How should I calculate weighted average in this case?

Question

If there is a product, whose price depends on its quality, and these price values are: [0.6; 1.1; 1.3; 2.0; 4.2; 21.6] Lets assume that there is some kind of a lottery, and I can by a ticket for it for 3.0 units of money and I can win that product with random quality. I want to calculate if buying the ticket will worth its price, but if I calculate the average of all qualities, the price of the finest will have a great impact on that average, however the chance of getting it is pretty low. If I do not count in the highest priced quality item, the value that I will get is somewhat compromised. How can I determine that buying a ticket is really worth it? Does this make sense?

Answer 1

Let's call the universe of possible rewards $R \in \{0.6, 1.1, 1.3, 2.0, 4.2, 21.6\}$. As the cost of a ticket is $3$, your profit is a random variable $P = R - 3$. If the expected value of $P$ is greater than $0$, then, in the long run, it payes to but the ticket. If it less than $0$, then, in the long run, you expect to lose money, and if it is equal to $0$, this is what is called a "fair game."

The possible values of $P$ are $\{-2.4, -1.9, -1.7, -1, 1.2, 18.6\}$. Therefore, as each choice has equal possibility, on average, you expect to make $\$2.133$ and you should by the ticket. However, as you note, you only have a 1/3 chance any one time of winning, it's just that when you win, 1/6 of the time you have a windfall, and when you lose, you never lose that badly (similar to a lottery).

To compare situations like this, what is often used is a measure that relates the risk and the reward, similar to a Sharpe ratio. For example, while the expected profit in this case is $\frac{32}{15}$, the variance of the profit is $\frac{500}{9}$. The profit per unit standard deviation (the reciprocal of the coefficient of variation) is about $0.286$. Compare this to the case where the rewards are instead $R' \in \{0.13333, 2.13333, 4.13333, 7.13333, 8.13333, 9.13333\}$. Subtracting the $3$ cost, the expected profit $P' = R' - 3$ is the same 2.13333. However, the profit per unit standard deviation is now $0.653$, almost $2.3$ times higher. Also, your probability of winning is twice as great, but your maximum windfall is three times lower ($6.13$ vs $18.6$).

Deciding which is a better choice for you enters into discussions of utility theory. Human beings are not always completely rational, nor do they have the same perception of risk (bungee jumping, anyone?). Moreover, depending on someone's financial situation, it may be better to take more risk (desperate straits may make betting all on a windfall situation the only way out of a jam). You, as the purchaser, have to decide what you think is more important (maximal upside, minimal probability of loss, etc.) Knowing the expectation and the distribution allows you to identify the various risk metrics and make that decision intentionally instead of by default.