In a game for two players, a fair coin is tossed repeatedly. Each player
is assigned a sequence of heads and tails and the player whose sequence appears first wins.
Four players, \(A\), \(B\), \(C\) and \(D\) take turns to play the game.
Each time they play, \(A\) is assigned the sequence
TTH (i.e.~Tail then Tail then Head), \(B\) is assigned THH,
\(C\) is assigned HHT and \(D\) is assigned~HTT.
- \(A\) and \(B\) play the game.
Let
\(p_{\mathstrut\mbox{\tiny HH}}\),
\(p_{\mathstrut\mbox{\tiny HT}}\),
\(p_{\mathstrut\mbox{\tiny TH}}\) and
\(p_{\mathstrut\mbox{\tiny TT}}\)
be the probabilities of \(A\) winning the
game given that the first two tosses of the coin show HH, HT, TH and TT, respectively.
Explain why
\(p_{\mathstrut\mbox{\tiny TT}} = 1\,\),
and why
$p_{\mathstrut\mbox{\tiny HT}} = {1 \over 2} \,
p_{\mathstrut\mbox{\tiny TH}} +
{1\over 2} \,
p_{\mathstrut\mbox{\tiny TT}}\,$.
Show that
$p_{\mathstrut\mbox{\tiny HH}} =
p_{\mathstrut\mbox{\tiny HT}}
= {2 \over 3}$
and that
\(p_{\mathstrut\mbox{\tiny TH}} = {1\over 3}\,\).
Deduce that the probability that A wins the game is \({2\over 3}\,\).
- \(B\) and \(C\) play the game.
Find the probability that \(B\) wins.
- Show that if \(C\) plays \(D\), then \(C\) is more likely to win than \(D\),
but that if \(D\) plays \(A\), then \(D\) is more likely to win than \(A\).