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2015 Paper 2 Q12
D: 1600.0 B: 1500.0
probability conditional probability tree diagram sequences Markov chain recurrence relation coin tossing Penney's game

Four players \(A\), \(B\), \(C\) and \(D\) play a coin-tossing game with a fair coin. Each player chooses a sequence of heads and tails, as follows: Player A: HHT; Player B: THH; Player C: TTH; Player D: HTT. The coin is then tossed until one of these sequences occurs, in which case the corresponding player is the winner.

  1. Show that, if only \(A\) and \(B\) play, then \(A\) has a probability of \(\frac14\) of winning.
  2. If all four players play together, find the probabilities of each one winning.
  3. Only \(B\) and \(C\) play. What is the probability of \(C\) winning if the first two tosses are TT? Let the probabilities of \(C\) winning if the first two tosses are HT, TH and HH be \(p\), \(q\) and \(r\), respectively. Show that \(p=\frac12 +\frac12q\). Find the probability that \(C\) wins.

Solution:

  1. The only way \(A\) can win is if the sequence starts HH, if it does not start like this, then the only way HHT can appear is after a sequence of THH...H, but then THH has already appeared and \(B\) has won. Therefore the probability is \(\frac14\)
  2. If HH appears before TT then either \(A\) or \(B\) will win. If HH appears first, then \(A\) has a \(\frac14\) probability of winning. So \(A\): \(\frac18\), \(B:\), \(\frac38\), \(C:\), \(\frac18\), \(D: \frac38\)
  3. If the first two tosses are TT then \(C\) will win. If the first two tosses are HT, then either the next toss is T and \(C\) wins, or the next toss is H, and it's as if we started TH. ie \(p = \frac12 + \frac12 q\). If the first two tosses are TH, then either the next toss is H and \(C\) losses or the next toss is T and it's like starting HT. So \(q = \frac12 p\). Therefore \(p = \frac12 + \frac14p \Rightarrow p = \frac13\) If the first two tosses are HH, then eventually a T appears, and it's the same as starting HT. Therefore the probability \(C\) wins is: \(\frac14 + \frac14 \cdot \frac13 + \frac14 \cdot \frac16 + \frac14 \cdot \frac13 = \frac{11}{24}\)

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