Puzzle about throwing two darts at the first uncountable ordinal. This is something I read a while ago that I can't find the source of, but the source credited Axioms of symmetry: throwing darts at the real number line (Freiling 1986).
Two people, each throws a dart at the set of the first uncountable ordinal. The first dart hits $n_1$, the second dart hits $n_2$. The one who hits the higher ordinal number wins.
Since in the first uncountable ordinal, there are countably many ordinals less than $n_1$, but uncountably ordinals greater than $n_1$, the probability that the second player wins is 1.
But if the two players don't interfere with each other, the order of throwing the dart "shouldn't" matter, so the probability that the second player wins "should" be $1/2$.
What is the probability of the second player winning?
 A: Uncountability doesn't matter here. You can make the same argument about throwing darts at $\omega$-there are only finitely many lower than the first, but infinitely many higher. In both cases, you can't define a uniform density on the ordinal. Once you define a density, you will find the chance of each winning is $1/2$ less half the ties.
A: As Ross Millikan points out, the choice of measure is important. However, there is a probability measure on $\omega_1$ that acts just like the intuition using the counting measure.
Let a set be "small" if it is countable. Obviously $\omega_1$ is not small, and countable unions of small sets are small. We let these sets have probability zero, and the complements (co-countable sets) have probability one. I leave it as an exercise to show that this is a countably additive $\{0,1\}$-measure on $\omega_1$.
There are a significant number of nonmeasurable sets in this measure, though, and this will be the source of the paradox, which is essentially a failure of Fubini's theorem.
Let $\alpha,\beta$ be the darts thrown by players one and two respectively. If we fix $\alpha$, the conditional probability $P(\beta>\alpha|\alpha)=\mu([\alpha+1,\omega_1))$ is $1$. Thus
$$\int\int{\bf 1}_{\beta>\alpha}\,d\beta\,d\alpha=\int1\,d\alpha=1.$$
At the same time, we can fix $\beta$ and consider the set of $\alpha$ that make player two win. This is the set $[0,\beta)$ which has measure zero, so:
$$\int\int{\bf 1}_{\beta>\alpha}\,d\alpha\,d\beta=\int0\,d\beta=0.$$
The problem is that $\{(\alpha,\beta)\in\omega_1^2\mid\alpha<\beta\}$ is a nonmeasurable subset of $\omega_1\times\omega_1$, so there is no consistent choice of probability that can be given to the event.
If we are given a different measure on $\omega_1$ in which $\{(\alpha,\beta)\in\omega_1^2\mid\alpha<\beta\}$ is measurable, then Fubini's theorem applies, so the probability that player one wins equals the probability of player two winning (and both of these are $1/2$ if $\{(\alpha,\alpha)\mid \alpha\in\omega_1\}$ has measure zero).
