What is the probability that a sample value from a Binomial distributions is smaller than a sample value from another?

Let $X \sim \mathrm{Binomial} (n, p_1)$ and let $Y \sim \mathrm{Binomial} (n, p_2)$ with $n, p_1$ and $p_2$ known and $p_1 < p_2$.

What is the probability that a sample value $x$ drawn from $X$ is smaller than a sample value $y$ drawn from $Y$? That is, what is $P(X<Y)$?

• Yes; that a sample drawn from X is smaller than another sample drawn from Y. Commented Dec 5, 2012 at 23:16
• The trick here is to use the law of total probability; if you know that $Y=y$, then you can calculate $P(X<y)$, by summing $X$'s probability mass function, $P(X=x) = \binom{n}{x} p_{1}^{x} (1-p_1)^{n-x}$, over $x=0,...,y$. Then you just take the expected value of that quantity, with respect to $Y$. Btw, the result is probably not going to look pretty :) Commented Dec 5, 2012 at 23:16
• And if $Y=y$ can take any value, one would multiply $P(Y=y)$ with $P(X<y)$. Is this correct? Commented Dec 5, 2012 at 23:20
• Yes, and sum over the whole domain $y=0,1,...,n$. This is what I meant by taking the expected value with respect to $Y$ Commented Dec 5, 2012 at 23:22
• $P(X<Y)$ is equivalent to $P(X-Y<0)$; in turn, that's $P(X+(-Y)<0)$ which equals $P(X+(n-Y)<n)$, and $(n-Y)\sim \rm{Bin}(n,1-p_2)$; each can be useful. If $n$ is not really large you can do the last by numerical convolution; I've done some decent-sized ones in R on negative binomials in the past; binomials are even better suited to it. If $n$ is not small and the $p$'s are not really close to the boundaries (the two are linked), you can approximate this probability pretty well using normal distributions (I'd suggest using a continuity correction on the difference $X-Y$ and comparing to 0). Commented Dec 6, 2012 at 0:43

Although an explicit formula might be tricky, it's quite easy to evaluate numerically. The key is to realize the $X$ and $Y$ are independent.
Start by pretending you know $Y=y$. You know that if $X \sim \mathcal{B}(n_1,p_1)$ then: $$P[X=x] = \binom{n_1}{x}p_1^x(1-p_1)^{n_1-x}$$
From there you can easily find: $$P[X \le y] = \sum_{x=0}^{y} \binom{n_1}{x}p_1^x(1-p_1)^{n_1-x}$$
Because $X$ and $Y$ are independent, we know: $$P[X<Y] = \sum_{y=0}^{n_2}P[X<y] * P[Y=y]$$ $$= \sum_{y=0}^{n_2}\left(\sum_{x=0}^{y} \binom{n_1}{x}p_1^x(1-p_1)^{n_1-x}\right)\binom{n_2}{y}p_2^y(1-p_2)^{n_2-y}$$