Recalibration of probability estimate when considering multiple future outcomes We are given a coin of unknown biasing, and allowed to perform experiments with it. Say we have the following series of outcomes of 20 Bernoulli trials, represented by $X_i, \  i \in \mathbb{N}$ :
$$
  T, T, T, H, H, H, H, H, H, H, T, H, H, T, T, T, H, H, H, T
$$
There are $12 \  H \equiv H_{12}$ and $8 \  T \equiv T_8$. 
While I don't have the complete derivation, I know that (and have cross-checked my intermediate form),
$$
  P\left( X_{21} = T \right) = \frac{8 + 1}{12+8+2} = \frac{9}{22}
$$
Then what is the probability that $P\left( \left( X_{21} = T \right) \cap \left( X_{22} = T \right) \right) = P(S)$ (suppose) ?
Clearly 
$$
  P(S) = P\left( X_{21} = T \right) \cdot P \left( X_{22} = T \right) = \frac{9}{22} \cdot P \left( X_{22} = T \right)
$$
But what is $P \left( X_{22} = T \right)$? Is it the same as $P\left( X_{21} = T \right)$, as $X_{21}$ hasn't actually happened? Or do we re-calibrate given the new outcome of $T$?
So is 
$$
  P(S) = {\left( \frac{9}{22} \right)}^2 = \frac{81}{484}
  \text{ or }
  P(S) = \frac{9}{22} \cdot \frac{10}{23}= \frac{90}{506}
$$
 A: $P(X_{21}=T)=\frac9{22}$ results from applying the rule of succession. At How to prove the rule of succession without calculus? I gave a combinatorial proof that doesn’t require integration. We can apply the same idea to derive $P(X_{21}=T\cap X_{22}=T)$.
Assuming a flat prior for the bias of the coin, we draw one number uniformly randomly from $[0,1]$ for the probability $p$ that a toss yields heads and then $22$ further numbers $u_i$ for the $22$ tosses, with toss $i$ showing heads exactly if $u_i\lt p$. The fact that the first $20$ tosses yielded $12$ heads and $8$ tails means that the rank of $p$ among $p$ and the first $20$ $u_i$ is $9$. If we now draw the next number for the $21$st toss, it’s equally likely to be inserted into the order of those $21$ numbers in any of the $22$ possible slots, $9$ of which are above $p$, whence $P(X_{21}=T)=\frac9{22}$. If it’s inserted above $p$ and we insert another one, there are now $23$ slots and $10$ of them are above $p$, so the probability that we get yet another tails is $P(X_{22}=T\mid X_{21}=T)=\frac{10}{23}$, and thus
$$
P(X_{21}=T\cap X_{22}=T)=P(X_{21}=T)P(X_{22}=T\mid X_{21}=T)=\frac9{22}\cdot\frac{10}{23}=\frac{45}{253}\;.
$$
The equation that follows “Clearly” in the question is wrong; the events $X_{21}=T$ and $X_{22}=T$ aren’t independent, so you can’t just multiply their probabilities. 
