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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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