Two players face off in a combat game with dice. The attacker will roll three dice and the defender, two. The attacker will defeat the defender in just one dice roll if and only if the following two conditions are met:

i) The highest die of the attacker is greater than the highest die of the defender.

ii) The second highest die of the attacker is greater than the second highest die of the defender (we agree that the "second highest die" can be equal to the highest die, in case two or more dice tie for the highest value).

Considering that all dice are fair with equally probable outcomes, calculate the probability of the attacker winning with the defender rolling:

a) 2 fives;

b) 1 five and 1 four.
Solution

a) To win, we need to roll at least two 6s. The probability will be equal to rolling:

- three sixes: $P_{1}=\left(\frac{1}{6}\right)^{3}$; or
- two sixes and another number less than 6: $P_{2}=3 \cdot\left(\frac{1}{6}\right)^{2} \cdot \frac{5}{6}=\frac{15}{6^{3}}$.

Therefore, the probability is $\frac{1}{6^{3}}+\frac{15}{6^{3}}=\frac{16}{6^{3}}=\frac{2}{27}$.
b) To win, we need to roll at least one greater than 5 and another greater than 4. The probability will be equal to rolling:

- three sixes: $P_{1}=\left(\frac{1}{6}\right)^{3}$; and
- two sixes and another any $(<6): P_{2}=3 \cdot\left(\frac{1}{6}\right)^{2} \cdot \frac{5}{6}=\frac{15}{6^{3}}$;
- one six and two fives: $P_{2}=3 \cdot \frac{1}{6} \cdot\left(\frac{1}{6}\right)^{2}=\frac{3}{6^{3}}$; or
- one six, one five, and another any ( $<5$ ): $P_{2}=3!\cdot \frac{1}{6} \cdot \frac{1}{6} \cdot \frac{4}{6}=\frac{24}{6^{3}}$;

Therefore, we have

$$
\frac{1}{6^{3}}+\frac{15}{6^{3}}+\frac{3}{6^{3}}+\frac{24}{6^{3}}=\frac{43}{216}
$$