A *paradox* is a seemingly valid piece of reasoning leading to an absurdity.

**Table of Contents**

- Barber Paradox
- Bertrand’s Box Paradox
- Ellsberg Paradox
- Hypothetical Syllogism
- Liar Paradox (Epimenides Paradox)
- Lottery / Preface Paradox
- Newcomb’s Paradox
- Raven Paradox (Confirmation Paradox, Hempel’s Paradox)
- Russell’s Paradox
- Simpson’s Paradox
- Sorites Paradox (Paradox of the Heap)
- Surprise Exam (Hangman Paradox)
- Twin Paradox
- Zeno’s Dichotomy Paradox

##### Surprise Exam (Hangman Paradox)

- A teacher announces a surprise exam for the following week. The exam is scheduled for Monday, Wednesday or Friday and is to be a surprise in the sense that no student knows beforehand the exam is on that day. One student reasons as follows:
- The teacher can’t wait till Friday to give the exam because then I’ll know beforehand the test is on that day and it won’t be a surprise. So the exam must be either Monday or Wednesday. But the same reasoning applies to Wednesday. Having ruled out Friday, if the teacher doesn’t give the exam on Monday, I’ll know it’s on Wednesday and so won’t be a surprise. This leaves Monday. But it can’t be a surprise on Monday since I know that’s the only remaining possibility.

- Much to the student’s surprise, the teacher gives the exam the following week.
- Where did the student’s reasoning go wrong?

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##### Hypothetical Syllogism

*Hypothetical Syllogism*is the valid argument-form in standard Propositional Logic.- A→B
- B→C
- So A→C

- But here’s counterexample:
- If I win the lottery, I will give half my annual income to charity.
- If I give half my annual income to charity, I will not have enough to live on.
- Therefore, if I win the lottery, I will not have enough to live on.

- Another argument-form valid in Propositional Logic:
- A→B
- So (A&C)→B

- But:
- If I strike the match, it will light
- Therefore, if I strike the match and it’s sopping wet, it will light.

- Is Propositional Logic wrong?

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##### Sorites Paradox (Paradox of the Heap)

- The following propositions all seem true:
- There are heaps of sand.
- Heaps consist of a finite number of grains of sand.
- If you remove one grain of sand from a heap of sand, the result is still a heap.

- A single grain of sand is not a heap.

- But the propositions can’t all be true. If you keep removing individual grains of sand from a heap you’ll eventually wind up with single grain of sand.
- Which of the four proposition is false?

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##### Ellsberg Paradox

- An urn contains 90 balls: 30 red and 60 either yellow or black. The number of yellow balls is unknown; likewise the number of black balls. You’re given a choice between two gambles:
- You receive $1,000 if you draw a red ball
- You receive $1,000 if you draw a black ball.

- Which gamble should you choose?

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##### Lottery / Preface Paradox

- The defendants in 10,000 criminal trials were justifiably found guilty beyond a reasonable doubt.
- As part of a blinded, academic study some defendants’ DNA was taken and tested. An unknown defendant was found innocent.
- Hence, there’s reasonable doubt that all 10,000 defendants are guilty.
- Yet, for each defendant, it’s still beyond a reasonable doubt that he/she is guilty based on trial evidence.
- How can this be?

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##### Newcomb’s Paradox

- There are two boxes, A and B. You can either
- open both boxes and keep their contents, or
- open just box B and keep its contents.

- There is $1,000 in box A. In box B there is either $1,000,000 or nothing, depending on a
*Predictor’s*prediction. If the*Predictor*has predicted you will pick just box B, he has put $1,000,000 in the box. If the*Predictor*has predicted you will pick both boxes, he has put nothing in box B. The Predictor has always correctly predicted your choices. - Which box should you open?

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##### Zeno’s Dichotomy Paradox

- To move from
*Point A*to*point B*you have to traverse the distance between A and*Point 1/2*. To pass from*Point 1/2*to B you have to traverse the distance between*Point 1/2*and*Point 3/*4. To pass from*Point 3/*4 to B you have to traverse the distance between*Point 3/*4 and*Point*7/8. And so on*ad infinitum*. But it is manifestly impossible to traverse infinitely many non-overlapping finite distances in a finite period of time. So, it is impossible to move from A to B. Or to move at all.

