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# The Beauty of Paradoxes (Part 1)

Updated: Apr 5

A paradox is a set of logical steps that reach a contradictory or unexpected conclusion.

Humans have been obsessed with these nonintuitive and perplexing statements since as early as the 6th century BCE in Ancient Greece. From Zeno's dichotomy paradox to Plutarch's Ship of Theseus, our minds seem to crave the initial (or possibly perpetual) confusion that these paradoxes incite. Some of these "thought experiments", such as that of Gabriel's Horn, have been solved with the advent of more advanced mathematics (calculus, in this case) hundreds of years after the problem was proposed. Others, such as Russell's paradox, still defy our human intuition and create bewildering contradictions.

In our third "Beauty of" installment, we will explore some of these paradoxes and related applications over the course of several blog posts.

## Part 1 - Probability

Consider a family with two children. Let’s assume that the probability of a single child being a girl or being a boy is half and half. Given that the older sibling is a girl, what is the chance that the younger child is also a girl? This answer to this question is simple enough: the gender of the younger child does not depend on the gender of the older sibling. Thus, the probability that the younger child is also a girl is simply one-half. Now consider another two-child family. Given that at least one of the siblings is a boy, what is the probability that both children are boys? This doesn’t seem too tough to solve: there is only one extra layer of complexity from the previous question. There are four possibilities for older sibling-younger sibling pairs: girl-girl, boy-girl, girl-boy, or boy-boy, all with equal chances of occurring. After factoring in the condition that at least one child is a boy, we can eliminate the girl-girl case. Therefore we have reduced this question down to asking: what is the chance that a family is boy-boy if the only possibilities are girl-boy, boy-girl, and boy-boy. Since all three cases are equally likely, the probability of having two boys is going to be one-third. This seems easy enough... However, what happens if we take another approach to solve this problem? Suppose you visit a two-child household and a boy answers the door. Since you see the boy answer the door, you know there is at least one boy in the family. Then, the probability that the other child is also a boy is obviously going to be just one-half. In other words, the probability of having two boys given that at least one of the children is a boy is one-half. Wait, how did we arrive at two different answers for one question? Both approaches are perfectly reasonable and neither explanation has any fallacies, so what is going on here? It turns out that this question is ambiguous. Depending on the interpretation of the question and the method of data collection, two distinct solutions can be argued for this question. So much for math problems always having one, distinct solution…

### Conditional Probability

Another mind-boggling probability problem arises with goats and cars. Imagine you are on a game show where you are presented with three closed doors. Behind two of the doors is a goat and behind the last one is a brand new car. First, you are asked to choose one of the three doors. Then, out of the remaining two doors, the host opens one of them to reveal a goat. The host then gives you the opportunity to switch your choice to the other unopened door. Afterward, the door you ended up choosing is opened and you win the prize behind that door (either a car or a goat). As cool as it would be to have a pet goat, let’s assume you are trying to maximize your chances of winning the car. What should your strategy be? Do you switch doors after a goat is revealed? Or do you stick with your original choice? Your initial intuition may be that it does not matter whether you switch doors or stay with your initial decision. After all, the probability is still one-third, right? It turns out that knowing the extra information of where a goat is actually changes the probability. To tackle this problem, we must first develop a sense of intuition for conditional probability. Let P(A) mean “the probability of event A occurring” and P(A|B) mean “the probability of event A occurring given that B already occurred.” If there are one hundred people in a room, some of which have blue eyes, some of which have blond hair, and some of which have both traits. The probability that a randomly selected individual has blue eyes given they have blond hair will be the same as the probability of randomly selecting a blue-eyed person out of the pool of blond-haired people. In other words, P(blue eyes | blond hair) is equal to the number of people with blue eyes and blond hair divided by the number of people with blond hair. Dividing both the numerator and denominator by the total number of people (100 in this case), we can reach the conclusion that P(blue eyes | blond hair) is equal to P(blue eyes and blond hair) divided by P(blonde hair). From this example, we can derive the general formula: P(A|B)=P(A&B)/P(B). ### Monty Hall Problem

Now let us apply our new tool of conditional probability to address our game show problem. Let us define three equally likely cases: in case 1, the car is behind door one, in case 2, the car is behind door two, and in case 3, the car is behind door three. Without loss of generality, let’s assume you always start by picking door one. In case 2, the host will reveal door three, and in case 3, the host will reveal door 2. In case 1, doors two and three each have an equal probability of being opened. Thus, given that the host reveals, say door two, the probability that you win by not switching doors—aka P(case 1 | door two revealed)—is P(case 1 and door two revealed) divided by P(door two revealed). The probability of being case 1 and the host revealing door two is one-third times one-half, which is one-sixth. Meanwhile, the host opens door two half of the time in case 1 and all of the time in case 3, meaning the probability door 2 is revealed is one-sixth plus one-third, or one-half. Combining all of this information so far, we get that P(case 1 | door two revealed) is one-third. This means that sticking to your initial choice after the host reveals a door with a goat behind it yields a one-third chance of winning the car. Naturally, switching to the other closed door after the host reveals a goat has a two-thirds chance of winning. We have now shown that you should always switch doors after the host reveals one of the goats, as you will win two times as often using this strategy compared to keeping with your original decision. This does not necessarily mean that you will always win a car if you switch, but it significantly increases your odds of winning if you apply this strategy. From our exploration into probability-pertaining paradoxes, we have seen how a vaguely-worded question can produce two equally reasonable, yet contradictory answers, as well as how game show probabilities can defy our intuition. Perhaps you may find yourself on one of these game shows; if you do, remember to do the math and show your work!

Stay tuned for part 2!

Author: Jerry Zhang