If the scores in a cup football match are equal at the end of the normal
period of play, a 'penalty shoot-out' is held in which each side takes up to
five shots (from the penalty spot) alternately, the shoot-out being stopped if
one side acquires an unassailable lead (i.e. has a lead greater than its
opponents have shots remaining). If the scores are still level after the
shoot-out a 'sudden death' competition takes place.
In sudden death each side takes one shot and the competition is over if one
side scores and the other does not; if both score, or both fail to score, a
further shot is taken by each side, and so on. Team 1, which takes the first
penalty, has a probability \(p_{1}\), which is independent of the player
involved, of scoring and a probability \(q_{1}\left(=1-p_{1}\right)\) of
missing; \(p_{2}\) and \(q_{2}\) are defined likewise.
Define \(\operatorname{Pr}(i: x, y)\) as the probability that team \(i\) has
scored \(x\) goals after \(y\) attempts, and let \(f(M)\) be the probability that
the shoot-out terminates after a total of \(M\) shots.
(a) Prove that the probability that 'sudden death' will be needed is
$$
f(11+)=\sum_{r=0}^{5}\left({ }^{5} C_{r}\right)^{2}\left(p_{1}
p_{2}\right)^{r}\left(q_{1} q_{2}\right)^{5-r}
$$
(b) Give reasoned arguments (preferably without first looking at the
expressions involved) which show that
$$
f(M=2 N)=\sum_{r=0}^{2 N-6}\left\\{\begin{array}{l}
p_{2} \operatorname{Pr}(1: r, N) \operatorname{Pr}(2: 5-N+r, N-1) \\
+q_{2} \operatorname{Pr}(1: 6-N+r, N) \operatorname{Pr}(2: r, N-1)
\end{array}\right\\} for \(N=3,4\)
(c) Give an explicit expression for \(\operatorname{Pr}(i: x, y)\) and hence
show that if the teams are so well matched that \(p_{1}=p_{2}=1 / 2\) then
$$
\begin{aligned}
f(2 N) &=\sum_{r=0}^{2 N-6}\left(\frac{1}{2^{2 N}}\right) \frac{N !(N-1) !
6}{r !(N-r) !(6-N+r) !(2 N-6-r) !} \\
f(2 N+1) &=\sum_{r=0}^{2 N-5}\left(\frac{1}{2^{2 N}}\right) \frac{(N !)^{2}}{r
!(N-r) !(5-N+r) !(2 N-5-r) !}
\end{aligned}
$$
(d) Evaluate these expressions to show that, expressing \(f(M)\) in units of
\(2^{-8}\), we have
$$
\begin{array}{lllllll}
M & 6 & 7 & 8 & 9 & 10 & 11+ \\
f(M) & 8 & 24 & 42 & 56 & 63 & 63
\end{array}
$$
Give a simple explanation of why \(f(10)=f(11+)\).
$$
for \(N=3,4,5\) and
$$
f(M=2 N+1)=\sum_{r=0}^{2 N-5}\left\\{\begin{array}{l}
p_{1} \operatorname{Pr}(1: 5-N+r, N) \operatorname{Pr}(2: r, N) \\
+q_{1} \operatorname{Pr}(1: r, N) \operatorname{Pr}(2: 5-N+r, N)
\end{array}\right\\}
$$
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