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##### Raven Paradox (Confirmation, Hempel’s Paradox)

- Observing a black raven tends to confirm the generalization that all ravens are black. Likewise observing a non-black non-raven tends to confirm the generalization that anything not black is not a raven. What tends to confirm a generalization also tends to confirm any logically equivalent generalization.
*All ravens are black*is logically equivalent to*anything not black is not a raven*. Thus observing a non-black non-raven tends to confirm that all ravens are black. This seems wrong.

- FYI, the following are logically equivalent in Predicate Logic:
- (x)(Rx → Bx)
- For any x, if x is a raven then x is black

- (x)(~Bx → ~Rx)
- For any x, if x is not black then x is not a raven

- (x)(Rx → Bx)
- In Predicate Logic neither statement logically entails
- (Ex)(Rx & Bx)
- There is an x such that x is a raven and x is black

- (Ex)(Rx & Bx)

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##### Liar Paradox (Epimenides Paradox)

- Consider the statement:
- This statement is false

- If the
*statement*is true, it’s false. And if it’s false, it’s true. - Therefore, what?

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##### Barber Paradox

- A barber in a village shaves every man in the village who doesn’t shave himself. He also shaves only these men. Does the barber shave himself? If he doesn’t, he does, because he shaves everyone who doesn’t shave himself. If he does, he doesn’t, because he shaves only men who don’t shave themselves.
- So does he shave himself or not?

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##### Russell’s Paradox

- Let X be the set of sets not members of themselves. Is X a member of itself? If it were then, according to the definition of X, it would not be a member of itself. So X is not a member of itself. On the other hand, if X were not a member of the set of sets not members of themselves, it would be a member of itself. Thus X is both a member and not a member of itself. But this is impossible.

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##### Twin Paradox

- The Scenario
- One of two identical twins blasts off into space, travels for several years near the speed of light, and returns. Einstein’s Special Relativity predicts she will be years younger than her stay-at-home sister.

- The Paradox
- Special Relativity’s prediction seems to apply to both twins.
- Special Relativity predicts that time runs slower in a reference frame moving relative to another.
- Motion is relative: if an object moves relative to a second, the second moves relative to the first. Since the spacecraft moves relative to Earth, the Earth moves relative to the spacecraft.
- Therefore, Special Relativity should predict that the Earth-bound twin will be younger than the traveling twin. So each twin would be younger than the other.

- Special Relativity’s prediction seems to apply to both twins.

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##### Simpson’s Paradox

- If David Justice had more hits than Derek Jeter in each of 1995, 1996, and 1997, then he had more hits in 1995, 1996, and 1997 combined.
- But if Justice had a higher batting average than Jeter in each of 1995, 1996, and 1997, it doesn’t follow he had a higher batting average in 1995, 1996, and 1997 combined.
- How come?

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##### Bertrand’s Box Paradox

- A box has three drawers.
- Drawer A has two two gold coins
- Drawer B has two silver coins
- Drawer C has one gold and one silver coin.

- A drawer is chosen at random and a coin randomly selected from the drawer. The selected coin is gold. What’s the probability the other coin is gold?
- First Solution: The coin was from one of two drawers: A or C. A has a second gold coin; C doesn’t. So the probability is ½.
- Second Solution: The coin was either gold coin #1 in drawer A, gold coin #2 in drawer A, or the gold coin in drawer C. In the first two instances, the other coin is gold. So the probability is ⅔.

- Which solution is correct?

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##### Resolutions

###### Surprise Exam (Hangman Paradox)

- The three-day surprise exam is logically analogous to the one-day case:
- “There will be an exam tomorrow but you won’t know beforehand that it’s tomorrow.”

