This isn't necessarily subtle, but...
I've encountered many students who mistakenly conclude that if an implication is true, then the converse must be true.
- That is, mistakenly concluding that if $p \rightarrow q$, then $q \rightarrow p$.
- The same error in reasoning comes when there is a chain of right-directional implications, and then assuming that then shows the equivalence of the original claim and the conclusion at the end of the chain of implications.
- A bit more subtly: I often encounter the erroneous conclusion that if $p\rightarrow q$, then $\lnot p\rightarrow \lnot q$.
- Also, when asked to prove a biconditional "if and only if" statement like $p \iff q$, some stop after proving only $p \rightarrow q$, thinking they are done.
And more subtly, when starting with an equation, then operating on each side of the equation (resulting in an equation), students often assume that whatever is the case about the end result also holds for the original. E.g. when given something of the form, $$\begin{eqnarray} y &=& f(x)\tag{1} \\ \text{So} \;\;y^2 &=& (f(x))^2\tag{2}\end{eqnarray}$$ and then (mistakenly) concluding solutions to $(2)$ are solutions to $(1)$. Or, e.g., given $$\begin{eqnarray}y^2 &=& x^2\tag{3} \\ \text{So} \;\;\sqrt{y^2} &=& \sqrt{x^2}\tag{4}\end{eqnarray},$$ then (mistakenly) concluding that the only solutions to $(3)$ are the solutions to $(4)$.
Additionally, one possible pitfall is not correctly applying DeMorgan's laws:
- As it relates to the distribution of negation over conjunction and the distribution of negation over disjunction: Mistakenly equating $\lnot (p \land q)$ with $ \lnot p \land \lnot q$ or $\lnot (p \lor q)$ with $\lnot p \lor \lnot q$.
- As it relates to containment in the complement of a union of sets and containment in the complement of the intersection of sets: E.g. Making the mistake of equating $\neg(A \cup B)$ with $\neg A \cup \neg B$ (and similarly in the case of the complement of an intersection).
Also, the negation of a quantified proposition seems to be problematic for some: making the mistake of equating $\lnot \forall x, P(x)$ with $\forall x, \lnot P(x)$, and similarly, when negating an existentially quantified statement.
The most famous counter-intuitive probability theory example is the Monty Hall Problem
- In a game show, there are three doors behind which there are a car and two goats. However, which door conceals which is unknown to you, the player.
- Your aim is to select the door behind which the car is. So, you go and stand in front of a door of your choice.
- At this point, regardless of which door you selected, the game show host chooses and opens one of the remaining two doors. If you chose the door with the car, the host selects one of the two remaining doors at random (with equal probability) and opens that door. If you chose a door with a goat, the host selects and opens the other door with a goat.
- You are given the option of standing where you are and switching to the other closed door.
Does switching to the other door increase your chances of winning? Or does it not matter?
The answer is that it does matter whether or not you switch. This is initially counter-intuitive for someone seeing this problem for the first time.
- If a family has two children, at least one of which is a daughter, what is the probability that both of them are daughters?
- If a family has two children, the elder of which is a daughter, what is the probability that both of them are the daughters?
A beginner in probability would expect the answers to both these questions to be the same, which they are not.
Math with Bad Drawings explains this paradox with a great story as a part of a seven-post series in Probability Theory
Nontransitive Dice
Let persons P, Q, R have three distinct dice.
If it is the case that P is more likely to win over Q, and Q is more likely to win over R, is it the case that P is likely to win over R?
The answer, strangely, is no. One such dice configuration is $(\left \{2,2,4,4,9,9 \right\},\left \{ 1,1,6,6,8,8\right \},\left \{ 3,3,5,5,7,7 \right \})$
Sleeping Beauty Paradox
(This is related to philosophy/epistemology and is more related to subjective probability/beliefs than objective interpretations of it.)
Today is Sunday. Sleeping Beauty drinks a powerful sleeping potion and falls asleep.
Her attendant tosses a fair coin and records the result.
- The coin lands in Heads. Beauty is awakened only on Monday and interviewed. Her memory is erased and she is again put back to sleep.
- The coin lands in Tails. Beauty is awakened and interviewed on Monday. Her memory is erased and she's put back to sleep again. On Tuesday, she is once again awaken, interviewed and finally put back to sleep.
In essence, the awakenings on Mondays and Tuesdays are indistinguishable to her.
The most important question she's asked in the interviews is
What is your credence (degree of belief) that the coin landed in heads?
Given that Sleeping Beauty is epistemologically rational and is aware of all the rules of the experiment on Sunday, what should be her answer?
This problem seems simple on the surface but there are both arguments for the answer $\frac{1}{2}$ and $\frac{1}{3}$ and there is no common consensus among modern epistemologists on this one.
Ellsberg Paradox
Consider the following situation:
In an urn, you have 90 balls of 3 colors: red, blue and yellow.
30 balls are known to be red. All the other balls are either blue or yellow.
There are two lotteries:
- Lottery A: A random ball is chosen. You win a prize if the ball is red.
- Lottery B: A random ball is chosen. You win a prize if the ball is blue.
Question: In which lottery would you want to participate?
- Lottery X: A random ball is chosen. You win a prize if the ball is either red or yellow.
- Lottery Y: A random ball is chosen. You win a prize if the ball is either blue or yellow.
Question: In which lottery would you want to participate?
If you are an average person, you'd choose Lottery A over Lottery B and Lottery Y over Lottery X.
However, it can be shown that there is no way to assign probabilities in a way that make this look rational. One way to deal with this is to extend the concept of probability to that of imprecise probabilities.
Best Answer
Here's a counterintuitive example from The Cauchy Schwarz Master Class, about what happens to cubes and spheres in high dimensions:
Consider a n-dimensional cube with side length 4, $B=[-2,2]^n$, with radius 1 spheres placed inside it at every corner of the smaller cube $[-1,1]^n$. Ie, the set of spheres centered at coordinates $(\pm 1,\pm 1, \dots, \pm 1)$ that all just barely touch their neighbor and the wall of the enclosing box. Place another sphere $S$ at the center of the box at 0, large enough so that it just barely touches all of the other spheres in each corner.
Below is a diagram for dimensions n=2 and n=3.
Does the box always contain the central sphere? (Ie, $S \subset B$?)
Surprisingly, No! The radius of the blue sphere $S$ actually diverges as the dimension increases, as shown by the simple calculation in the following image,
The crossover point is dimension n=9, where the central sphere just barely touches the faces of the red box, as well as each of the 512(!) spheres in the corners. In fact, in high dimensions nearly all of the central sphere's volume is outside the box.