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.

You don't understand the question.

It is: When given the counts of heads and tails resulting from the flips, what is the probability that the order of the results has heads in the first three places?

Notice: We do not have to worry about the probability of any of the flips resulting in heads or tails. The coin does not even need to be fair; as long as the same one used each time (the flips have identical and independent distributions), bias has no impact on this question.

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?

An equivalent problem is: When 5 red and 5 black cards are fairly shuffled, what is the probability that the first three will be red?

There are $\binom{5}{3}$ (equiprobable) ways to select three of the five red cards out of $\binom{10}{3}$ ways to select any three of all ten cards.

$$\frac{\dbinom{5}{3}}{\dbinom{10}{3}}=\cfrac{\;\cfrac{5!}{3!2! }\;}{\;\cfrac{10!}{3!7!}\;}=\dfrac{5! \; 7!}{2! \; 10!} =\frac{1}{12}$$

Alternatively: there are $\binom{7}{2}$ ways to order the cards/coins such that the first three are red/head, out of $\binom{10}{5}$ ways to order them in total. Divide and calculate to obtain the same result.

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.

There are $\frac{10!}{5!5!}$ ways of getting an equal number of heads and tails. The first three are fixed at heads so we need 2 more heads and 5 tails in the remaining seven tosses this can be done in $\frac{7!}{2!5!}$ ways. The probability is therefore $$ {\frac{7!}{2!5!}\over\frac{10!}{5!5!}} = \frac{1}{12} $$