Assume that the coin is fair and is tossed an infinitely number of times, independently.
In the original version of this question, the former event is "9 out of the first 20 coin tosses are heads", and given that fact, the probability of the latter event is indeed lower. I thought it was because the 9 heads, being distributed amongst the 20 coin tosses, would make it less likely for a large portion of them, namely 7, to be present in the first 10 tosses.
If so, I think it'd make sense that knowing that 14 out of the first 20 coin tosses are heads would have no effect on the probability of the desired event. But the Bayes' theorem doesn't seem to say so. Below is my working out in details of the problem:
Let $Y$ be the event that exactly 7 out of the first 10 tosses are heads, and $Z$ the event that exactly 14 out of the first 20 tosses are heads. Then: $$P(Y) = {10 \choose 7} \frac{1}{2^{10}} \approx 0.117$$ $$P(Z) = {20 \choose 14} \frac{1}{2^{20}}$$ $$\text{and } P(Z|Y) = {10 \choose 7} \frac{1}{2^{10}} \text{, }$$ as given that exactly 7 out of the first 10 tosses are heads, $P(Z|Y)$ is just the probability that exactly 7 out of the last 10 tosses are heads. Then: $$P(Y|Z) = \frac{P(Y\cap Z)}{P(Z)} = \frac{P(Z|Y)P(Y)}{P(Z)} = \frac{{10 \choose 7}{10 \choose 7}}{20 \choose 14} \approx 0.372 \neq P(Y)$$
Where did my intuition/calculation go wrong?