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

What's the probability that Alice is lying?

  • $\begingroup$ What is $p{{}}$? $\endgroup$ – Wojowu Nov 1 '15 at 16:35
  • $\begingroup$ I don't think this puzzle makes sense. $\endgroup$ – Crostul Nov 1 '15 at 16:39
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    $\begingroup$ It depends. Is Alice a Bayesian or frequentist? $\endgroup$ – Paul Nov 1 '15 at 16:43
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    $\begingroup$ @fleablood I'm not seeing a paradox when $p=1$: that seems unambiguously a lie since it implies that some probability is $>1$. $\endgroup$ – Erick Wong Nov 1 '15 at 16:59
  • $\begingroup$ I think this puzzle makes perfect sense. $\endgroup$ – user285523 Nov 1 '15 at 22:31

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 $q=0$.


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.


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?

  • $\begingroup$ Does "I'm lying" refer to her overall probability of lying, or to this particular statement? I would have thought it was the latter. Only in puzzles do we encounter people whose lies are iid. $\endgroup$ – Robert Israel Nov 3 '15 at 22:14
  • $\begingroup$ @RobertIsrael: I've never understood statements related to probabilities without the possibility of doing an experiment with independent trials. So, "The prob that I lie is greater than $p$" is meaningful if there is an experiment. I've created such an experiment. Out of 1000 E. T. Jaynes' probably 0 would agree with me. $\endgroup$ – zoli Nov 3 '15 at 22:24

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.


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.

  • $\begingroup$ If $p>0$ then Alice is asserting a positive probability that she is lying. Why do you say it's obvious that she is not lying? $\endgroup$ – Erick Wong Nov 1 '15 at 17:52
  • $\begingroup$ @Erick Wong In a continuous(non-discrete) space, $P(x=0)$ (in fact $x$ equal to any number) is always $0$ and $P(p>0)$ is always the same as $P(p\geq 0)$. $\endgroup$ – cr001 Nov 2 '15 at 1:49

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