Sample two numbers from 1 to 10; maximize the expected product Assume you sample two numbers, randomly drawn from 1 to 10; you could choose two strategies: 1) pick with replacement and 2) pick without replacement. Which strategy would you prefer to maximize the expected product?
I encountered this problem while preparing for an interview. My intuition is picking with replacement would be better. However, I found it too complex to use the brute-force method to prove it. Could anyone suggest a better approach?

There are three facts that led me to the conclusion that replacement gives a higher expected value:
1. Products result in right-skewed distributions. When you multiply numbers together, you tend to get results clustered mostly in small numbers, with a few large results.
2. Among right-skewed distribution, increasing the variance tends to increase the mean. Since extremes above the median are more extreme than the extremes below it, getting more extreme increases the result on average. For instance, in the question you present, we can quickly estimate the median as being somewhere around $5\times6$ (the product of the two middle numbers). The minimum is $1\times1$, and the maximum is $10\times10$. So getting an extremely low number costs us at most $30$, while an extremely high number can give us as much as $70$ over the median.
3. Allowing replacement increases the variance. With replacement, you can get something consisting entirely of the most extreme numbers. Without replacement, you can only get one instance of the maximum and the other one can be at most second highest, and similarly for the minimum.
This isn't a rigorous proof, but understanding how asymmetry, variance, and replacement interact is important for working with statistics.
As a side note, I don't think is really game theory. If simply phrasing it as a choice between options made it game theory, then all optimization problems would be game theory (and pretty much everything can be phrased as an optimization problem).