- To which the natural reply is:
- “But you just told us”

- A person’s asserting something is normally a basis for believing what’s asserted.
- When a teacher announces a test for the following day, the students are in a position to know there’s a test the following day. If the teacher announces a surprise test for the following day, the students should be in a position to know that:
- there’s a test a test tomorrow
- they don’t know there’s a test tomorrow

- But they can’t know both these things:
- K(T & ~K(T)) is contradictory since it entails both K(T) and ~K(T), where
- K = the students know that
- T = there’s a test tomorrow
- & = and
- ~ = it’s false that

- K(T & ~K(T)) is contradictory since it entails both K(T) and ~K(T), where
- plato.stanford.edu/entries/epistemic-paradoxes/#SurTesPar
- wikipedia.org/wiki/Unexpected_hanging_paradox

###### Hypothetical Syllogism

- In Propositional (Sentence) Logic the conditional P→Q, called
*material implication*, is defined so that its truth value is a function of the truth values of P and Q: *Hypothetical Syllogism*is valid for P→Q.- But material implication captures only half the ordinary sense of
*if… then*: - The result is that some argument-forms are valid for P→Q but not for
*if P then Q*. - One such argument-form is
*Hypothetical Syllogism*- A→B
- B→C
- So A→C

- The counterexample was:
- If I win the lottery, I will give half my annual income to charity.
- If I give half my annual income to charity, I will not have enough to live on.
- Therefore, if I win the lottery, I will not have enough to live on.

- The conclusion is false. But, assuming I don’t win the lottery, the material implication is true:
- I win the lottery → I will not have enough to live on.

*Hypothetical Syllogism*can be modified, however, so the resulting argument-form is valid for the ordinary conditional.- A→B
- (A&B)→ C
- So, A→C

- The counterexample now fails, since the second premise is false:
- If I win the lottery, I will give half my annual income to charity.
- If I win the lottery and give half my annual income to charity, I will not have enough to live on.
- Therefore, if I win the lottery, I will not have enough to live on.

- The second argument-form considered was:
- A→B
- So (A&C)→B

- The counterexample:
- If I strike the match, it will light
- Therefore, if I strike the match and it’s sopping wet, it will light.

- The conclusion is false. But if I don’t strike the match, the material implication is true:
- (I strike the match & it’s soaking wet) → it will light.

- Like hypothetical syllogism, this argument-form can be modified so the resulting argument-form is valid:
- A→B
- A→C
- So, (A&C)→B

- The counterexample now fails, because of the second premise:
- If I strike the match, it will light
- If I strike the match, it’s soaking wet.
- Therefore, if I strike the match and it’s sopping wet, it will light.

- wikipedia.org/wiki/Material_conditional
- wikipedia.org/wiki/Paradoxes_of_material_implication
- oxfordreference.com/view/10.1093/oi/authority.20110803100139775
- plato.stanford.edu/entries/logic-relevance/

###### Sorites Paradox (Paradox of the Heap)

- The first premise is false, that if you remove one grain of sand from a heap of sand, the result is still a heap.
- But its falsity hides behind the vagueness of
*heap*. - As individual grains of sands are removed, the remaining grains are at first clearly a heap. Then they pass through a gray zone where it’s indeterminate whether they constitute a heap. Finally they become a single grain of sand, clearly not a heap.
- It’s false that one grain of sand always leaves a heap, though it’s impossible to specify which grain of sand.
- plato.stanford.edu/entries/sorites-paradox/
- wikipedia.org/wiki/Sorites_paradox
- britannica.com/topic/sorites-problem

View When human life begins is indeterminate

###### Ellsberg Paradox

- Wrong Answer: It doesn’t matter which gamble you choose because the probability of winning is the same, 1/3.
- But the gambles are different.
- Drawing a red ball is a matter of
*risk*, where the probabilities of the outcomes are calculable . - Drawing a black ball is a matter of
*uncertainty*, where the probabilities are indeterminate and incalculable.

- Drawing a red ball is a matter of

- But the gambles are different.
- Better Answer:
- Consider a choice between:
- An urn containing $1,000.
- An urn containing anywhere from $0 to $1,000,000 with unknown odds.

- It’s rational to prefer the first urn because it’s a certain $1,000 vs an unknown quantity.
- In the same way it’s rational to prefer betting on drawing the red ball because it’s a certain ⅓ probability vs an incalculable one.

