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, the game show host opens one of the two other doors which does not contain a car, necessarily revealing 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.

Birthday Problem

For me this was the first example of how counter intuitive the real world probability problems are due to the inherent underestimation/overestimation involved in mental maps for permutation and combination (which is an inverse multiplication problem in general), which form the basis for probability calculation. The question is:

How many people should be in a room so that the probability of at least two people sharing the same birthday, is at least as high as the probability of getting heads in a toss of an unbiased coin (i.e., $0.5$).

This is a good problem for students to hone their skills in estimating the permutations and combinations, the base for computation of a priori probability.

I still feel the number of persons for the answer to be surreal and hard to believe! (The real answer is $22$).

Pupils should at this juncture be told about quick and dirty mental maps for permutations and combinations calculations and should be encouraged to inculcate a habit of mental computations, which will help them in forming intuition about probability. It will also serve them well in taking to the other higher level problems such as the Monty Hall problem or conditional probability problems mentioned above, such as:

$0.5\%$ of the total population out of a population of $10$ million is supposed to be affected by a strange disease. A test has been developed for that disease and it has a truth ratio of $99\%$ (i.e., its true $99\%$ of the times). A random person from the population is selected and is found to be tested positive for that disease. What is the real probability of that person suffering from the strange disease. The real answer here is approximately $33\%$.

Here strange disease can be replaced by any real world problems (such as HIV patients or a successful trading / betting strategy or number of terrorists in a country) and this example can be used to give students a feel, why in such cases (HIV patients or so) there are bound to be many false positives (as no real world tests, I believe for such cases are $99\%$ true) and how popular opinions are wrong in such cases most of the times.

This should be the starting point for introducing some of the work of Daniel Kahneman and Amos Tversky as no probability course in modern times can be complete without giving pupils a sense of how fragile one's intuitions and estimates are in estimating probabilities and uncertainties and how to deal with them. $20\%$ of the course should be devoted to this aspect and it can be one of the final real world projects of students.

A famous example is St. Petersburg paradox:

Consider a game which you earn $2^n\: \$ $ if you get $n$ consecutive Heads in a fair coin toss. The fair entrance fee of this game is $\infty$

I particular like the triple-or-nothing game:

You start with $1$ sweet $^{[1]}$ in the pot. At each step, you can either choose to leave the game with all the sweets in the pot, or you can continue the game. If you continue, a fair coin is flipped, and if it comes up heads then the sweets in the pot are tripled, but if it comes up tails then the pot is emptied.

If you can play this game only once, how many sweets would you be willing to pay to play? And how should you play? (Assume that you want to get the most sweets possible.)

_{$^{[1]}$ Let's not be money-minded here...}

The naive (and incorrect) analysis is to compute that at any step if there are $x$ sweets in the pot and you continue the game then the expected number of sweets in the pot will become $1.5x$. Thus you should not stop. But that is stupid; if you never stop you will never get any sweets! So when to stop?

Worse still, a correct analysis will tell you that no matter how many sweets you pay, you can play in such a way that the expected number of sweets you leave with is more than what you paid! The (silly) conclusion is that you should be willing to pay any number of sweets to play!

If you think really carefully about it, you will realize that expectation is a very poor indicator of rationality of choice. Instead, everyone will have some risk aversity, more generally a mapping from probability distributions to favourability. One possibility is that a probability distribution is unfavourable iff its median is not positive (representing no net gain). Then clearly this game will never be favourable to anyone with this kind of risk aversity except if you commit to playing for exactly one step. In real life, people will evaluate distributions in a much more complicated way than just checking the median.

That said, a reasonable rule of thumb is that it is not worth to make a decision whose estimated benefit does not have both positive mean and positive median. Positive mean is necessary for rules of thumb, otherwise you will not benefit in the long run. Positive median will prevent other stupid decisions such as playing the triple-or-nothing game for more than one step or paying more than 1.5 sweets to play it. More risk-averse people will play for zero steps and just take the initial sweet and leave!

