Probability of defeating enemy (info on distributions added) I have 27 hit points and my opponent has 50, and the winner is the player that reduces the other player's hit points to 0 first.
My expected damage inflicted per round is 5.
My expected damage taken per round is 7/3.
I have the first attack.
What are my chances of winning?
edit: As pointed out, the question is unanswerable without further info on distributions, so here it is:
A round consists of multiple attacks. If I hit, I can attack again. My damage per hit is 5/6 1, and I have a 5/6 probability of hitting. I figured this makes a geometric series with $a = 5/6, r = 5/6$, so using $a / (1 - r)$ we get an expected damage of 5. Edit: Typo above. Damage is 1 point if I hit, so expected damage is 5/6. (Geometric series is still correct.)
When defending, similar rules apply. My damage taken is 7/6, and probability of being hit is 1/2. So a geometric series gives expected damage taken as 7/3. Edit to clarify: Actually I have a 1/6 chance of taking 3 damage, and 2/6 chance of taking 2 damage, which aggregates to 7/6.

So as a supplementary question, clearly this is a discrete
  distribution, but can we approximate to something linear and solve
  analytically? Or am I best off just doing a million trials by
  computer?

 A: You don't have enough information to answer.  A simple way to see that is to assume your opponent will always do 7/3 damage, so you will always get 12 hits.  If you always do 5 damage you will win every time.  If you do 5000 damage one time in 1000, you will win about 12/1000 (slightly less).
A: It depends on the actual damage distributions.  For example, if you always take 7/3 and deal 5, your odds of winning are 100%.  If you always deal 5, take 7 with 1/3 probability and 0 with 2/3 probability, your odds of winning are 12800/19683 or 65%.
A: If you're reduced to brute-forcing it by computer, you don't need a million trials. Just fill in a $50\times 27\times 2$ array with the probabilities of winning, given the point standings and who's currently attacking. Compute the lowest points first and work your way up to the $(50,27)$ point.
A: Let's try something practical, building on Henning Makholm's answer.
First, let's get rid of fractional hit points, by dividing hit points by the effect of hits:  you lose if you take 24 hits and you win if you inflict 60 hits.  
Now if you have taken $a$ hits and your opponent has taken $b$ hits, let $f(a,b)$ be the probability of you winning if you have the next attempt and $g(a,b)$ be the probability of you winning if your opponent has the next attempt. $f(24,b)=g(24,b)=0$ if $b<60$ and $f(a,60)=g(a,60)=1$ if $a<24$. Then $$f(a,b)=\tfrac{5}{6}f(a,b+1)+\tfrac{1}{6}g(a,b)$$ $$g(a,b)=\tfrac{1}{2}f(a,b)+\tfrac{1}{2}g(a+1,b)$$ but this is slightly circular so turn them into $$f(a,b)=\tfrac{10}{11}f(a,b+1)+\tfrac{1}{11}g(a+1,b)$$ $$g(a,b)=\tfrac{5}{11}f(a,b+1)+\tfrac{6}{11}g(a+1,b)$$ 
and calculate.  I think $f(0,0)$ may be about $0.95168$.
