Binomial distribution for two groups if success rate is not given Two groups of twelve statisticians are taught two different methods of Statistics. (Assume that a statistician in group one is matched in terms of their Statistics ability with a statistician in group 2 before the start of the study).What is the probability that at least 9 statisticians from one of the groups will obtain higher scores than every other statistician of the other group? What are the assumptions made?
How can we solve this problem if there is no success rate given?
 A: This is the setting of the Wilcoxon Rank Sum Test.  It can be solved with a minimal set of assumptions.
One such set of assumptions is formulated for the purpose of critically examining  the null hypothesis "the teaching had no effect."  This hypothesis is tantamount to supposing all 12+12 = 24 scores were randomly assigned to the two groups (which is how such a study should be carried out, by the way) by taking a simple random sample of 12 of them to put in one group (call this group "A") and leaving the remaining 12 in the other group.
The calculations are simplest and definite when, in addition, it is supposed that the test instrument is sufficiently precise that the ninth highest score is strictly greater than the tenth highest (out of all 24).  In this case, when the 24 scores are put in order from smallest to largest, there is no ambiguity concerning which are the nine highest.
What is the chance that the nine highest scores were randomly assigned to a common group?
By symmetry--the groups play equivalent roles in this setting--the answer is twice the chance that the nine highest scores were assigned to group A.  There are $\binom{24-9}{12} = 15\times 14\times 13 / 3! = 455$ equally likely ways that could occur out of the $\binom{24}{12} = 2704156$ distinct equiprobable configurations. The answer therefore is
$$2\frac{\binom{24-9}{12}}{\binom{24}{12}} = \frac{910}{2704156} \approx 0.0003365.$$

As a quick check--if you're uncertain of the validity of these combinatorial calculations--you could run a simulation.  Here is R code to simulate this situation a million times.  It returns the proportion of times in which the highest nine scores are all in the same group:
mean(replicate(1e6, {length(unique(sample(rep(c("A","B"), 12), 9)))==1}))

When I ran it (waiting about ten seconds for the calculations to complete), the output was 0.000308--indistinguishable from the computed answer, up to random variation in the simulation (which has a standard error of 0.000018).
