A riddle about a liar Say Alice says:'the probability that I'm lying is greater than p.' 

What's the probability that Alice is lying?
It's interesting to try to make sense of this problem.  We'll suppose there's a "real" probability that Alice is lying (either in general, or with respect to the particular statement she is making); call that $q$.  Take a Bayesian approach, and assume a prior distribution on $q$.  (Maybe we'll be lucky, and the answer won't depend on that prior, but, no, we won't be that lucky.)

Let $X$ be the event that Alice claims $q&gt;p$.  We are interested in the probability that Alice is lying given $X$, i.e. $P(q\le p \mid X)$.  Bayes' theorem gives:

$$\begin{align}
 &amp;P(q\le p\mid X) =P(q\le p\,\, \&amp;\,\,  X)/P(X)\\
 &amp;= \frac{P(X\mid q\le p) P(q\le p)}{P(X\mid q\le p) P(q\le p)+P(X\mid q&gt; p) P(q&gt; p)}\\
\end{align}
$$
Now, $P(X\mid q\le p)=q$, since Alice would be lying there, and $P(X\mid q&gt;p)=1-q$ since Alice would be telling the truth there.  Thus, given Alice's statement, the posterior probability that Alice is lying is
$$\frac{q\, P(q\le p)}{q\, P(q\le p)+(1-q)P(q&gt;p)}$$
where the probabilities are our Bayesian priors.

In general this depends on the priors, but in the particular case $p=1$ the probability simplifies to $1$.  Even when $p=0$ this probability may be nonzero, depending on whether our prior has a point mass at $q=0$.
Let $P$(Alice lying$)=q$, then there are $(1-q)$ chance that $q&gt;p$ and $q$ chance that $q\leq p$, so ${1-q\over q}={1-p\over p-0}$ which means $q=p$.

Some intuitive example: if $p=0$ then obiously alice is not lying. If $p=1$ then obviously alice is lying. If $p={1\over2}$ then there is  $1\over2$ chance she is lying.