Intuitively understanding the probability that one of the two coins is a head given that there is a tail Suppose that I flip two coins without showing you the result. Now, I tell you that one of the coins is T. The probability that the other coin is H is
$$
\mathbf{P}[HT, TH| \text{at least one } T] = \frac{\mathbf{P}[HT, TH]}{\mathbf{P}\text{[at least one } T]} = \frac{2}{3}
$$
But the thing is, a fair coin should always yield $50\%$ for each flip. I understand that the two coins are no longer independent, but can someone provide me of an intuitive explanation of why this is? 
 A: Just list the possibilities.  Initially you have $HH,HT, TH, TT.$  You have ruled out $HH$, so you have three possibilities left.  In two of them the non-tails coin is heads.
A: There are four possibilities: $HH, HT, TH, TT.$ With the information that one coin is tails, you've removed the first one and are left with $HT, TH, TT.$
Two out of three of these possibilities have the other coin being heads.
A: Let $\Omega = \{H,T\}^2$, $\mathcal F=2^\Omega$, and $\mathbb P$ be uniform on the atoms of $\Omega$. Define $X:\Omega\to\mathbb R$ by 
$$ X(\omega) = 
\begin{cases}
0,& \omega = (T,T)\\
1,& \omega = (T,H)\text{ or } \omega=(H,T)\\
2,& \omega = (H,H).
\end{cases}
$$
That is, $X$ is the number of heads in a given outcome, where the experiment is flipping a fair coin twice. Then 
\begin{align}
\mathbb P(X = 1\mid X\leqslant 1) &= \frac{\mathbb P(X=1, X\leqslant 1)}{\mathbb P(X\leqslant 1)}\\
&=\frac{\mathbb P(X=1)}{\mathbb P(X\leqslant 1)}\\
&= \frac{\mathbb P(\{T,H\}, \{H,T\} ) }{\mathbb P(\{(T,T),(T,H),(H,T)\}}\\
&= \frac{\mathbb P(\{T,H\})+\mathbb P(\{H,T\})}{\mathbb P(\{T,T\})+\mathbb P(\{T,H\})+\mathbb P(\{H,T\})}\\
&= \frac{2\cdot\frac14}{3\cdot\frac14}\\
&=\frac 23.
\end{align}
A: The paradox arises simply because there are two ways to flip one head and one tail (the most likely outcome of flipping two coins) but only one way to flip two tails.
As a generalisation, consider flipping 24 coins and then revealing that there was a tail. What is the probability that at least one other coin was a head? Out of the $2^{24}-1=16777215$ 24-coin flips that contain a tail, only one of them contains 24 tails, leaving the conditional probability that at least one coin is a head as $\frac{16777214}{16777215}$ – in other words, virtually certain.
