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?

The term you are looking for is independence.
In probability, independent events are ones where the occurrence of one event does not affect the probability of occurrence of the other.  For example, flipping heads on a coin does not make it more or less likely to roll a $6$ on a die.  We say that these events are independent.
To refute the first answer, we want to rephrase the question to "is the next flip more likely to be Heads or Tails [given the first nine flips were Heads]?"  We can now apply what we know about conditional probability.
