Probability that a character kills the boss in a 3 versus 1 fight In a RPG game, your three characters A, B and C fight a boss.
The boss has $1000$ hp.
The attacks are sequential: A attacks then B attacks then C attacks then A attacks again, etc.
Character A can do any integral amount of damage between $25$ and $50$ with equal probability.
Character B can do any integral amount of damage between $30$ and $70$ with equal probability.
Character C can do any integral amount of damage between $10$ and $80$ with equal probability.
Assuming that the boss is not strong enough to kill any of the characters before it dies, what is the probability that player A will be the one to deliver the final blow and kill the boss?
Same question for players B and C.
Unfortunately I don't even know how to get started on this problem, any hint would be helpful.

I think this problem can only be solved either approximately (asympotically exact) or numerically. Karl's approximation is the most natural answer, and I'd go for it. If you instead want an exact result, here's an efficient numerical procedure (computer assisted, though).
Letting $g_A(n)$ be the probability that player $A$ wins given that the boss has power $n$, we have the recursion
$$ g_A(n) = \sum_{j=n}^\infty p_A(j) + \sum_{j=1}^{n-1} p_{ABC}(j)g_A(n-j)$$
where $ p_A(j)$ is the probability mass function of the damage inflicted by player $A$ and $p_{ABC}(j)$ is the same for the three players summed.
Solved numerically in Octave (or Matlab) this gives
$$g_A(1000) =  0.28347$$
pretty near Karl's approximation ( $0.28302$)
The graph shows that that approximation is quite good after $N\approx 900$

Code:
pa = [1:100];
pa = pa >= 25 & pa <= 50;
pa = pa/sum(pa);
pb = [1:100];
pb = pb >= 30 & pb <= 70;
pb = pb/sum(pb);
pc = [1:100];
pc = pc >= 10 & pc <= 80;
pc = pc/sum(pc);
pab = [0, conv(pa,pb)];
pab = [0, conv(pa,pb)];
pac = [0, conv(pa,pc)];
pbc = [0, conv(pb,pc)];
pabc = [0, conv(pab,pc)];
N=1000;
pabc(N)=0;
ga = zeros(1,N);
ga(1)=1;
for n=2:N
  ga(n) = sum(pa(n:100)) +  flip(ga(1:n-1)) * pabc(1:n-1)';
endfor
ga(N)

For completeness, here are the other probabilities

gb = zeros(1,N); % prob of B winning, if B has next turn
gb(1)=1;
 for n=2:N
 gb(n) = sum(pb(n:100)) +  flip(gb(1:n-1)) * pabc(1:n-1)';
endfor
gbb=zeros(1,N);  % prob of B winning, if A has next turn
pa(N)=0;  % extend pa size
for n=2:N
 gbb(n)= flip(gb(1:n-1)) * pa(1:n-1)';
endfor
gcc = 1 - (ga+gbb);

Notice that the period of the quasioscillations is around the mean value of the whole turn : $(25+50+30+70+10+80)/2 = 132.50$
