The following resolutions of paradoxes deserve to be better known. (All of these paradoxes are resolved; you don’t need to worry about them any more.)

Consider the sentence “This sentence is false.” If it is true, then it is false, and if it is false, then it is true.

The Liar Paradox was resolved in Semantic Paradoxes as Equations by Lan Wen, The Mathematical Intelligencer, vol. 23, issue 1, December 2001, pp. 43–48. Professor Wen showed how to take any system of boolean equations that doesn’t have a solution and create a corresponding semantic paradox. He presents the following paradox:

The Three Cards Paradox. Consider three cards with the following three sentences:

- The sentence on the second card is true, and the sentence on the third card is false.
- Either the sentence on the first card is false, or the sentence on the third card is true.
- The sentences on the first and second cards are both true.

Note that there is no self-reference in the Three Cards Paradox, so the self-reference in the Liar Paradox is a red herring.

The Liar Paradox is the semantic equivalent of assuming that all systems of equations have solutions. To be more precise, the following algorithm is implicit in Prof. Wen’s article:

- Take any system of n boolean equations in the variables x
_{1}, …, x_{n}where equation i has x_{i}by itself on the left-hand side of the equation. - Construct n English sentences as follows: For sentence i, take the
right-hand side of boolean equation i and translate it into English by replacing each
x
_{j}with “Sentence j”, “∧” with “and”, “∨” with “or”, “→” with “if … then”, “↔” with “if and only if”, “¬” with “is false”, and the absense of “¬” with “is true”.

If the equations are complicated, the English may need some adjustment to be grammatical, but for many systems of boolean equations, this algorithm will produce English sentences that are intelligible. In particular, it works for the Liar Paradox, Three Card Paradox, Löb’s Paradox (also known as Curry’s Paradox), and the Truth-Teller, all of which are discussed and resolved in Professor Wen’s article.

If the system of boolean equations does not have a solution, then we get a paradox if we assume the
n English sentences are each either true or false. Just as the variables
x_{i} are neither true nor false, the English sentences are neither true nor false.

In other words, the semantic paradoxes are boolean equations. Starting with the English obscures this, but the algorithm above reveals where the English comes from. It is unnecesary (and incorrect) to invoke Kripke’s truth-value gaps, Tarski’s hierarchy theory, non-well-founded sets, etc. to resolve the paradoxes.

Two members of a criminal gang are arrested and imprisoned. Each prisoner is in solitary confinement with no means of communicating with the other. The prosecutors lack sufficient evidence to convict the pair on the principal charge. They hope to get both sentenced to a year in prison on a lesser charge. Simultaneously, the prosecutors offer each prisoner a bargain. Each prisoner is given the opportunity either to betray the other by testifying that the other committed the crime or to cooperate with the other by remaining silent. The offer is:

- If A and B each betray the other, each of them serves two years in prison.
- If A betrays B but B remains silent, A will be set free and B will serve three years in prison (and vice versa).
- If A and B both remain silent, both of them will only serve one year in prison (on the lesser charge).

The Nash equilibrium is for both prisoners to betray the other. So far, this isn’t a paradox. However, it is often stated or implied that the Nash equilibrium is the optimal strategy, even though both prisoners do better if they both remain silent.

The Prisoner’s Dilemma was resolved by Douglas Hofstadter. If both prisoners are rational, then they should not choose the Nash equilibrium. Because the situation is symmetric and because both prisoners are rational, whichever strategy prisoner A decides is best, prisoner B will also decide is best. Since both remaining silent is better than both betraying, each should decide to remain silent. The Nash equilibrium is not relevant because it considers situations that are impossible if both prisoners are rational (i.e., one prisoner remaining silent and the other betraying).

This is discussed in detail in Metamagical Themas: Questing for the Essence of Mind and Pattern by Douglas R. Hofstadter, Basic Books, March 1996, ISBN 978-0465045662. Also see my Prisoner’s Dilemma, Letter to the Editor, Notices of the American Mathematical Society, vol. 51, no. 7, August 2004, p. 735.

A highly superior being from another part of the galaxy presents you with two boxes, one open and one closed. In the open box there is a thousand-dollar bill. In the closed box there is either one million dollars or there is nothing. You are to choose between taking both boxes or taking the closed box only. But, there’s a catch.

The being claims that they are able to predict what any human being will decide to do. If they predicted that you would take only the closed box, then they placed a million dollars in it. But, if they predicted that you would take both boxes, they left the closed box empty. Furthermore, they have run this experiment with 999 people before, and have been right every time.

What do you do?

This is a paradox because there seem to be good arguments for both options. Considering the being’s accuracy in the past, the odds are that they will be right again, so you should take only the closed box. On the other hand, the money is already in the box, so you might as well take both.

This is basically just the Prisoner’s Dilemma again, but dressed up on one side to try to make the player think it very likely that both they and the being will choose the same strategy and on the other side to try to bring causality into it. Douglas Hofstadter’s resolution of the Prisoner’s Dilemma also resolves Newcomb’s Paradox. (I'm not sure if I read this resolution somewhere or thought of it myself. Let me know if you see it somewhere.)

Is it rational to vote? The chance of one vote affecting the election is small, so one could argue that it isn’t worth the trouble to vote. But people do vote.

There are counter arguments that say that it is rational to vote. See Voting as a rational choice: why and how people vote to improve the well-being of others by Aaron Edlin, Andrew Gelman, and Noah Kaplan, Rationality and Society, 2007, vol 19(3), pages 293–314. They argue that you should consider the benefit not just to the voter, but to the country. Also see posts on Andrew Gelman’s blog here and here. While some people may use this reasoning to justify voting, it is not a good resolution to the paradox.

Here is a good resolution to the paradox: If you believe that there are other rational voters, then Douglas Hofstadter’s resolution of the Prisoner’s Dilemma applies to voting. So, you don’t have just one vote, but rather all you rational voters are voting together. So, it is rational to vote (assuming you believe there are a significant number of rational voters).

Besides resolving the paradox, this is also very close to the arguments that many people give for why you should vote, e.g., “If no one voted, then democracy would not work.” (I have not seen this resolution anywhere else. Let me know if you see it somewhere.)

On Sunday evening a judge tells a condemned prisoner that they will be awakened and hanged on the morning of one of the following five days. The judge says that it will happen unexpectedly, i.e., the prisoner will be uncertain about when the hanging will occur until the moment the attendants arrive. But the prisoner’s attorney convinces the prisoner that no such hanging is possible. The first step in the attorney’s argument is to eliminate Friday as execution day: If the judge sets Friday as the morning of the hanging, the prisoner will know it on Thursday because they are still alive and realize that tomorrow is the final day of the execution period. So, Friday is ruled out. But then, by the process of elimination, so are Thursday, Wednesday, Tuesday, and Monday. Of course, on Wednesday, the prisoner is hauled out of bed, much to their surprise, and hanged.

The resolution is to realize that there is a difference between being able to guarantee something will happen and it happening because you got lucky. The prisoner is correct that the judge cannot guarantee that the hanging will be unexpected. (A judge's sentence is normally interpreted as a guarantee that something will happen.) But, this does not prevent the judge from getting lucky. An analogy would be if the judge said that the prisoner would be hanged in the morning and it would not rain for the whole day of the hanging. The judge cannot guarantee that it won't rain, but they might get lucky. (I have not seen this resolution anywhere else. Let me know if you see it somewhere.)

Resolutions of paradoxes should (usually) be short. If you need a whole book to explain your resolution, then it probably is not the correct resolution.

Page published 2016-09-11. Section on Liar Paradox expanded 2017-06-27.

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