Where is the gambler's fallacy in this coin flipping binomial distribution? Suppose we have a biased coin where:
$p(Heads) = 0.6$
$p(Tails) = 0.4$
If $X$ is the number of heads obtained in 10 flips, the binomial distribution says:
$p(X = 9) = 0.04$
$p(X = 10) = 0.006$
Suppose we are now flipping this biased coin, and we have flipped it 9 times so far. On all 9 flips, the coin landed on heads. Is heads or tails more likely on the next flip?
Answer 1: Since $p(X = 9) > p(X = 10)$, $X = 9$ is the more likely outcome. Therefore, the next flip is more likely to be tails.
Answer 2: Since $p(Heads) > p(Tails)$, the next flip is more likely to be heads.
I think the first answer is wrong because it looks like the gambler's fallacy, but I can't explain it in mathematical terms. Can someone explain how the reasoning in the first answer is faulty? How do I refute the reasoning given in the first answer?

Independence is of course important, but there is also dependence here that one should make note of. Once you know the result of the first $9$ coin flips, your probability distribution for $X$ gets skewed. For example, the likeliest value for $X$ is $6$, given no prior information. But once you see that the first 9 coins have been heads, you cannot still possibly hope that $X$ will be $6$ at the end. Similarly, even though $X=9$ is likelier than $X=10$, given no prior information, if you know that the first $9$ coins landed heads, the probability that $X=9$ and $X=10$ becomes equal.
