Yet another variant of the Borel-Kolmogorov paradox

Suppose that a point is uniformly randomly choosen in the unit square. Then you learn that either the point lies on the diagonal y = x (red), or it lies on the horizontal line y = 1/2 (blue). What probability should you assign to its lying on the diagonal?

Answer 1: The diagonal has length 2^1/2 and the horizontal line has length 1. Thus, the total length of the lines where the point might be is 2^1/2 + 1, and the probability that it’s on the diagonal is 2^1/2/(2^1/2 + 1) ≈ 0.59.

Answer 2: We can think of the uniform random choice of a point in the unit square as the choice of two independent coordinates, x and y. Suppose that x has been chosen. Then to be on the diagonal line, y has to equal x, while to be on the horizontal line, y has to equal 1/2. These two things are clearly equally likely, regardless of what x is, so the probability must be 1/2.

Both answers seem reasonable.

Suppose you are attracted to Answer 1 which gives 0.59. Then I can give you an argument for a third answer.

Answer 1b: Here is a way to uniformly choose a point in a square. I first uniformly choose a point in a rectangle whose height is twice its width, and then divide the y coordinate by a factor of two. Being on the diagonal or on the middle horizontal line of the point uniformly chosen in the square are respectively equivalent to being on the diagonal or on the middle horizontal line of the rectangle. But the length of the diagonal of the rectangle is 5^1/2, while the middle horizontal line has the same length 1 as in the square. So applying the reasoning behind Answer 1 to the rectangle case, the probability that the point is on the diagonal is 5^1/2/(5^1/2 + 1) ≈ 0.69.

Thus, if you are attracted to the “geometrical” reasoning behind Answer 1, there are infinitely many other answers available, corresponding to the infinitely many ways of generating a point uniformly in a square by generating it in a rectangle and squashing or stretching.

This might push you to Answer 2, since the reasoning behind Answer 2 seems much more determinate. But there are variants to Answer 2. Here is another way to generate a point uniformly on the unit square. Rotate the unit square by 45 degrees clockwise around the orgin to get a diamond whose size along the x and y axes is 2^1/2. Now choose x with a symmetric triangular probability density between 0 and 2^1/2, choose y₀ uniformly between −1 and 1, and then rescale y₀ to make its range fit within the diamond. Parallel reasoning to that used in Answer 2 will now generate a different answer, indeed an answer making the diagonal be more likely.

Note that while I put the paradox in terms of conditioning on a measure zero (the union of the two line segments), one can also put the paradox in terms of comparing probabilities if one likes to be able to compare zero probability sets.

Lesson: Either there are infinitely many different kinds of “uniform distributions of a point in a square”, or else we shouldn’t compare sets of zero measure.