The probability of selecting any given point on the (x, y) plane is 0, because there are uncountably many points in the plane. But what is the probability that the point you select will have a real number for its x coordinate? It seems like the answer should be 100%, but I'm having trouble understanding how this fits with the fact that the probability of selecting any given point is 0.
Here's another way of putting the question. Say I set up an idealized decision procedure that picks a point either on the complex or real plane. Suppose the procedure is set up in a way such that the probabilities of each being selected are equal. Before letting the decision procedure run, I should have a .5 credence that the result will contain only real numbers. Then I find out that the decision procedure has selected the real plane. Now my credence should be 1.
Is the answer that the question is badly formed? Is it that there is no way to make sense of the "probability that the selected point has only real numbers" given that there is no procedure that could actually pick such a number? This doesn't seem right, because any area of the real plane should have points with only real numbers.
First, you have to define a probability distribution on the plane before any question of probability will make sense. Usually, such distributions are defined in terms of integrals, in such a way that the probability of any set containing a single point, and indeed the probability of any set with $0$ area, is $0$. However, in order for this to actually be a probability distribution, one requirement is that the probability of the entire plane must be $1$.
There's no contradiction here. The same kind of thing is true of the real line, or of the interval $[0,1]$. On the interval $[0,1]$ with the uniform distribution, the probability of any interval $[a,b]\subseteq[0,1]$ is given by $\int_a^b\mathrm{d}x=b-a$. The probability of the singleton set $\{a\}$ is $\int_a^a\mathrm{d}x=0$.