Average of a game of dice best of three

Question

In a game of dice, I throw a 20 sided dice (values from 1 to 20) three times. My result is the best of three.

For example, if I get the values 4, 16, and 7, my result is 16. If I get the values 14, 1, and 3, my result is 14.

What is the average result (or "payoff" if I understand the notation correctly) of the game?

An explanation would be appreciated.

Answer 1

The probability that $n$ is the largest of three rolls is the probability that each roll doesn't contain numbers greater than $n$ multiplied by the probability $n$ has been selected at least once given there are no larger numbers:

$$P(\text{max}(\{N_1,N_2,N_3\}) \leq n) =P\left(\left(N_1\leq n\right) \land \left(N_2\leq n\right) \land \left(N_3\leq n\right)\right) \times P\left(\left(N_1 = n\right) \lor \left(N_2 = n\right) \lor \left(N_3 = n\right) |\left(N_1\leq n\right) \land \left(N_2\leq n\right) \land \left(N_3\leq n\right)\right)$$

$$\begin{align}P_1&=P\left(\left(N_1\leq n\right) \land \left(N_2\leq n\right) \land \left(N_3\leq n\right)\right) \\ &= \left(\frac{n}{20}\right)^3\end{align}$$

$$\begin{align} P_2 &= P\left(\left(N_1 = n\right) \lor \left(N_2 = n\right) \lor \left(N_3 = n\right) |\left(N_1\leq n\right) \land \left(N_2\leq n\right) \land \left(N_3\leq n\right)\right)\\ &= 1 - \left(\frac{n-1}{n}\right)^3 \end{align}$$

so for dice with $m$ sides,

$$\begin{align}E(\text{max}(\{N_1,N_2,N_3\}) &= \sum_{x=1}^{m}x\times\left(\frac{x}{m}\right)^3\left(1 - \left(\frac{x-1}{x}\right)^3\right)\\ &= \frac{(m+1) ( 3 m -1)}{4 m}, (\text{according to WolframAlpha})\\ \end{align}$$ for $m=20$, $$\begin{align}E(\text{max}(\{N_1,N_2,N_3\})&=\frac{1239}{80}\\ &=15.4875\end{align}$$

Disclaimer: I'm not 100% sure if this is correct, perhaps someone could correct me if it isn't.

Answer 2

A slightly different way to solve, but essentially the same result:

Using the Layer Cake theorem, $E[X] = \sum_{x=1}^{max(X)} P(X\geq x)$. Thus, we need to find $P(X \geq x)$. By the Principle of Inclusion and Exclusion, this is: $$\begin{align}P(X \geq x) &= P(X_1\geq x) + P(X_2\geq x) + P(X_3\geq x) \\&\quad- P(X_1\geq x)P(X_2\geq x) - P(X_1\geq x)P(X_3\geq x)-P(X_2\geq x)P(X_3\geq x) \\&\quad +P(X_1\geq x)P(X_2\geq x)P(X_3\geq x)\\ &= 3*\frac{21-x}{20}-3*\left(\frac{21-x}{20}\right)^2+\left(\frac{21-x}{20}\right)^3\\ &=3*\frac{21-x}{20}*\left(1-\frac{21-x}{20}\right)+\left(\frac{21-x}{20}\right)^3\\ &=3*\frac{21-x}{20}*\left(\frac{x-1}{20}\right)+\left(\frac{21-x}{20}\right)^3\end{align}$$

, where $X_i$ is the $i^{th}$ throw. Thus $E[X] = \sum_{x=1}^{20}3*\frac{21-x}{20}*\left(\frac{x-1}{20}\right)+\left(\frac{21-x}{20}\right)^3 = 15.4875$