Expected Value versus Long Run Median in Gambling Situations

Question

Almost always, we use the expected value to analyze a gambling problem. Since the expected value is effectively mechanically identical to the weighted mean, this would imply that when we use the expected value to gain insight into a gambling game, we essentially look at the average of the outcome distribution.

This may work well for symmetric outcome distributions, but when the outcome for some gambling game follows a skewed distribution, the arithmetic mean suddenly does not become a very reliable measure of the center of the outcome distribution; instead, the median does. So, why do we not look at the median of all outcomes when analyzing a gambling game and its betting strategies?

Consider a game whose expected value of profit when following a certain betting strategy is negative. However, that same strategy in the same game yields a median value of profit that is positive. Is this a game that is worth playing, just as a game whose expected value of profit is positive is worth playing? Or, what if a game has an expected value of profit that is positive but a median value of profit that is negative?

To be clear, I'm asking for some insight into the game examples above, as well as for some possible further reading about how to interpret the median of an outcome distribution when analyzing gambling scenarios and betting strategies.

All help is appreciated.

Answer 1

In your question you state:

but when the outcome for some gambling game follows a skewed distribution, the arithmetic mean suddenly does not become a very reliable measure of the center of the outcome distribution; instead, the median does.

I agree that the median has 50% of the area under the curve to its left, and the other 50% of the area under the curve to the right. But that doesn't mean that the median is the amount you can expect to have in your bankroll at the end of an infinite amount of gambling.

The arithmetic mean (rather than the median) tells you the expected amount of money you will have in your bankroll.

Lastly, you mention the scenario where the expected value of profit is negative, but the median value of profit is positive. I've never heard of a median value of profit.

You may be confusing mean, median, and expected values of a probability distribution.

Answer 2

In the long run the expected value will be the mean of the distribution. That is the content of the law of large numbers. Your "consider a game" is like the doubling strategy in US roulette. If you consider one game to be a run until you win or cannot play any more and you can afford to double $9$ times you have about a $\frac {612}{613}$ chance of winning $1$ and a $\frac 1{613}$ chance of losing $1023$. The median is $+1$ and the expected value is $-\frac {411}{613}$. Yes, if you play once, you have a very high chance of coming out ahead. If you play lots of times you will be behind. Each bet is a losing one, how can adding a bunch of negative numbers come out positive?