Paradoxes

This statement is false. Wait; if it is false, then the statement must be true since it says that it is false. So then is the statement true? But if it’s true, then according to the statement, it must be false. That is what is known as a paradox.

According to the Merriam-Webster dictionary, a paradox is “a statement that is seemingly contradictory or opposed to common sense and yet is perhaps true,” or “a self-contradictory statement that at first seems true.” Though the dictionary definition may seem a bit dry, paradoxes are a fun subject to explore!

One of the most famous paradoxes is the Monty Hall Problem. The Monty Hall Problem begins like so: there are three doors; behind one of them is a million-dollar car, and behind the other two are goats. You don’t know which door hides that car, but clearly, the car is better than the goats. The host asks you to pick a door; you’ll get whatever is behind that door. Say you pick door number 1. The host then reveals that one of the goats is behind door 2. Should you switch your pick from door 1 to door 3? It seems like there is a 50-50 chance of getting the car either way since there are two possibilities and both are equally likely. 

However, according to the problem, you are twice as likely to get the car if you switch. Preposterous! That must be a typo! It seems contradictory to common sense, but it might be true. Consider this: If your first pick is a goat, and the host reveals the location of the other goat, then you SHOULD switch, because the remaining option must be the car. Now, you don’t know whether or not your first pick was a goat. But there IS a 2 in 3 chance that it was – pretty good odds! Which means you should switch. 

This defies common sense! After all, it’s not like the goat and car switch around after you’ve made your choices. But if you decide to switch, you actually have a ⅔ chance of ending up with a car while only a ⅓ chance of ending up with a goat. Next time you’re on a game show with cars and goats, keep the Monty Hall Problem in mind.

Up next is my favorite paradox by far. However, it’s a little mathy… So if you don’t like math, you might want to skip it. Introducing: Bertrand’s Paradox.

Let there be an equilateral triangle with each vertex touching the circumference of a circle. What is the probability that the length of a randomly selected chord is greater than the length of the equilateral triangle? (A chord is a line with both endpoints on the circumference.) One might immediately think: suppose we have one point on one of the vertices of the equilateral triangle. (This can be true for any point on the circumference; just mentally rotate the triangle.) Then we notice that any line segment that goes through the triangle is longer than the length of the triangle’s sides. (Refer to Fig. 1) Whenever the line segment does not go through the equilateral triangle, its length is less than the triangle. Because each portion of the circumference between two points of the triangle is equal to one-third of the total circumference, all the chords that go through the triangle represent one-third of the total possible chords. Therefore, the probability that the chord is longer must be ⅓. 

Others might think of it differently. They might think to choose a chord from a set of parallel chords. Why? Well, every chord is part of exactly one set of parallel chords. (See Fig. 2 above to convince yourself of this.) So, if we look at just one of these sets, the probabilities are going to be the same as for every other set, since they’re all identical but rotated. (Refer to Fig. 2) Looking at a single set, one such chord would be the base of the triangle. 

You may notice that if you draw a line from the center of the circle to the circumference perpendicular to the side of the triangle it is passing through, then that side of the triangle divides the radius into equal lengths. (Take our word on this fact, although proving it may be a fun exercise!) Now, do this on the other side of the triangle as well. As you can see, this forms a diameter, with the middle section of the diameter being the portion of the chords that are longer than a side of the triangle. This means that there is a ½ chance that the randomly selected chord is longer than the side length of the triangle. 

There is one more proof that the probability must also be ¼, but we will skip it for the sake of concision. Notice that we started with something quite reasonable, but we reasoned our way to multiple contradictory conclusions.

There is another probability problem that is a little easier to understand: The Sleeping Beauty Paradox. It goes like this: Sleeping Beauty is asked to participate in an experiment. She is first put to sleep on Sunday. Then a coin is flipped. If the coin comes up heads, she is woken up once on Monday and then put back to sleep. If the coin comes up tails, she is woken up once on Monday, put back to sleep, woken up again on Tuesday, and then put back to sleep. Each time she wakes up, she has no recollection of whether she has been woken up or what day it is. That is to say, each time she wakes up, it is the same experience; there is no way for her to distinguish whether or not she has woken up before. Each time she wakes up, she is asked, “What is the probability that the coin came up heads?”
Take a moment to think about this. Once you have an answer, you are probably fairly certain you are correct. However, there have been many heated debates on this topic, with many famous mathematicians writing multiple papers on it. Some think it is ⅓ while others believe it is ½. I believe that it is ⅓ simply because there are three cases: either she woke up on Monday with heads; Monday with tails; or Tuesday with tails. The three are all equally likely. 

Adam Elga explained it this way: suppose when Sleeping Beauty wakes up, she is certain that the coin flip landed on tails. Since she can’t distinguish between Monday tails and Tuesday tails, the two events must be equally likely. Now, suppose that when Sleeping Beauty wakes up, she is certain that she woke up on a Monday. It is reasonable that the probability of Monday heads must equal the probability of Monday tails. So we have that P(TuesdayTails) = P(MondayTails) = P(MondayHeads), where P(E) is the probability that the event E occurs. Since the probabilities must add up to 1, each of the events must have a probability of ⅓. 

People who argue it is ½ say that a coin flip has a ½ chance of coming up heads. Why should it change if Sleeping Beauty wakes up twice? Before the experiment, she knows the coin must have a ½ chance of coming up heads. Why should that change if she gets no information when she wakes up from the experiment? Now that you’ve seen both sides of the argument, what do you think?

We’ll end with a ridiculous one. The interesting number paradox attempts to sort all natural numbers into interesting and uninteresting numbers, then show that all natural numbers are interesting. It is a proof by contradiction. Assume that NOT all natural numbers are interesting. Then the smallest uninteresting number is interesting because it is the smallest uninteresting number. Thus, we have a contradiction, and we have shown that all natural numbers are interesting. As I said, ridiculous.