First flip given many heads

Condition on the exact head count and note that, given the count, each position is equally likely to be a head.

1. The sequences with $8$, $9$ and $10$ heads number $45$, $10$ and $1$, a total of $56$.
2. Given $k$ heads among $10$ flips, the first flip is heads in a fraction $\frac{k}{10}$ of the sequences.
3. The sequences with first flip heads number $45\times \frac{8}{10}=36$, $10\times \frac{9}{10}=9$ and $1$, a total of $46$.
4. The conditional probability is $\frac{46}{56}$, which reduces to $\frac{23}{28}$.

Answer: $\frac{23}{28}$