I don't know that you can argue that they're lying.

The probability of sampling any particular number exactly is $0$. Presumably, there is some rounding done, so "$0$" might be any number in some specified interval around $0$. Once that interval is specified, you can work out the probability that a given sample is in the interval, and then the probability that $k$ independently chosen samples are all in that interval.

@lulu, the numbers are exact.

No, they aren't. Specifying a normally chosen real number would require specifying infinitely many digits. If you are using random number generator or something like that, it will have a specified precision level.

@lulu, This is a mathematical question, not the engineering question. The numbers are exact.

"In fact, if we look at PDF, then the sequence 0,...,0 has the highest likelihood among all sequences." why is this sentence true?

@Randall, what I meant that is that $p(x_1) \cdot p(x_2) \cdots p(x_k)$ is the largest when $x_1=x_2=\ldots=x_k=0$, where p is the PDF.

But as lulu has pointed out, that's not relevant since $P(X=x_k)=0$ no matter what $x_k$ is.

@Randall, I also point it out in the question itself. The question is "what is relevant".

You could frame this as a hypothesis test (essentially all the commonly used tests e.g. Kolmogorov-Smirnov would reject that this sample came from a standard normal distribution en.wikipedia.org/wiki/Normality_test), but it's worth noting there's a post hoc testing issue here.

@user51547, as I understand, the tests will produce the result of form "the probability that the data is sampled from $\mathcal N(0,1)$ is at most $\epsilon$", for some positive $\epsilon$. What I want is the result of form like "the probability that the data is sampled from $\mathcal N(0,1)$ is $0$". Again, I'm not sure what's the correct mathematical statement.

Here is a very simple test. The null is that the data is from an absolutely continuous distribution. You reject the null when $x_i =x_j$ for any $i\neq j$. This test has size $0$, albeit poor power. But in this case, it is perfectly sufficient.

@Andrew, Thanks, I think it works in this case. But I don't see how to generalize it to other scenarios. E.g. assume that $x_i = i$. In this case, I similarly want to say that this has probability 0 (and, more generally, having any rational difference between $x_i$'s has probability 0).

@Andrew, maybe this itself is the argument: "If $x_1,\ldots,x_k$ are i.i.d. from $\mathcal N(0,1)$, then the probability that $x_i - x_j$ is rational for some $i$ and $j$ is $0$"? But again, I can make the same argument for any fixed difference between points.

Yes, that would be another test as well. Fix a Borel set $B$ with Lebesgue measure zero. If any arithmetic operations on the $X_i$ end up in $B$, reject the null.

Your concern about being able to make the same argument for any fixed difference is a valid one. The problem with these types of arguments is that you're coming up with a test post hoc/after seeing the data. One way you could get around this issue is to first to decide on a test, ask your friend to sample more points, and test on that new sample (but maybe that is against the spirit of your question).

In a continuous distribution the probability of a simple point is zero, so it dont have sense if they arent in a nonzero measure interval: as example, if $P(x_k=0) =0$ then $P(x_k\neq 0)=1-P(x_k=0) =1$, so if are independent $P(x_1\neq 0\ |\ x_2\neq 0\ |\cdots|\ x_k\neq 0)=P(x_1\neq 0)\cdot P(x_2\neq 0)\cdots P(x_k\neq 0) = 1\cdot 1\cdots 1=1$ then the probability of everyone equal zero is $1-P(x_1\neq 0\ |\ x_2\neq 0\ |\cdots|\ x_k\neq 0)=0$...since for other values it will be also 0, a better way is thinking in the probability of being positive $= 1/2$ if the mean is zero, and take $(1/2)^k$