Conditional Probability for Coin Flip Data

Question

I am reading the Tutorial of Fisher Information and it is mentioned that the say we have a random variable $X^n$ where $n$ refers to number of tosses so the RV $X^2$ could be {$1,1$} or {$0,1$} then the number of head given $\theta$ is given by the following binomial distribution formula, here $\theta$ determines coint is fair or biased:

$$p(y|\theta)=\begin{pmatrix}n\\y \end{pmatrix}\theta^y(1-\theta)^{n-y}\ , \textrm{where y is amount of heads}$$ $$ \begin{pmatrix}n\\y \end{pmatrix}=\dfrac{n!}{y!(n-y)!}$$

Then it mentions

Observe that the conditional probability of the raw data given Y = y is equal to $$P(X^n|Y=y,\theta)={1\over{\mathbin{}\begin{pmatrix}n\\y \end{pmatrix}}}$$

Now what I dont understand why is the above conditional probability of raw data independent of $\theta$?

If theta was different then we would have seen different combination of data, say $\theta$ is 1, (Biased towards heads) we would observe Y=1,1,1,1 for 4 tosses.

Answer 1

Intuitively, if you know that there are $y$ heads in $n$ tosses, no arrangement of 1s and 0s are superior to another, so each of the arrangements has the same probability, e.g. $$P(X^3=(1,1,0))=P(X^3=(1,0,1)=P(X^3=(0,1,1)))$$

which makes $$P(X^n|Y=y,\theta)=\frac{1}{n\choose y}$$ since there are $n \choose y$ equally likely situations. We can also find the same result using Bayes Rule, but intuition is simpler I suppose.