Let 0<=p<=1, I want the value of q1 and q2 where 


$q1=\sum\_{k=0}^n [C(n,k)p^k(1-p)^{n-k}\*\sum\_{i=0}^{k-1} C(m,i)p^i(1-p)^{m-i}]$ 
$q2=\sum\_{k=0}^n [C(n,k)p^k(1-p)^{n-k}\*\sum\_{i=k}^{m} C(m,i)p^i(1-p)^{m-i}]$ 


where C(m,i) is the number of i-combination of a set of m elements.


Obviously q1+q2=1.


For special m,n, for example, n=8, m=4, p=0.5, q1 is about 0.75, q2 is about 0.25. I guess that q1 will be much greater than q2 when n>>m. 


So, I want an approximate estimation on q1 and q2.


If we let p=0.5, then we have 


$q1=p^{n+m}\sum\_{k=0}^n [C(n,k)\*\sum\_{i=0}^{k-1} C(m,i)]$ 


This transformation may make the problem easier. 



asymptotic or approximate formula for a combination expression
If independent variables $X,Y$ are distributed Binom$(n,p)$, Binom$(m,p)$, respectively, then $q\_1$ is the probability that $X>Y$. If $mp,np$ are large and the line $X=Y$ is not too far from the point $(np,mp)$, then the normal approximation of $X$ and $Y$ will give a reasonable answer since $X-Y$ has a 1-dimensional normal distribution. Namely, $X-Y\sim{}$N$(\mu,\sigma^2)$ where $\mu=(n-m)p$ and $\sigma^2=(n+m)p(1-p)$. If the normal approximation is not good, the same mean and variance are true so you can still get a fair idea.


