Sums of normal random variables Consider a sample of n independent normal rvs. I would like to identify a systematic way of calculating the probability of having the sum of a subset of them larger than the sum of the rest of rvs.
An example case:
Population of fish. Mean: 10 kg, stdv: 3 kg.
I fish five fish (n=5). What is the probability of having two fish weighing more than the rest of the three fish?
The steps which can be followed is to calculate the prob for every combination of fish and then use the inclusion exclusion formula for their union. Is there anything smarter?
Note: if four fish were considered the probability of having two of them heavier than the other two should be one. How could this be computed immediately?
Thanks for the answers.
 A: Your example suggests that not only are the $n$ variables $X_1,X_2,\ldots,X_n$ independent, they also have the same Normal distribution.  Let its parameters be $\mu$ (the mean) and $\sigma^2$ (the variance) and suppose the subset consists of $k$ of these variables.  We might as well index the variables so that $X_1,\ldots, X_k$ are this subset.
The question asks to compute the chance that the sum of the first $k$ variables equals or exceeds the sum of the rest:
$$p_{n,k}(\mu,\sigma) = \Pr(X_1+\cdots+X_k \ge X_{k+1}+\cdots+X_n ) =  \Pr(Y \le 0)$$
where
$$Y = -(X_1+\cdots+X_k) + (X_{k+1}+\cdots+X_n).$$
$Y$ is a linear combination of independent Normal variables and therefore has a Normal distribution--but which one?  The laws of expectation and variance immediately tell us
$$E[Y] = -k\mu + (n-k)\mu = (n-2k)\mu$$
and
$$\operatorname{Var}(Y) = k \sigma^2 + (n-k)\sigma^2 = n\sigma^2.$$
Therefore $$Z=\frac{Y - (n-2k)\mu}{\sigma\sqrt{n}}$$ has a standard Normal distribution with distribution function $\Phi,$ whence the answer is

$$p_{n,k}(\mu,\sigma) = \Pr(Y \le 0) = \Pr\left(Z \le -\frac{(n-2k)\mu}{\sigma\sqrt{n}}\right) = \Phi\left(-\frac{(n-2k)\mu}{\sigma\sqrt{n}}\right).$$

In the question, $n=5,k=2,\mu=10,$ and $\sigma=3,$ whence
$$p_{5,2}(10,3) = \Phi\left(-\frac{(5-2(2))10}{3\sqrt{10}}\right)\approx 0.0680186.$$

Generalization
Little needs to change in this analysis even when the $X_i$ have different normal distributions or are even correlated: you only need to assume they have an $n$-variate Normal distribution to assure their linear combination still has a Normal distribution.  The calculations are carried out in the same way and result in a similar formula.

Check
A commenter suggested solving this with simulation.  Although that wouldn't be a solution, it's a decent way to check a solution quickly.  Thus, in R we might establish the inputs of the simulation in some arbitrary way as
n <- 5
k <- 2
mu <- 10
sigma <- 3
n.sim <- 1e6 # Simulation size
set.seed(17) # For reproducible results

and simulate such data and compare the sums with these two lines:
x <- matrix(rnorm(n*n.sim, mu, sigma), ncol=n)
p.hat <- mean(rowSums(x[, 1:k]) >= rowSums(x[, -(1:k)]))

The post-processing consists of finding the fraction of simulated datasets in which one sum exceeds the other and comparing that to the theoretical solution:
se <- sqrt(p.hat * (1-p.hat) / n.sim)
p <- pnorm(-(n-2*k)*mu / (sigma * sqrt(n)))
signif(c(Simulation=p.hat, Theory=p, `Z-score`=(p.hat-p)/se), 3)

The output in this case is

Simulation     Theory    Z-score 
    0.0677     0.0680    -1.1900


The agreement is close and the small absolute z-score allows us to attribute the discrepancy to random fluctuations rather than any error in the theoretical derivation.
