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?

Here is an intuitive approach to the problem. Suppose the first number we pick is a 1 - we'd obviously be better off picking the second number without replacement, in order to eliminate the chance of getting another 1. Suppose the first number we pick is a 10 - we'd obviously be better off picking the second number with replacement, to allow for the possibility of getting another 10. You can extend this logic to see that if we pick a number above the average of 5.5, we would prefer to pick with replacement, but if we pick below the average, we would prefer to pick without replacement.
There are 5 numbers above average and 5 below, so an equal number of instances where we'd prefer one strategy versus the other, all of which are equiprobable. But the difference is in the value of the potential product. If the first number is a 1, it doesn't really matter a whole lot what the second number is, since the score is multiplied by just 1 - with replacement, the score will be between 1 and 10, instead of between 2 and 10 without. But if the first number is a 10, there is a greater impact due to the fact that it is multiplied by 10, giving us a possible score between 10 and 100 with replacement, instead of 10 to 90 without.  If we adopt the "with replacement" strategy, the "gain" from getting double 10's is larger than the "loss" we get from picking double 1's. You can extend this symmetrically to see that the gain from double 9's is greater than the loss from double 2's, and so forth. We always gain more at the upper end with the "with replacement" strategy than we lose at the lower end, so the "with replacement" strategy is preferable overall.
