Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2004 / Paper 2 / 5

Use the intended model of independent Bernoulli trials, with the same scoring probability $p$ for every shot. Equal marginal scoring probabilities alone would not establish the stated formula without independence. Let $N_1,N_2$ count misses in the two groups of five. They are independent random variables with binomial distribution $\mathrm{Bin}(5,1-p)$. A tie occurs precisely when $N_1=N_2=n$ for one of the mutually exclusive possibilities $n=0,\dots ,5$. Therefore the tie probability for independent binomial counts isAt $p=1/2$, Vandermonde's identity gives ${\sum }_n{\left(\genfrac{}{}{0ex}{}{5}{n}\right)}^2=\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252$, soExchanging the two teams preserves the distribution and interchanges the two winning events. Their total probability is $1-\alpha$, so the first team wins at the end of the initial period with probability $193/512$. In this model the shooting order gives no advantage.

Each additional pair of independent shots resolves the tie with probability $r=2p(1-p)$, since exactly one team must score. It preserves the tie with probability $q=p^2+(1-p{)}^2=1-r$. Finishing after $10+2k+2$ shots means an initial tie, then $k$ tied pairs, then a decisive pair. Independence and geometric continuation after an initial tie giveBecause $0<p<1$, $r>0$ and $q<1$. The total eventual winning probability is the sum of winning in the first ten shots and winning later:Equivalently, conditional on the initial tie, the probability of still being tied after $m$ further pairs is $q^m\to 0$.