First I'll phrase the question as a riddle, and than as a general math problem.

We have 12 lettered vases $(A,B,...,L)$, in each vase there are 30 numbered balls (1-30). In each ball there is some random amount of money between 1-1000 dollars (the distribution of the money in the balls is some IID). Now we have two options:

1) We can ask which number (1-30) contains in average the largest amount of money, and than we get 12 balls of that number, we open them all and we take the ball with the highest amount of the 12.

2) We can ask which vase (A-L) contains the largest sum of money, and than we get 30 balls from that vase, we open them all and we take the ball with the highest amount of the 30.

What is the better strategy, assuming that we want a lot of money...

Now to phrase it more generally:

Let $ A \in\mathbb{N}^{m\times n} , n>m$,

Let $x_{ij}$ be i.i.d. (real) random variables with mean 0 and variance 1

Let $X_1=\max_{1\leq i\leq m}{\sum_{1\leq j\leq n}a_{ij}}$

and let $X_2=\max_{1\leq j\leq n}{\sum_{1\leq i\leq m}a_{ij}}$

What is larger (in average, or the expected value of) $X_1$ or $X_2$? That is, what we should take first, the maximum value of the rows and than the highest value in that row, or the maximum value of the columns and than the highest value in that column?

**Another question** (more combinatorial): We can ask in a case where all of the values in the matrix are different and between 1 to $m\cdot n$. So we have a pure combinatorial question about the possible $m\cdot n!$ permutations of the numbers in the matrix.

The riddle is of course like this:

Let $ A \in\mathbb{N}^{12\times 30}$ ,$1\leq a_{ij} \leq 1000$ (by i.i.d distibution)

Let $X_1=\max_{1\leq i\leq 12}{\sum_{1\leq j\leq 30}a_{ij}}$ (Option 1)

and let $X_2=\max_{1\leq j\leq 30}{\sum_{1\leq i\leq 12}a_{ij}}$ (Option 2)

Thanks! David

Edited. Thanks for the comments of Yemon, Gerry and James.