= Solution
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> $\operatorname{Bin}(5,1-p)$. A tie occurs precisely when $N_1=N_2=n$ for one of the mutually exclusive possibilities $n=0,\ldots,5$. Therefore the <tie probability for independent binomial counts> is
$$
\boxed{\alpha=\sum_{n=0}^5\binom5n^2(1-p)^{2n}p^{10-2n}.}
$$
At $p=1/2$, <Vandermonde's identity> gives $\sum_n\binom5n^2=\binom{10}5=252$, so
$$
\boxed{\alpha=\frac{252}{1024}=\frac{63}{256}.}
$$
Exchanging the two teams preserves the distribution and interchanges the two winning events. Their total probability is $1-\alpha$, so \b[the first team wins at the end of the initial period with probability $193/512$]. In this model the shooting order gives no advantage.
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