Betting on a sample from a known distribution This was an interview question. Given a known distribution, sample a value from it with replacement for many times. Two people A and B bet on the sample with their own guesses, and the one closer to the actual sample value wins. What's your betting strategy?
The interviewer claimed the median value is better than mean. I didn't get it. Can someone explain? Thanks.

The median minimizes expected absolute distance between observations sampled from the distribution and a single value. That is:
$$
\underset{m}{\text{min}}\text{ }\mathbb E\left[
\left\vert X - m\right\vert
\right] = \text{median}
$$
This is in contrast to the mean, which minimizes expected squared distance.
$$
\underset{m}{\text{min}}\text{ }\mathbb E\left[
\left( X - m\right)^2
\right] = \text{mean}
$$
I think this is what the interviewer wanted to test if you knew.