This rule will show (reasonably) not only that it is not worth to pay even 2 sweets to play the triple-or-nothing game only once, but also that it is not worth to offer the game for others to play! Any application of probability to real-life decisions should be able to deal with such situations.

[Further remarks...]

My claim about the rule of thumb being reasonable is that it should work quite well in real life. Whether it agrees with various mathematical models of human rationality is irrelevant. Secondly, my rule of thumb is merely for determining whether a single option is worth taking or not. To compare between multiple choices of which you must pick one, you would have to extend the rule of thumb. One possible way is to define the value of each choice to be the minimum of the mean and median benefit. Then you of course pick the choice with the maximum value. Thirdly, different people will of course have different ways to evaluate a choice based on its benefit's probability distribution (assuming it can even be translated to some real number). A very risk averse person might take the minimum of the 1st percentile (roughly speaking the minimum benefit you believe you will gain in 99% of the cases) and the mean (average benefit). Someone else may combine the percentiles and mean in a different fashion, such as taking $-\infty$ as the value if the 5th percentile is below some threshold (such as representing serious hurt), but taking the mean otherwise.

I find that almost anything about probability is counter-intuitive to my college students on first encounter. Possibly this may depend on your audience. Here are a few examples:

$1.$ Question: "If a certain event has a $40\%$ chance of success, and we run $50$ experiments, then how many would you expect to succeed?" The most common responses I usually get are "all of them" and "none of them". This is after an hour-long lecture on the subject.

$2.$ Question: "Interpret this probability statement: There is a $30\%$ chance of rain today in the New York area." I usually only get about a $65\%$ successful response rate on this on a multiple-choice quiz, even after the hour-long lecture on the subject. Once I had a student so bamboozled by it that she called up the national meteorology service for a consultation.

$3.$ Question: "We have a hand of four cards $\{A, 2, 3, 4\}$, and pick out two at random; what is the probability we get the $A$ or $2$ ?" Common responses are $25\%$, $50\%$, and $75\%$. I've never had anyone in a class intuit the correct answer on first presentation.

$4.$ Question: "If you drive to school on a given day, you either get in an accident or you don't. Are these equally likely outcomes?" At least half of any class answers "yes" on the first presentation. This can be repeated with the same result with similar follow-up questions.

A while back, the xkcd blog posted this problem, which I found fascinating. Usually when I re-tell it, I do so slightly differently from the original author:

I have selected two numbers from $\mathbb{R}$, following some unknown and not necessarily independent distribution. I have written each number in a separate envelope. By fair coin toss, I select one of these two envelopes to open, revealing that number. I then ask the question "Is the number in the other envelope larger than this one?". You win if you guess correctly.

Can you win this game with probability $>\frac{1}{2}$? Note, that is a strict inequality. Winning with probability $=\frac{1}{2}$ is obviously easy.

Now, the solution to this starts out with a double-integral, so depending on the level of the class you're teaching it may not be appropriate.

Strongly related with your example is this consequence of the Arc sine law for last visits. Let's assume playing with a fair coin.

Theorem (false) In a long coin-tossing game each player will be on the winning side for about half the time, and the lead will pass not infrequently from one player to the other.

The following text is from the classic An Introduction to Probability Theory and Its Applications, volume 1, by William Feller.

• According to widespread beliefs a so-called law of averages should ensure the Theorem above. But, in fact this theorem is wrong and contrary to the usual belief the following holds:

  With probability $\frac{1}{2}$ no equalization occurred in the second half of the game regardless of the length of the game. Furthermore, the probabilities near the end point are greatest.

The boy or girl paradox already mentioned by Agnishom has an interesting variation:

''Suppose we were told not only that Mr. Smith has two children, and one of them is a boy, but also that the boy was born on a Tuesday: does this change the previous analyses?'' (for the question ''what is the probability that both children are boys'')?

Using some elementary computations with Bayes formula, the seemingly useless information that a child was born on Tuesday, changes the results.

