Dice With Same Numbers

Let's calculate the expected return per outcome. Let's define the following:

• $E_n$ is the expected return when $n$ dice are the same.
• $P_n$ is the probability that $n$ dice are the same.

Let's start with the case where 3 dice are the same, where you earn €16. The first die can be any number, but the other two dice need to be the same number. So, \begin{equation} P_3 = 1 * \frac{1}{6} * \frac{1}{6} = \frac{1}{36} \end{equation} \begin{equation} E_3 = P_3 * 16 = \frac{16}{36} \end{equation} Next, consider the case where 2 dice are the same, where you earn €14. The first die can be any number, the second die needs to be the same, and the third die needs to be a different number. You can order this in $\binom{3}{2} = 3$ ways. So, \begin{equation} P_2 = 1 * \frac{1}{6} * \frac{5}{6} * \binom{3}{2}  = \frac{5}{12} \end{equation} \begin{equation} E_2 = P_2 * 14 = \frac{210}{36} \end{equation} Finally, consider the case where no dice are the same, where you lose €4. The first die can be any number, the second die needs to be another number and the third die needs to be another number again. So, \begin{equation} P_0 = 1 * \frac{5}{6} * \frac{4}{6}  = \frac{5}{9} \end{equation} \begin{equation} E_0 = P_0 * -4 = - \frac{80}{36} \end{equation} The expected return of this game is defined as: \begin{equation} E = E_3 + E_2 + E_0 \end{equation} \begin{equation} E = \frac{16}{36} + \frac{210}{36} - \frac{80}{36} \end{equation} \begin{equation} E = \frac{146}{36} = \frac{73}{18} \approx 4.06 \end{equation}