $y$ respectively, the player with $x$ going next. 
Conditioning on the result of the first player's turn,
$$A(x,y) = 1 + \sum_{i < y} p_i A(y-i, x)$$
You're interested in $A(n,n)$, which this recursion will determine as a rather complicated rational function in the $p_i$.  For example, I get
$$ A({2,2})={\frac {{p_{{0}}}^{3}-{p_{{0}}}^{2}p_{{1}}-{p_{{0}}}^{2}+{p
_{{1}}}^{2}-p_{{0}}+p_{{1}}+1}{ \left( p_{{0}}+1 \right)  \left(1 -  p_{{0
}} \right) ^{3}}}
$$
while $A(3,3)$ is of the form $P(p_0, p_1, p_2)/((1+p_0)^3 (1-p_0)^5)$
where $P$ is a polynomial in $p_0, p_1, p_2$ of total degree $7$ with $39$ terms.  The condition $A(n,n) = t$ then determines a surface in $(p_0, \ldots, p_{n-1})$ space.  
EDIT:
For example, here are the curves $A(2,2) = t$ in $(p_0,p_1)$ space for $t = 2$ to $10$ (from left to right).  Note that in this case any damage $\ge 2$ is immediately lethal, and 
$\sum_{i \ge 2} p_i = 1 - p_0 - p_1$.  For $A(2,2) = 1$ you'd need $p_0 = p_1 = 0$.

