Examples of simple but highly unintuitive results?

Question

QUESTION: What are some simple math problems whose answers are highly unintuitive, and what makes them so?

There are plenty of unintuitive and frankly baffling results in math, like the Banach-Tarski paradox and this crazy MSE question which uses the axiom of choice to predict real numbers. However, these are pretty esoteric, and a layman might have trouble even understanding what exactly the question is asking. I’m more interested in examples like the Potato paradox:

Fred brings home $100$ kg of potatoes, which (being purely mathematical potatoes) consist of $99\%$ water. He then leaves them outside overnight so that they consist of $98\%$ water. What is their new weight? The surprising answer is $50$ kg.

I think I can explain why this answer seems unintuitive. Our intuition tells us that a small change in the water percentage should result in a small change in the mass of the potatoes. However, this heuristic is misleading in this case, in part because of the fact that $1/x\to \infty$ as $x\to 0$ and $1/x$ makes large “jumps” in value for $x$ close to $0$.

What are some other examples of simple problems with unintuitive answers? (I expect that there are plenty of examples that have to do with probability, since humans have terrible probabilistic intuition, and plenty of examples involving infinity, since people have a hard time conceptualizing the infinite.)

Also, please try to articulate exactly why you think your problem has an unintuitive answer, as I’ve attempted to do for the Potato paradox.

Answer 1

Here's another problem about the unintuitive effect of $\frac1x$.

You want to drive point A to point B and back at an average speed of 60 mph. However, on the way from A to B, there was traffic, which slowed you down to 30 mph. How quickly do you have to drive from B to A so that your average speed is 60 mph (over the entire round trip)?

A reasonable first guess is 90 mph, and then you might wonder if the true answer is a bit different, but actually the answer is a lot different:

It's impossible! If A and B are $\ell$ miles apart, then an average speed of 60 mph means going the $2\ell$ miles from A to B and back in only $\frac{2\ell}{60} = \frac{\ell}{30}$ hours. However, going from A to B at 30 mph already took $\frac{\ell}{30}$ hours, so the return trip would have to be done in zero time.

Exponential functions have an even more unintuitive effect (though we're currently all getting a crash course in those), and there's the traditional problem:

A population of algae is introduced into a lake on day 1. The algae grows very quickly, doubling in population (and in area covered) every day. On day 30, half of the lake is covered. At this rate, when will the algae cover the entire lake?

Maybe our first guess (because we expect all functions to be linear) is day 60 or 59 or something, but actually

The answer is day 31; doubling "half of the lake" just once gives us the entire area of the lake.

It's traditional to give three examples, so here's the birthday paradox. The math here is a bit fancier (though I'm giving the version that requires less calculation), but the statement is easy for anyone to understand:

A professor teaches a lecture class of about the same size every year. The roster lists everyone's birthday. The professor notices that on average, there is one pair of students per year with the same birthday. About how large are the professor's lectures?

There are $365$ days in most years, so we might expect that a sizable fraction of the year needs to be covered. However, the answer is only:

About $27$ or $28$ students per class. With $27$ students, there are $\binom{27}{2} = 351$ pairs of students; with $28$ students, there are $\binom{28}{2} = 378$. To get the average number of pairs that share a birthday, divide by $365$ (under the assumption that birthdays are uniform and February 29 doesn't exist, which is not far from the truth).

This paradox gets more surprising if we replace "birthday" with some other statistic that's uniformly spread over even more values, but I can't think of one that's also a reasonable piece of data for the professor to collect from the students.

Answer 2

Expected first entry time of a Brownian motion!

Let $B$ be a Brownian motion with start in $0$,

$a \in \mathbb{R}_{> 0}$ and $\tau_a$ be the first entry time of $a$, i.e.

$$ \tau_a := \inf\{t \geq 0 \mid B_t = a\}.$$

The value $\tau_a$ is almost surely finite, i.e. almost every path of the Brownian motion hits $a$ in finite time, but the expected value of $\tau_a$ is actually infinite!