To understand the intuition behind, consider an extreme case where you knew that one boy was born on December $30$. Then it is very unlikely that the other child is born on that date too, so one child is ''specified'' by the date-information. This reduces the question to ''is the other child a boy''? and changes the probability from $\frac13$ to approximately $\frac12$.

However, I do not recommend to use this example for teaching, as there are many interpretations of this paradox (that partially depend on language nuances of the formulation) and it can add more confusion then clarify something.

Airplane Seating

$100$ people are boarding a plane in a line and each of them is assigned to one of the $100$ seats on a plane. However, the first person in line forgot his boarding pass and as a result decided to sit down in a random seat. The second person will do the following:

1. Sit in her seat if it still available.
2. If her seat is not available, choose a random seat among the seats remaining and sit there.

Each following person sits according to the same rules as the second person. What is the probability the $100^{th}$ person will be able to sit in her assigned seat?

Most people think the probability is very small and think there is a tiny chance of the 100th person's seat being left after all the people move around. But the actual probability ends up being $\frac{1}{2}$.

Base rate fallacy

If presented with related base rate information (or generic information) and specific information (information only pertaining to a certain case), the mind tends to ignore the former and focus on the latter.

Example:
A group of police officers have breathalyzers displaying false drunkenness in 5% of the cases in which the driver is sober. However, the breathalyzers never fail to detect a truly drunk person. One in a thousand drivers is driving drunk. Suppose the police officers then stop a driver at random, and force the driver to take a breathalyzer test. It indicates that the driver is drunk. We assume you don't know anything else about him or her. How high is the probability he or she really is drunk?

Intuitive first answer might be as high as 0.95, but the correct probability is about 0.02.

Solution : Using Bayes's theorem.

The goal is to find the probability that the driver is drunk given that the breathalyzer indicated he/she is drunk, which can be represented as $${\displaystyle p(\mathrm {drunk} |D)}$$
where "D" means that the breathalyzer indicates that the driver is drunk.

Bayes's theorem tells us that

$$ {\displaystyle p(\mathrm {drunk} |D) = {\frac {p(D|\mathrm {drunk} )\, p(\mathrm {drunk} )}{p(D)}}} $$

We were told the following in the first paragraph:

$${\displaystyle p(\mathrm {drunk} )=0.001} $$ $${\displaystyle p(\mathrm {sober} )=0.999} $$ $${\displaystyle p(D|\mathrm {drunk} )=1.00} $$ $${\displaystyle p(D|\mathrm {sober} )=0.05} $$

As you can see from the formula, one needs p(D) for Bayes' theorem, which one can compute from the preceding values using
$${\displaystyle p(D)=p(D|\mathrm {drunk} )\,p(\mathrm {drunk} )+p(D|\mathrm {sober} )\,p(\mathrm {sober} )} $$

which gives $$ {\displaystyle p(D)=(1.00\times 0.001)+(0.05\times 0.999)=0.05095} $$

Plugging these numbers into Bayes' theorem, one finds that $$ {\displaystyle p(\mathrm {drunk} |D)={\frac {1.00\times 0.001}{0.05095}}=0.019627 \approx 0.02 } $$

A more intuitive explanation: on average, for every 1,000 drivers tested, 1 driver is drunk, and it is 100% certain that for that driver there is a true positive test result, so there is 1 true positive test result 999 drivers are not drunk, and among those drivers there are 5% false positive test results, so there are 49.95 false positive test results.
Therefore, the probability that one of the drivers among the $$1 + 49.95 = 50.95 $$positive test results really is drunk is $$ {\displaystyle p(\mathrm {drunk} |D)=1/50.95\approx 0.019627} $$

Bertrand's Paradox

Given two concentric circles ($S_1$, $S_2$) with radii $R_1=r$ and $R_2=\frac{r}2$, what is the probability, upon choosing a chord $c$ of the circle $S_1$ at random, that $c\:\cap\: S_2 \neq \phi$ ?

