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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.

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1 Answer 1

up vote 1 down vote accepted

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.

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