A coin is flipped ten times. What is the probability that the first three are heads if an equal number of heads and tails are flipped? I understand the question but I am not sure how to solve it. For example, if we flip HHHTTTTT then the next three must be heads because of the question. This however seems counterintuitive. I believe that there are $2^{10}$ possible strings, but I am unsure of how to count all possible strings that begin with HHH.

We do a formal conditional probability calculation.
Let $A$ be the event the first $3$ tosses are heads, and let $B$ be the event we have an equal number of heads and tails in the $10$ tosses. We want $\Pr(A|B)$. By the definition of conditional probability, we have
$$\Pr(A|B)=\frac{\Pr(A\cap B)}{\Pr(B)}.$$
We calculate the two probabilities on the right. 
First we calculate $\Pr(B)$. The probability of $5$ heads and $5$ tails in $10$ tosses is $\frac{\binom{10}{5}}{2^{10}}$.
Next we calculate $\Pr(A\cap B)$. The probability the first $3$ tosses are heads is $\frac{1}{2^3}$. Given that the first $3$ tosses were heads, the probability of $5$ heads and $5$ tails is the probability of $2$ heads in the last $7$ tosses. This is $\frac{\binom{7}{2}}{2^7}$. It follows that $\Pr(A\cap B)=\frac{\binom{7}{2}}{2^{10}}$. 
Finally, divide. 
