Probability for Fighting Fantasy! What if I have the same skill but more life? so I've been rereading my old fighting fantasy books and I thought of a problem I'm unsure of how to solve.
In the book you fight various opponents by rolling two 6 sided dice and adding your skill (just a number), then you roll the dice for them and add their skill. If your total is higher than theirs you take off 2 of their stamina and visa versa if they beat you. You keep fighting rounds until one of you runs out of stamina and dies.
Its easy to see the probability of winning an individual round if you have more, less or the same skill as your opponent. But how does having more or less stamina affect the outcome of the entire fight? Obviously having more stamina than your opponent means you will have a better chance of winning but how do you do the sum?
If you need any clarification please ask!
Thanks
 A: To simplify things, let us change it so that the losing party takes one damage instead.
Also, let us ignore ties completely by conditioning on that damage actually occurs. (Take for example, rolling one die each, player one has +1 skill, player two has +0.  The probability of player 1 damaging his opponent will be $\frac{21}{36}$, the probability of player 2 damaging his opponent will be $\frac{10}{36}$, and the probability of a tie is $\frac{5}{36}$.  Instead of working with these, we will instead use $\frac{21}{31}$ and $\frac{10}{31}$, i.e. the probabilities given that damage actually occurs).
Let $p$ be the probability that player $1$ scores a hit and $q=1-p$ be the probability that player $2$ scores a hit (given that damage occurs).  Let $M$ be the amount of health player $1$ has and $N$ be the amount of health that player two has.
Suppose also for simplification, that after a player has been defeated, we continue rolling the dice for a little while after anyways.  The reason should become clear momentarily.
We see that after $M+N-1$ successful attacks, exactly one of the players will be dead and the other will still be alive.  Specifically, player $1$ will be dead if player $2$ successfully hit player $1$ at least $M$ times.  On the other hand player $2$ will be dead if player $1$ successfully hit player $2$ at least $N$ times (equivalently player $2$ hit player $1$ at most $M-1$ times).
The probability of player $1$ hitting player two exactly $k$ times follows a binomial distribution and occurs with probability 
$$\binom{M+N-1}{k}p^kq^{M+N-1-k}$$
player $1$ hitting player two at least $N$ times then occurs with probability
$$\sum\limits_{k=N}^{M+N-1}\binom{M+N-1}{k}p^kq^{M+N-1-k}$$

Example1:  Both players have equal skill, player one has $10$ health and player two has $15$ health.  We see that $p=q=0.5$.
Player 1 in this case has probability $\sum\limits_{k=15}^{24}\binom{24}{k}0.5^{24} \approx 0.1537$ of winning.
Example 2:  Both players have equal skill, player one has $10$ health and player two has $20$ health.
Player 1 in this case has probability $\sum\limits_{k=20}^{29}\binom{29}{k}0.5^{29}\approx 0.0307$ of winning.
Example 3:  Player 1 has probability $0.7$ to damage player $2$ and player $2$ has probability $0.3$ to damage player $1$.  (For example, player one having +2 skill while player two having +0 skill.  Specific calculations of $p$ and $q$ are tedious, but I can explain that as well if desired).  Further, player $1$ has $10$ health and player $2$ has $15$ health.
Player 1 in this case has probability $\sum\limits_{k=15}^{24}\binom{24}{k}0.7^k0.3^{24-k}\approx 0.847$ chance of winning.
Example 4:  Player 1 again has probability $0.7$ to damage player $2$, player one has $10$ health, and player $2$ has $20$ health.
Player 1 in this case has probability $\sum\limits_{k=20}^{29}\binom{29}{k}0.7^k0.3^{29-k}\approx 0.636$ chance of winning.

To fix the simplification made at the beginning, just set $M$ and $N$ to be the number of successful hits needed to knock each respective player out.  I.e. $M=\left\lceil\frac{P_1\text{health}}{2}\right\rceil$
