I flip M coins, my opponent flips N coins. Who has more heads wins. Is there a closed form for probability? In this game, I flip M fair coins and my opponent flips N coins. If I get more heads from my coins than my opponent, I win, otherwise I lose. I wish to know the probability that I win the game.
I came to this:
$$P(victory) = \sum\limits_{i=1}^M \left(\binom{M}{i}\left(\frac{1}{2}\right)^i\left(\frac{1}{2}\right)^{M-i} 
\times 
\sum\limits_{j=0}^{i-1} \left(\binom{N}{j}\left(\frac{1}{2}\right)^j\left(\frac{1}{2}\right)^{N-j} \right)
\right)$$
My questions would be: Is the above formula correct? And can a closed form formula exist, and if not, is there a simple proof?

Your formula is basically right, but you can simplify the $1/2$ factors. Letting $V\equiv M>N$ (victory), using conditional probabilities:
$$P(V)= \sum_M P(V , M) = \sum_M P(M) P(V | M) = \sum_{m=0}^M \frac{1}{2^M}{M \choose m} \sum_{j=0}^{m-1} \frac{1}{2^N}{N \choose j}=\\=\frac{1}{2^{M+N}}\sum_{m=0}^M {M \choose m} \sum_{j=0}^{m-1} {N \choose j}$$
and you must also impose the convention ${N \choose j}=0$ if $j>N$ 
I doubt that this has a simpler expression. 
