Let $X~Bin(10,0.5)$ where $X$ is the number of heads observed in $10$ trials. Given that $X\ge9$ that is I know I will observe 9 heads or more in 10 tosses. I get $P(X=9|X\ge9) = 10/11$, While $P(X=10|X\ge9) = 1/11$. Using Bayes Rule. I know this is correct but I cant understand this intuitively. Here is what is bothering me
1) I am doing $10$ iid trials, $10$ coin flips
2) I know that I got at least $9$ heads.
3) Given this information, why cant I just conduct one additional trial and determine if I will get $10$ heads or $9$, that is $X=10$ or $X=9$.
$P(Head)=0.5$ and $P(Tail)=0.5$ So,
$$P(X=9|X\ge9) = 0.5$$ $$P(X=9|X\ge9) = 0.5$$
Because what you describe is the probabilities under the condition that: the first nine trials were all heads.
However, what you want is the probabilities under the condition that: at least nine of the ten trials are all heads.
They are quite different conditions.
Let's simplify the example to 3 trials and the condition that at least 2 are heads. The list of (equally likely) possible outcomes is:
$$\rm \overbrace{\underbrace{HHH, HHT,}_{\text{first 2 are heads}} HTH, THH}^{\text{at least 2 are heads}}$$