Coin toss - winning by tossing $k$ heads first Two players A and B take turns throwing a fair coin. 
The players that tosses $k$ heads first wins. Let player A begin. What is the likelihood p that players A wins? 
For a given $k$ the solution is very straightforward, as it can be done in several steps by first computing the likelihood of Player A winning in $m$ turns, doing this for all $m$ and then adding everything up. 
Thus the solution for the general case might be: (because of possible typos or errors, I won't claim this being correct!)
$ p =  \sum_{m=k}^{\infty} \frac{1}{2^{2m-1}} \frac{(m-1)!}{(m-k-1)!(k-1)!}  \sum_{i = 0}^{k-1}  \frac{(m-1)!}{(m-i)!i!} $
This is so ugly. And personally I hope there is a better way of doing that, by using generating functions or whatever. I still lack the ability doing such computations in reasonable fashion.
Therefore I would be very happy, if someone had a way to make life easier and show how to do it. 
As always thanks for any constructive comment, answers. 
 A: If they don't take turns but throw simultaneously, then they have equal probability of winning, except that they may draw. If one wins, that same one wins in the original game, but if they draw, then A wins in the original game. So it only remains to find the probability of a draw. It should be easy for you now!
Okay as implicitly requested by ByronSchmuland, here is how to find the probability of a draw. We find the probability of a draw that ends at the $n$-th move, in which case both must throw heads on the $n$-th move and must have thrown exactly $(k-1)$ heads prior to that. The probability of that is just $\binom{n-1}{k-1}^2 (\frac{1}{2})^{2n}$. The total probability of a draw is then a simple summation over $n$ from $1$ to $\infty$. I doubt there is a closed form.
A: Not a complete solution either, but attacking the problem from a different angle:
We use the probability $P(n, m)$ that I toss $n$ heads before my opponent tosses $m$ heads. There are a few simple values: $P(0, m) = 1$, $P(n>1, 0) = 0$
and note that $P(0, 0) =1$ in particular.
To find $P(n, m)$ evaluate one coin toss for each player to get:
$$P(n, m) = (P(n, m) + P(n-1, m) + P(n, m-1) + P(n-1, m-1)) /4\text{, or}$$
$$P(n, m) = (P(n-1, m) + P(n, m-1) + P(n-1, m-1))/3$$
While this is still not a direct formula, the recursive calculation is easy to understand. Unfortunately, it is rather slow ($O(n\cdot m)$) to calculate and takes quite a bit of space ($O(n+m)$).
Doing some calculation by hand gives: 
$$P(1, 1) = 2/3$$
$$P(2, 2) = 16/27$$
$$P(3, 3) = 138/243$$
A: Following the hint in user21820's answer, and using the identity here, the probability of a tie can be written as a finite sum 
 $$t(k):={1\over 3^{k}}\sum_{j=0}^{k-1} {k-1+j\choose j}{k-1\choose j}{1\over 3^{j}},$$
and the probability that player A wins is ${t(k)+1\over 2}.$