- Consider a choice between:
- wikipedia.org/wiki/Knightian_uncertainty
- plato.stanford.edu/entries/decision-theory-descriptive/#IssuProbBeli
- plato.stanford.edu/entries/imprecise-probabilities/#EllDec
- plato.stanford.edu/entries/rationality-normative-utility/#MaxExpUtiImp
- wikipedia.org/wiki/Ellsberg_paradox

###### Twin Paradox

View Twin Paradox

###### Liar Paradox

- Therefore, what?
- Therefore, the
*Law of Excluded Middle*is false, that:- Every proposition is either true or false.

- For the
*statement*in question is neither true nor false:- If it’s true, it’s both true and false
- If it’s false, it’s both true and false
- So if the
*statement*is either true or false, it’s both true and false, a contradiction. - Hence the
*statement*must be neither true or false.

- wikipedia.org/wiki/Liar_paradox
- plato.stanford.edu/entries/liar-paradox/
- britannica.com/topic/number-game/Logical-paradoxes#ref396232

###### Barber Paradox

- The resolution is that there can be no such barber. Were there, he would both shave and not shave himself, a contradiction.
- wikipedia.org/wiki/Barber_paradox
- plus.maths.org/content/mathematical-mysteries-barbers-paradox
- britannica.com/topic/number-game/Logical-paradoxes#ref396232

###### Lottery / Preface Paradox

- The resolution is that the
*Principle of Agglomeration*is false, the principle that:- If it’s beyond a reasonable doubt that P and it’s beyond a reasonable doubt that Q, then it’s beyond a reasonable doubt that P&Q.

- For if the principle were valid, the following argument would establish a falsehood
- For each of the 10,000 defendants D
_{1}, D_{2}, … D_{10000}, it’s beyond a reasonable doubt that D_{n}is guilty. - Thus, per the
*Principle*, it’s beyond a reasonable doubt that defendants D_{1}and D_{2}are guilty, and it’s beyond a reasonable doubt that defendants D_{1},D_{2 }and D_{3}are guilty, and so on. - Therefore, it’s beyond a reasonable doubt that all 10,000 defendants are guilty.

- For each of the 10,000 defendants D
- The conclusion is false because an unknown defendant was found innocent.
- The
*Principle of Agglomeration*can be patched:- If it’s beyond a reasonable doubt that P and it’s beyond a reasonable doubt that Q, then
*other things being equal*it’s beyond a reasonable doubt that P&Q.

- If it’s beyond a reasonable doubt that P and it’s beyond a reasonable doubt that Q, then
- wikipedia.org/wiki/Lottery_paradox
- plato.stanford.edu/entries/epistemic-paradoxes/

###### Russell’s Paradox

- Russell’s Paradox has the same form as the Barber Paradox. In the latter, the existence of a certain barber (a barber who shaves all and only those men who don’t shave themselves) leads to a contradiction and therefore is logically impossible. In the former the existence of a certain set (the set of sets not members of themselves) leads to a contradiction and therefore is logically impossible. There is no such barber. There is no such set.
- The significance of Russell’s Paradox is that it refutes a natural assumption of set theory, that every predicate defines a set. For example, “is red” defines the set of red things and “is an integer” defines the set of integers. But there is no set defined by “is not a member of itself.”
- wikipedia.org/wiki/Paradoxes_of_set_theory
- plato.stanford.edu/entries/settheory-early/
- britannica.com/topic/Russells-paradox

###### Newcomb’s Paradox

- First View
- The fact the
*Predictor*has always correctly predicted your choices is strong evidence he has correctly predicted your choice in this case. Therefore, if you open just box B, the*Predictor*will have predicted you would do so and put $1,000,000 in the box.

- The fact the
- Second view:
- The
*Predictor*has already acted. Right now there’s either $1,000,000 lying in box B or the box is empty. Your choice has no effect on which fact is the case. You should therefore open both boxes — at least you’ll get $1,000.