You can proof this by considering $\inf\{t \geq 0 \mid B_t \in \{a, -a\}\}$.

image credit

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Answer 3

The "paradox" with the name "Achilles and the Turtle" is very confusing to a lot of people. It goes as follows:

Archimedis and a turtle have a race. The Achilles moves 10x faster than the Turtle, the ladder gets a head Start of 100 meters though. The race begins, Achilles quickly runs the 100m - the turtle however moves 10m in this time, so it is still 10m ahead. Achilles tries to run the 10m to the turtle, but in that time the turtle again crawls foreward 1m - so it is still 1m ahead. This continues forever. In the end - can the faster Achilles ever overtake the turtle?

This is confusing since it seems like Achilles is getting arbitarily close to the turtle but can't ever pass it, although he should be able to.

A relevant factor, namely time, is not mentioned. The time spans we consider are getting smaller by a factor of 10 every iteration, so if we sum the observed time up we obtain a geometric series with q= 1/10. This implies that the total time we can witness with this kind of observation is finite, since the geometric series converges. This is the reason that in this perspective Achilles can't ever pass the turtle - he is never given quite enough time! With linear time however, he would simply pass the turtle at the 111.11111... meter mark.

This unintuitive, because in our world time goss by linearly, so it seems logical to assume this must hold for this problem as well - a fallacy! Also, some people have a misleading intuition of mathematical series because they don't consider that infinitely many things can add up to something finite.

Answer 4

Regions of a circle

One of my favorites:

Choose n points around the circumference of a circle, and join every point to every other with a line segment. Assuming that no three of the line segments concur, how many regions does this divide the circle into?

There's a rather obvious pattern, that breaks down at $n=6$.

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Bayes' theorem

Related to the example given in the question is the following:

There is a rare disease that only $0.1\%$ of the population have. Suppose you have a test that can determine if someone has this rare disease at a $99\%$ rate of accuracy. If you test positive for the disease, what is the probability that you have the disease?

Seems pretty obvious, $99\%$ right?

Suppose there are $1,000,000$ people in the population. $999,000$ don't have the disease, which means $9,990$ people falsely test positive. $1,000$ people have the disease, and $990$ correctly test positive. So out of all of the people who have tested positive, $\frac{990}{990+9990} = \frac{1}{11}$ actually have the disease!

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Simpson's Paradox

Imagine you have two bins: $A$ and $B$.

$A$ contains $5$ white balls, $6$ black balls

$B$ contains $3$ white balls, $4$ black balls

You want to pull a white ball, but you can only pull once from a bin of your choice at random. Which bin would you pull from? Clearly, $A$ gives you better odds.

Imagine you have two bins: $C$ and $D$.

$C$ contains $6$ white balls, $3$ black balls

$D$ contains $9$ white balls, $5$ black balls

Now which bin would you choose? Clearly, $C$ gives you better odds.

Let's combine bins $A$ and $C$, and combine bins $B$ and $D$. Would you pull from the bin with $A$ and $C$, or the bin with $B$ and $D$? Seeing as how $A$ and $C$ were both the better choices, their combination must still be the correct choice right?

$AC$ contains $11$ white balls, $9$ black balls

$BD$ contains $12$ white balls, $9$ black balls

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Gabriel's Horn

There exists a shape that has infinite surface area but finite volume. The fact that such a shape can even exist may be pretty unintuitive to begin with.

Even more baffling is the idea that you can paint an infinite surface area in a finite amount of time and paint. Simply fill the horn with an amount of paint equal to its volume (which is finite), pour all of the paint out, and the entire interior of the shape has now been painted!

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Partial differential equations

"In order to eat as much as possible in a day, one should not eat as much as possible all day."

Sounds confusing and unintuitive at first, right?

