A riddle about a liar

Question

Say Alice says:'the probability that I'm lying is greater than p.'

What's the probability that Alice is lying?

Answer 1

Let $P$(Alice lying$)=q$, then there are $(1-q)$ chance that $q>p$ 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.

Answer 2

Below described is a situation in which it makes sense that Alice claims that she lies with probability greater than $p$. The following story requires, however, a sequence of experiments. (BTW interpreting a probability related question in reality always require a series of experiments.)

Alice generates independent random numbers between zero and one with uniform distribution. Let these random variables be denoted by $p_1,p_2,...p_n,...$

The $n^{th}$ experiment is as follows

Alice says: "I lie with probability greater than $p_n$."

The probability that $p_n\ge 0.5$ is $0.5$. If $p_n>0.5$ then Alice lies. So, there is no contardiction: Alice lies with probability $0.5$ and what she lies, time to time, is that she lies with a greater probability than $0.5$ while the truth is that she lies only with probability $0.5.$

I then answered the following question:

Let $p$ be a random variable uniformly distributed over $[0,1]$. Alice says: "The probability that I'm lying is greater than $p$." What is the probability that she lies?

Or was this not the question?

Answer 3

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 $q>p$. We are interested in the probability that Alice is lying given $X$, i.e. $P(q\le p \mid X)$. Bayes' theorem gives:

$$\begin{align} &P(q\le p\mid X) =P(q\le p\,\, \&\,\, X)/P(X)\\ &= \frac{P(X\mid q\le p) P(q\le p)}{P(X\mid q\le p) P(q\le p)+P(X\mid q> p) P(q> p)}\\ \end{align} $$ Now, $P(X\mid q\le p)=q$, since Alice would be lying there, and $P(X\mid q>p)=1-q$ since Alice would be telling the truth there. Thus, given Alice's statement, the posterior probability that Alice is lying is $$\frac{q\, P(q\le p)}{q\, P(q\le p)+(1-q)P(q>p)}$$ 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 $q=0$.

Answer 4

Assume that Alice has an a priori lying probability $x\in[0,1]$. Confronted with a value $p\in[0,1]$ Alice could say "$x<p$" or "$x>p$". The probability that she says st: "$x>p"$ computes to $$P[{\rm st}]=P[x<p]\> x + P[x>p]\>(1-x)\ ,$$ and the probability that this statement is actually false is just $$P[{\rm st}\wedge{\rm false}]=P[x<p]\> x\ .$$ It follows that the a posteriori probability that the statement we heard is false is given by $$P[{\rm false}\>|\>{\rm st}]={P[{\rm st}\wedge{\rm false}]\over P[{\rm st}]}={P[x<p]\> x\over P[x<p]\> x + P[x>p]\>(1-x)}\ .\tag{1}$$ We now make the essential assumption that Alice had choosen her lying probability $x$ uniformly distributed in $[0,1]$. Then $(1)$ becomes $$P[{\rm false}\>|\>{\rm st}]={px\over 1-p+(2p-1)x}\ .$$ Integrating this over $0\leq x\leq1$ we finally obtain $$P[{\rm Alice\ was\ lying\ just\ now}]=\ldots\ ,$$ a complicated expression in terms of $p$ and $\log p$, whereby the cases $p<{1\over2}$, $p={1\over2}$, an $p>{1\over2}$ have to be distinguished.

Answer 5

I presume "I'm lying" refers to this particular statement she's making, rather than her overall probability of lying on every statement she ever makes. Now one problem is that there really is no such thing as "the probability" that this statement is a lie: in order to define probabilities you must set up a probability model, and no such model has been specified. But you also need to have a consistent notion of "lying", and in the presence of self-referential statements that is problematic.

I'll assume $0 < p < 1$.

From one point of view, a statement of fact that doesn't refer to the future or anything else that is unknown is either true or false: there is no probability involved. So the actual "probability" $q$ that her statement is a lie can only be either $0$ or $1$. If we accept this, her statement is equivalent to "$q = 1$", which is a lie if $q = 0$ and is not a lie if $q = 1$. Thus in this case we are reduced to the classic Epimenides paradox.