"At least" type probability question. Recently, I asked a question: Team A has more Points than team B
Though I ultimately got the right answer, it took extreme casework, and long computations. My question is: suppose the question was restated:

Team A plays $5$ matches with Team $B, C, D, E, F$. Every team has probability $\frac{1}{2}$ of any match it plays. What is the probability that Team A wins at least two of the matches it plays?

It would take casework like:
$$\frac{\binom{5}{2} + \binom{5}{3} + \binom{5}{4} + \binom{5}{5}}{1024}$$

Is there another, easier/efficient way to do this?

 A: I'm not sure what generalizations of your problem are of interest,
whether you are willing to use normal approximations, or whether
you envision using software. Here are several approaches, along
with brief explanations, and computations in R. 
The availability
of software, where appropriate, tends to change one's view of
what computations are easy what computations are too tedious
to contemplate. With several styles of computation on display
below, I will leave it to you to decide which methods are
helpful for whatever models you may have in mind.
The first method works only if Team A has probability $p = 1/2$
of winning. The other methods can be generalized to any $p$ between
0 and 1.
1) Use binomial coefficients as in your question.
You want either 
$$\left[{5 \choose 2}+{5 \choose 3}+{5 \choose 4}+{5 \choose 5}\right]/2^5
= [10 + 10 + 5 + 1]/2^5 = 26/32 = 0.8125$$
or
$$ 1 - \left[{5 \choose 0}+{5 \choose 2}\right]/2^5 = 1 - [1 + 5]/32 = 0.8125.$$
In R, the first of these could be computed as
 i = 2:5;  sum(choose(5,i))/2^5
 ## 0.8125

(2) Use the probability function (PDF or PMF) of $Binom(5, 1/2).$
$$\sum_{i=2}^5 {5 \choose i} (.5)^i (1-.5)^{5-i} = 
\sum_{i=2}^5 {r \choose i}(.5)^5 = 0.8125.$$
In R this is
 sum(dbinom(2:5, 5, .5))  # 'dbinom' is the binomial PMF
 ## 0.8125

(3) Use the cumulative distribution function (CDF) $F$ of $X \sim Binom(5, 1/2).$ Many books from the pre-computer age have extensive tables,
and a few still include them.
$$ P(X \ge 2) = 1 - P(X \le 1) = 1 - F(1; 5, .5) = 0.8125.$$
In R this is
 1 - pbinom(1, 5, .5)  # 'pbinom' is the binomial CDF
 ## 0.8125

(3) Approximate using a normal distribution with mean $\mu = 5(.5) = 2.5$ and standard deviation $\sigma = \sqrt{5(.5)(1-.5)}.$ Using
the continuity correction, this becomes
$$ P(X \ge 2) = 1 - P(X \le 1) = 1 - P(X \le 1.5)
= 1 - P(Z \le (1.5 - \mu)/\sigma)\\ = 1 - \Phi(-0.8944) \approx 0.8145,$$
where $Z$ is a standard normal random variable, $\Phi$ is its CDF,
and the answer is obtained by numerical integration or use of
printed tables of $\Phi.$ This particular example is right at
the edge of most rules of thumb for use of the normal approximation.
Yet the answer is accurate to two decimal places, which is about
all one can expect from normal approximations to binomial in general.
In R, it is not necessary to do an explicit 'standardization' because
 CDFs are available for all members of the normal family.
 1-pnorm((1.5 - 5/2)/sqrt(5/4))   # CDF of NORM(0,1)
 ## 0.8144533
 1 - pnorm(1.5, 5/2, sqrt(5/4))   # CDF with specified mean and SD
 ## 0.8144533

