Flip a coin 6 times. What is probability of at least 4 heads? I can figure out the much simpler case of the probability of getting at least 2 heads in 3 coin flips: There are 8 (2^3) ways to flip a coin 3 times: HHH, HHT, TTT, TTH, HTH, HTT, THT, THH. 4 of these contain 2 or more heads. Therefor the probability of at least 2 heads in 3 coin flips is 4/8. How could I have done this without writing out all the possibilities and counting the ones with 2 or 3 H's?
Today the Indiana Pacers won the first game in the 4 out of 7 series with the Miami Heat. Assuming that either team has a 50% chance of winning each of the remaining games, what is the probability of the Heat winning 4 of the remaining games?
 A: 
Today the Indiana Pacers won the first game in the 4 out of 7 series with the Miami Heat. Assuming that either team has a 50% chance of winning each of the remaining games, what is the probability of the Heat winning 4 of the remaining games?

There are $6$ remaining games.  The desired criteria is that Heat wins at least $4$, when given that Heat lost the first 1.  This is a binomial distribution; so named because of the use of the binomial coefficient to count number of permutations of outcomes that match the desired criteria.
The probability of exactly $k$ successes in $n$ trials with probability $p$ of success in any trial, is: $${n\choose k}p^k(1-p)^{n-k} \;=\; \,^n\mathrm{\large C}_k\;p^k(1-p)^{n-k} \;=\; \frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}$$
So: $\mathbb{\large P}(\text{win at least }4\text{ more of }6) = {6\choose 4}\left(\frac 12\right)^4\left(\frac 12\right)^2+{6\choose 5}\left(\frac 12\right)^5\left(\frac 12\right)^1+{6\choose 6}\left(\frac 12\right)^6\left(\frac 12\right)^0$.
$$\therefore \mathbb{\large P}(\text{win at least }5\text{ more of }6) = \frac 1{2^6}\left(\frac{6!}{4!2!}+\frac{6!}{5!1!}+\frac{6!}{6!0!}\right)$$
A: You rightly suspect that there may be some simpler, formulaic method to find these combinations.
It's given by:
$$\dfrac{n!}{k!(n-k)!}$$
Where $!$ denotes the factorial ($n!=n\cdot(n-1)\cdot(n-2)\cdot ...\cdot 2\cdot1$), $n$ is the 'number of things', and $k$ is 'how many things are chosen'.
So in this specific case, $n=6, k=4$. That gives us the number of ways we can get 4 heads from 6 flips.
You can probably see how to extend this to your sports example, and lose the restriction that their odds are only as good as a coin-flip.