Simply speaking, your task is to

choose a chord of the larger circle at random and find the probability that it will intersect the smaller circle.

Surprisingly, Bertrand's Paradox offers three distinct yet valid solutions.

The same problem can also be stated as:

Given an equilateral triangle inscribed in a circle, find the probability of randomly choosing a chord of the circle greater than the length of a side of the triangle.

The counter-intuition steps in when you understand that the answer to the stated problem is $\frac12,\:\frac13,$ and even $\frac14$ at the same time, and all three answers are perfectly valid.

The crucial reason why there are three solutions to this in different cases is that the methods of selection of random variables are different in each case.

Here's the Wikipedia page for details on how each value is obtained and through what steps.

I remember that a professor had begun my high-school level probability class using Bertrand's Paradox as an introductory example.

It's not counter intuitive but it's amazing for teaching in class.

Pick $a,b \in [n]$ randomly. $\mathbb{P}[gcd(a,b)=1]$ tends to $\frac{6}{\pi^2}$ as $n$ goes to infinity.

Also, there is some other interesting problem whose answers have $\pi , e ,...$

I flip two coins. Given that one is heads, what's the probability the other one is heads?

Surprisingly, it's not $\frac12$.

I think the most stunning example are the non-transitive dice.

Take three cubic dice with the following numbers on their sides:

• Die $A:$ $3 \: 3 \: 3 \: 3 \: 3 \: 6$
• Die $B:$ $2 \: 2 \: 2 \: 5 \: 5 \: 5$
• Die $C:$ $1 \: 4 \:4 \: 4 \: 4 \:4$

Now I offer you the following game: You choose a die as you like, then I choose another die, and then we are rolling and the highest number wins.

No matter which die you choose, I can choose another one that wins more often than loses against your choice.

Perhaps Parrondo's Paradox would be interesting. One can combine losing propositions into a winning proposition.

Simpson's Paradox is also interesting. (And actually occurred in a court case.)

The secretary's problem (which has other names).The secretary has $n$ letters ($0<n<\infty$) and $n$ pre-addressed envelopes but puts the letters into the envelopes randomly, one letter per envelope. What is the chance $C(n)$ that NO letter gets into the right envelope?

The answer is $C(n)=\sum_{j=0}^n(-1)^j/j!,$ which converges to $1/e$ as $n\to \infty$. I think the method of solution is instructive.

One counter-intuitive result is that $C(n)$ is not monotonic in $n.$

Also many people would be inclined to guess that $C(n)>1/2$ for large $n.$

Another version of this is to take two shuffled decks, each with $n$ playing cards, and ask for the chance that no card occupies the same position in both decks.

I first saw this in "101 Great Problems In Elementary Mathematics" by H. Dorrie.

Someone mentioned non-transitive dice, and that reminded me of this one:

Suppose there are two unweighted six-sided dice that you cannot examine, but which you can direct a machine to roll and inform you of the sum. You can do this as often as you like, and the distribution of the sum is exactly what you would expect of a pair of ordinary six-sided dice.

Are they, in fact, a pair of ordinary six-sided dice?

Not necessarily.

Then someone mentioned a secretary problem, which turned out to be about derangements. I had in mind a different secretary's problem, which is also called the sultan's dowry:

You have $100$ candidates, upon which there exists a total order. The candidates have appointments with you, one at a time, in a random order. From each interview, you can tell exactly how that candidate ranks amongst those you have already examined. At that point, you may either accept or reject the candidate. Any acceptance or rejection is permanent; no candidate, once rejected, may be reconsidered. Your objective is solely to accept the best candidate. What is the strategy for maximizing your probability of doing so, and what is that probability?

As often happens in probability puzzles, the answer is $1/e$*, which many people find surprisingly high.

*Approximately, with the approximation getting better and better as the number of candidates increases without bound.

