# Probability questions that have answer $\frac{1}{2}$ but resist intuitive explanation.

My question is: What are some examples of probability questions that have answer $$\frac{1}{2}$$ but resist intuitive explanation?

## Context

Some probability questions have answer $$\frac{1}{2}$$, and - as you might expect - have an intuitive explanation, i.e. an explanation that requires little calculation, instead utilizing a clever framing of the question that makes the answer obvious.

For example:

"Flip seven unbiased coins; what is the probability of getting at least four heads?"
Some people will calculate $$P=\frac{\binom{7}{4}+\binom{7}{5}+\binom{7}{6}+\binom{7}{7}}{2^7}=\frac{35+21+7+1}{128}=\frac{64}{128}=\frac{1}{2}$$, but there is an intuitive explanation: getting at least four heads means getting more heads than tails, which, by symmetry, has probability $$\frac{1}{2}$$.

Another example:

Another example, this one about geometric probability:

"Break a stick at two random points. The probability that the longest piece is at least twice as long as each of the other pieces is $$\frac{1}{2}$$. Why?" Here is a (somewhat) intuitive explanation.

On the other hand, I have found that some probability questions have answer $$\frac{1}{2}$$ but do not seem to have an intuitive explanation. For example:

I find these later kinds of questions interesting, because the answer, $$\frac{1}{2}$$, is the simplest probability of all (unless you count $$0$$ or $$1$$, but those are sort of degenerate), but there does not seem to be a simple explanation. It's as if the question is making fun of us: "You can't find an elegant solution!" Or maybe there is no elegant solution, and the answer is $$\frac12$$ by coincidence.

I would be interested in more of these kinds of questions (they don't have to be related to geometry).

## Remarks

I require that the answer is exactly $$\frac{1}{2}$$, not any other number, to avoid a slippery slope of what numbers are acceptable.

Let's avoid ad hoc questions, for example: "Find the probability that two uniformly random points inside a unit square are within $$d$$ distance of each other, where $$d$$ is the smallest positive root of $$\frac12x^4-\frac83x^3+\pi x^2-\frac12=0$$ ($$d\approx0.5120$$)." The answer is $$\frac12$$ because $$d$$ is the median distance. Obviously anyone asking this question already knows that the answer is $$\frac12$$.

I am using the "big-list" tag, whose description reads "Questions asking for a "big list" of examples, illustrations, etc. Ask only when the topic is compelling..." I think my question might provide some insight on probability and intuition.

• Could you describe more the term intuitive explanation in the OP? I know that it cannot be defined, and it is a relative concept, but can be clarified more.
– Amir
Commented Apr 3 at 7:25
• @Amir An "intuitive explanation" typically involves a smart way of simplifying the question, making the answer obvious. It might make use of some kind of symmetry. It does not involve a lot of calculation (so no complicated integrals).
– Dan
Commented Apr 3 at 7:46
• You may add the problem Taking Seats on a Plane to the list given in the OP.
– Amir
Commented Apr 3 at 8:31
• A triangle's vertices are three uniformly random points on a unit circle. Simulations suggested a beautiful conjecture: $P(\text{perimeter}>10\times\text{area})=\frac12$. But the probability is actually $\frac{8}{\pi^2}\int_{\arccos\frac{5+\sqrt{15}}{10}}^{\arccos\frac{5-\sqrt{15}}{10}}\arccos\left(\cos x+\frac{1}{10\cos x}\right)dx\approx0.49994$.
– Dan
Commented Apr 8 at 6:53

## 4 Answers

Regarding the "Intuition is silent"-problem:

The smallest possible circle has the diameter equal to the distance of the two points furthest from each other. In the case that the circle is defined by three points, said diameter will be arbitrary close to said distance, as with infinite points, both cases converge to each other.

With $$n\rightarrow\infty$$ they both will be arbitrary close to the disk's edge. Looking only at the second point, it will be arbritrary close to the diameter of the disk through the first point. As the distances are arbitrary small, the arc segment becomes a straight line and it forms two right angles with the diameter intersecting it. Now, wether or not the circle will be on the disk depends on the distances of the point to the disk's edge and to the diameter. If the former is larger than the latter, the circle will be on the disk. If not, it won't. Geometrically, this probability corresponds to the point being on one or the other side of the angle bisectors of the right angles. So it is $$\frac{1}{2}$$.

• "The smallest possible circle has the diameter equal to the distance of the two points furthest from each other." That's true if the smallest circle goes touches two of the points, but it's not true if the smallest circle touches three of the points. Also, I don't understand the part that says "If the former is larger than the latter, the circle will be on the disk. If not, it won't."
– Dan
Commented Apr 3 at 9:37
• With $n\rightarrow\infty$, the circle's diameter will be arbitrary close to the the distance of the two points furthest apart from each other. The second issue is hard to explain (for me). Intuitively, the circle will intersect the disk's circumference, if the second point is further from the disk's diameter through the first, than it is from the disk's edge.
– Max
Commented Apr 3 at 9:51
• Either of the two points with the greatest distance between them. EDIT: the comment this one was an answer to is gone.
– Max
Commented Apr 3 at 10:06
• If the two most-separated points lie strictly inside the smallest circle (which is possible when the smallest circle touches three points), then their location relative to the bisectors does not affect the position of the smallest circle.
– Dan
Commented Apr 3 at 10:26
• Yes, but with infinite points, the three point case and the two point case converge. As stated above, the smallest circle's diameter will be arbitrary close to the distance of the two points furthest from each other.
– Max
Commented Apr 3 at 10:44

A triangle's vertices are uniformly random points on a circle. The side lengths, in increasing order, are $$a,b,c$$.

We have $$P(a^\color{red}{1}+b^\color{red}{1}>c^\color{red}{1})=1$$, which is just the triangle inequality.

We also have $$P(a^\color{red}{2}+b^\color{red}{2}>c^\color{red}{2})=\frac14$$, which has an intuitive explanation.

So what is $$P(a^\color{red}{\sqrt2}+b^\color{red}{\sqrt2}>c^\color{red}{\sqrt2})$$ ?

The answer turns out to be $$\frac12$$, but an intuitive explanation seems to be lacking.

For almost all real numbers, their binary digits are evenly distributed i.e. the chance of a random digit to be $$0$$ (or $$1$$) is $$1/2$$. More rigourously, the probability of a uniformly randomly chosen digit from the first $$n$$ binary digits of the number is $$0$$ (or $$1$$) approaches $$1/2$$ as $$n$$ approaches $$\infty$$.

A short way to say this is that almost all real numbers are normal in base 2. A more general theorem is that it is true for all bases. I chose this problem because normality, despite intuitive as it may sound, is exceptionally hard to prove for real numbers. We currently don't even know if $$\sqrt{2}$$ is normal or not.

The vertices of a triangle are three uniformly random points on a circle. The side lengths are, in random order, $$a,b,c$$.

The triangle inequality tells us that $$P(a+b.

But what is $$P\left(a+b<\left(\frac{a}{b}\right)c\right)$$ given that $$\frac{a}{b}>1$$ ?

The answer is $$\frac12$$, but an intuitive explanation seems to be lacking.