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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.