Consider the $d-$dimensional sphere, then as $d$ goes to infinity the mass concentrates on the equator $x_1=0$

In contract bridge, there is the principle of restricted choice. It's always seemed counterintuitive to me.

https://en.m.wikipedia.org/wiki/Principle_of_restricted_choice

Lake Wobegon Dice

Find a set of $n$ dice (each with $s$ sides, numbered appropriately), in which each die is more likely to roll above the set average on that roll than below the set average. Given $n$, find the Lake Wobegon Optimal set, in which that probability is maximum.

"Lake Wobegon Dice," by Jorge Moraleda and David G. Stork, College Mathematics Journal, 43(2):152--159 (2012)

Abstract:

• We present sets of $n$ non-standard dice—Lake Wobegon dice—having the following paradoxical property: On every (random) roll of a set, each die is more likely to roll greater than the set average than less than the set average; in a specific statistical sense, then, each die is “better than the set average.”

  We define the Lake Wobegon Dominance of a die in a set as the probability the die rolls greater than the set average minus the probability the die rolls less than the set average. We further define the Lake Wobegon Dominance of the set to be the dominance of the set’s least dominant die and prove that such paradoxical dominance is bounded above by $(n-2)/n$ regardless of the number of sides $s$ on each die and the maximum number of pips $p$ on each side. A set achieving this bound is called Lake Wobegon Optimal. We give a constructive proof that Lake Wobegon Optimal sets exist for all $n \ge 3$ if one is free to choose $s$ and $p$. We also show how to construct minimal optimal sets, that is, that set that requires the smallest range in the number of pips on the faces.

  We determine the frequency of such Lake Wobegon sets in the $n = 3$ case through exhaustive computer search and find the unique optimal $n = 3$ set having minimal $s$ and $p$. We investigate symmetry properties of such sets, and present equivalence classes having identical paradoxical dominance. We construct inverse sets, in which on any roll each die is more likely to roll less than the set average than greater than the set average, and thus each die is “worse than the set average.” We show the unique extreme “worst” case, the Lake Wobegon pessimal set.

One of the most puzzling results in probability is that the probability of randomly (and with uniform probability) picking a rational number among the set of reals is zero, even though this is not impossible in an experimental situation (throwing a dart). This is nicely explained here.

The set of rational numbers, for instance in the $\Omega=[0,1]$ interval is the countable union of disjoint singletons, and each one of these singletons has a probability of zero. Here is the proof:

A singleton, $\{b\}$, is a Borel measurable set with a Lebesgue measure of zero. The proof is as follows:

$$\Pr\left(\{b\}\right)=\Pr\left(\bigcap_{n=1}^\infty\left(b-\frac{1}{n},b + \frac{1}{n}\right]\cap \Omega\right)$$

is the probability of nested decreasing sets, allowing the use of the theorem of continuity of probability measures $(*)$ to re-write it as:

$$\Pr\left(\{b\}\right)=\lim_{n\rightarrow \infty}\,\Pr\left(\left(b-\frac{1}{n},b + \frac{1}{n}\right]\cap \Omega\right)$$

The probability of $$\Pr\left(b-\frac{1}{n},\,b + \frac{1}{n} \right)\leq \lim_{n\rightarrow \infty}\frac{2}{n}=0$$

Therefore, by countable additivity of measures $(**)$, the probability for the whole set of $\mathbb Q$ is zero:

$$\Pr(\mathbb Q\;\cap \Omega) = 0$$

The apparent paradox is that even though a rational number can be indeed be chosen under experimental conditions, and despite the infinity number of rational numbers in the $[0,1]$ interval, the probability of randomly choosing a rational is strictly zero.

The source is this great explanation here.

$(*)$ If $B_j, j = 1, 2,\cdots,$ is a sequence of events decreasing to $B$, then $\displaystyle\lim_{n\rightarrow \infty} \Pr \{B_n\} = \Pr \{B\} .$

$(**)$ For all countable collections $\{E_i\}_{i=1}^\infty$ of pairwise disjoint sets in a sigma algebra: $$\mu\left( \bigcup_{k=1}^\infty \, E_k \right)=\sum_{k=1}^\infty \mu(E_k).$$