If someone wanted to maximize their food consumption, rather than continuously consume food the entire day, it may be optimal to consume 3 large meals or 5 small meals instead. This is understood through partial differential equations: the rate of digestion may be dependent on various other factors such as the amount of food in the stomach or appetite.

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Function asymptotic growth

The wheat and chessboard problem is very famous, and is a go-to example educators use to demonstrate the monstrous unexpected growth of exponential functions.

But without the understanding of the growth rate of functions, some other results are also surprising:

There are more possible chess games than there are atoms in the observable universe.

There are approximately $10^{80}$ atoms but approximately $10^{120}$ possible chess games.

All of the digits of the number $$9^{9^{9^9}}$$ cannot possibly be contained within the observable universe.

How could four simple $9$'s create such a large number?

A little bit more esoteric would be examples like Kruskal's tree theorem and $TREE(3)$, or certain Diophantine equations like the positive solutions of $\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = n$.

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Infinite ordinals

Goodstein sequences, when evaluated naively, seem like they would not only grow extremely quickly, but grow forever.

However, a very basic understanding of infinite ordinals is enough to directly map the sequence to an ordinal sequence, making the fact that the sequence must eventually terminate to $0$ quite unsurprising and obvious.

Someone in the comments of the question mentioned the hydra game, which can be understood with the same idea.

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There may be a couple more good examples from this thread: Examples of patterns that eventually fail

Answer 5

The birthday paradox. Quote from Understanding the Birthday Paradox:

In a room of just 23 people there’s a 50-50 chance of at least two people having the same birthday. In a room of 75 there’s a 99.9% chance of at least two people matching.

Put down the calculator and pitchfork, I don’t speak heresy. The birthday paradox is strange, counter-intuitive, and completely true. It’s only a “paradox” because our brains can’t handle the compounding power of exponents. We expect probabilities to be linear and only consider the scenarios we’re involved in (both faulty assumptions, by the way).

Answer 6

The wheat and chessboard problem is probably one of the simplest. Its is an old one too, going back at least eight hundred years or so.

If you put one grain of wheat or rice in the first square of a chessboard, two in the second, four in the third, and so on, doubling the number at each step, how many grains of wheat or rice do you have?

This is a very simple problem, one that only needs pen and paper (doubling and summing) to do; it's just that the immense scale of the result is completely unintuitive, $$N = \sum_{i=1}^{64} 2^{i-1} = 2^{64} - 1 = 18,446,744,073,709,551,615$$ which is on the order of a couple of hundred times the yearly production of wheat on the entire planet.

Answer 7

In a far future, humanity has settled on Mars and Venus. Now Earth gets visits both from Martians and from Venusians, and some of them happen to be criminals.

In particular, of all the crimes done by visitors from other planets, $3/4$ are done by Martians, while only $1/4$ are done by Venusians.

Therefore a Martian visitor is three times as likely to be a criminal than a Venusian visitor, right?

Wrong! With the data given above it is impossible to say whether Martian or Venusian visitors are more likely to be criminals.

The conclusion would only be right if there happen to be as many Martian visitors as Venusian visitors, and moreover Martian and Venusian criminals do on average the same number of crimes on visit.

If there happen to be three times as many Martian visitors as Venusian visitors, both have the same crime rates. Should the number of Martian visitors be five times as high as the number of Venusian visitor, the same numbers would even mean that Martian visitors are doing less crimes on average than Venusian visitors. On the other hand, if Venusian visitors outnumber Martian visitors, it means that Martian visitors are even more criminal than those numbers suggest.

Also, if each Martian criminal on average tends to do three times as many crimes as a Venusian criminal, then even with equal numbers of visitors it means that the probability of a random Martian visitor being a criminal is the same as the probability of a random Venusian visitor being a criminal. In that case the Martian criminals do more crimes not because there are more of them, but because the criminal ones are more active.

This is unintuitive because most people don't even notice that they make hidden assumptions when making that wrong conclusion.