Player $A$ rolls one die. Player $B$ rolls two dice. If $A$ rolls a number greater or equal to the largest number rolled by $B$, then $A$ wins, otherwise $B$ wins. What is the probability that B wins?
1. You roll a die 3 times. What is the probability that at least one roll is greater than 2?
2. You roll two dice. What is the probability that their sum is less than 7?
3. A coin is flipped 3 times, displaying either heads ($H$) or tails ($T$). What is the probability that you do not get $\text{H H H}$?

Let $a$ represent the number rolled by $A$. What is the probability that $B$ wins in terms of $a$?

$A$ wins if both of $B$'s dice roll are smaller than $a$.

Consider the scenario from the previous hint. What is the probability that $B$ does not win for a given $a$?

How may we map the expression from the previous hint to all possible $a$?

Let $a \in \{1,2,3,4,5,6\}$ represent the number rolled by $A$. $B$ wins if at least one of $B$'s rolls is higher than $a$. This would be $1$ minus the probability that both of $B$'s rolls are smaller than or equal to $A$'s.

For a given $a$, there exists $a$ numbers smaller or equal to $a$ that may be rolled. Therefore, for a given $a$ the probability of $B$ winning is $1-(\frac{a}{6})^2$. Now, consider all possible values of $a$. The probability of any value of $a \in \{1,2,3,4,5,6\}$ being rolled is $\frac{1}{6}.$ So, the overall probability that $B$ wins is $\sum_{a=1}^{6} \frac{1}{6}\big(1-(\frac{a}{6})^2\big) = \frac{125}{216}.$

See properties of summations here to aid in evaluating the sum.