 # Practice Paper 1 Question 9

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?

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## Warm-up Questions

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 $$\text{H H H}$$?

## Hints

• Hint 1
Let $$a$$ represent the number rolled by $$A$$. What is the probability that $$B$$ wins in terms of $$a$$?
• Hint 2
$$A$$ wins if both of $$B$$'s dice roll are smaller than $$a$$.
• Hint 3
Consider the scenario from the previous hint. What is the probability that $$B$$ does not win for a given $$a$$?
• Hint 4
How may we map the expression from the previous hint to all possible $$a$$?

## Solution

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.

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