- The
- The key unknown is the basis for the
*Predictor’s*predictions. If the*Predictor*has figured out the laws of nature governing human behavior, he may be able to predict a person’s choice with just as much certainty as Newton’s theory predicts the orbits of the planets. On the other hand, his predictions may be no more than the ordinary predictions we make about people’s behavior. - wikipedia.org/wiki/Newcomb%27s_paradox
- plato.stanford.edu/entries/decision-causal/#NewcProb

###### Zeno’s Dichotomy Paradox

- Zeno put forth his paradox to prove that motion was impossible.
- Since we know things move, it’s more rational to reject the assumption underlying the paradox, that it’s impossible to traverse infinitely many non-overlapping finite distances in a finite period of time.
- Let the distance between A and B be one yard.
- We know things travel a yard in a finite amount of time.
- A yard consists of infinitely many non-overlapping finite distances, per Zeno’s Paradox.
- Therefore, it’s possible to traverse infinitely many non-overlapping finite distances in a finite period of time.

- Mathematics has no problem computing the infinite sum:
- 1/2, 1/4, 1/8, 1/16, 1/32, 1/64, ….

- wikipedia.org/wiki/Zeno%27s_paradoxes
- plato.stanford.edu/entries/paradox-zeno/
- britannica.com/topic/paradoxes-of-Zeno

###### Raven Paradox (Confirmation, Hempel’s Paradox)

- Suppose 10 things are in a box: 5 black, 5 red, and an unknown number of ravens. An object from the box is randomly selected.
- The selection of a non-raven is some evidence there are no ravens and therefore tends to confirm that all ravens are black.
- The non-existence of ravens logically entails all ravens are black.
- The greater the number of non-ravens selected, the more likely there are no ravens.

- The selection of a black raven proves there are ravens and tends to confirm that all ravens are black.

- The selection of a non-raven is some evidence there are no ravens and therefore tends to confirm that all ravens are black.
- Suppose it’s known there are ravens. An object from the box is randomly selected.
- The selection of a non-raven is irrelevant to the existence of ravens and makes it only slightly more likely that all ravens are black.
- The greater the number of non-ravens selected, the less likely there are any more ravens in the box.

- The selection of a black raven makes it more likely that all ravens are black.

- The selection of a non-raven is irrelevant to the existence of ravens and makes it only slightly more likely that all ravens are black.
- wikipedia.org/wiki/Raven_paradox
- plato.stanford.edu/entries/confirmation/#TwoParOthDif

###### Simpson’s Paradox

- Suppose 110 men and 110 women apply to graduate school. 20 women are accepted vs 90 men. That’s 18% vs 82%.
- Surely there’s sex discrimination.
- However, broken down by Sciences and Humanities, the data are:
- That is
- A higher percentage of women are admitted in the Science.
- A higher percentage of women are admitted in the Humanities.
- But a much larger percentage of men are admitted in Sciences and Humanities combined.

- How can this be?
- The answer:

- Many more men apply to the Sciences but women have a slightly higher acceptance rate.
- Many more women apply to the Humanities but men have a slightly lower acceptance rate.

- Simpson’s Paradox shows that the following form of argument is deductively invalid:
- The percentage of W’s is greater than the percentage M’s in each of the groups G
_{1}, G_{2}, G_{3},…G_{n}. - Therefore, the percentage of W’s is greater than the percentage M’s in G
_{1}, G_{2}, G_{3},…G_{n}combined.

- The percentage of W’s is greater than the percentage M’s in each of the groups G

- Links
- wikipedia.org/wiki/Simpson%27s_paradox#Kidney_stone_treatment
- math.kent.edu/~darci/simpson/bballexamples.html
- wikipedia.org/wiki/Simpson%27s_paradox
- plato.stanford.edu/entries/paradox-simpson/
- britannica.com/topic/Simpsons-paradox
*Coronavirus vaccines work. But this statistical illusion makes people think they don’t*, Jordan Ellenberg, WaPo

###### Bertrand’s Box Paradox

- If the probability were 1/2, the relative frequency of 2nd gold coins y against trials would approach 50%
- The relative frequency doesn’t approach 50%.
- Therefore the probability is not 1/2.

View Monty Hall Problem