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2011 Paper 2 Q12
D: 1600.0 B: 1484.0

Xavier and Younis are playing a match. The match consists of a series of games and each game consists of three points. Xavier has probability \(p\) and Younis has probability \(1-p\) of winning the first point of any game. In the second and third points of each game, the player who won the previous point has probability \(p\) and the player who lost the previous point has probability \(1-p\) of winning the point. If a player wins two consecutive points in a single game, the match ends and that player has won; otherwise the match continues with another game.

  1. Let \(w\) be the probability that Younis wins the match. Show that, for \(p\ne0\), \[ w = \frac{1-p^2}{2-p}. \] Show that \(w>\frac12\) if \(p<\frac12\), and \(w<\frac12\) if \(p>\frac12\). Does \(w\) increase whenever \(p\) decreases?
  2. If Xavier wins the match, Younis gives him \(\pounds1\); if Younis wins the match, Xavier gives him \(\pounds k\). Find the value of \(k\) for which the game is `fair' in the case when \(p =\frac23\).
  3. What happens when \(p = 0\)?


Solution:

  1. We can be in several states.
    1. No points played
    2. Y just won the last point
    3. X just won the last point
    4. Y won the game
    5. X won the game
    The probability \(Y\) wins from any of these states are: \begin{align*} &&P_{-} &= p P_X + (1-p) P_Y &= w \\ &&P_Y &= p + (1-p)P_X \\ &&P_X &= (1-p)P_Y \\ \\ \Rightarrow &&& \begin{cases} P_Y - (1-p)P_X &= p \\ (1-p)P_Y -P_X &= 0\end{cases} \\ \Rightarrow && P_Y &= \frac{1}{1-(1-p)^2} \cdot p \\ &&&= \frac{1}{2-p} \\ && P_X &= \frac{1-p}{2-p} \\ && w &= \frac{p(1-p) + (1-p)}{2-p} \\ &&&= \frac{1-p^2}{2-p} \end{align*}
  2. If \(p = \frac23\) then \(w = \frac{1-\frac49}{2-\frac23} = \frac{5}{12}\). The game is fair if \(\mathbb{E}(result) = 0\), ie \(\frac{5}{12} \cdot k - \frac{7}{12} 1 \Rightarrow k = \frac{7}{5} = 1.4\)
  3. If \(p = 0\) then they will keep playing forever, since no-one can win two points in a row.