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?

If you toss $M$ coins, there are ${M \choose m}$ ways to get exactly $m$ heads.
Then, the number of ways you can win when your opponent tosses $N$ coins, given that you've tossed $m$ heads, is
$${M \choose m}\sum_{i=0}^{m-1}{N \choose i}.$$
Now, just sum over all possible values of $m$, and divide by the total number of outcomes:
$$P_{win}(M, N) = 2^{-(M+N)}\sum_{m=0}^{M}{M \choose m}\sum_{i=0}^{m-1}{N \choose i}.$$
This looks pretty close to what you have.
