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Introduction

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Abstract

Paradoxes are robust, widespread intellectual illusions in which seemingly compelling reasoning generates an absurd or contradictory conclusion. Paradoxes are charming, fun, and may reveal deep confusions about important philosophical matters. A solution should dispel the illusion, so that the paradoxical reasoning no longer seems compelling. The following chapters offer solutions to ten difficult philosophical paradoxes.

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Notes

  1. 1.

    Similarly, Sainsbury (2009, p. 1) defines a paradox as “an apparently unacceptable conclusion derived by apparently acceptable reasoning from apparently acceptable premises”. But see below in the text for further conditions on paradoxicality.

  2. 2.

    This is not exactly how the real game worked, but pretend the game works this way for purposes of the problem. In the real game, Monty was not required to show the contestant a goat or offer the chance to switch, and usually he did not do so (Tierney 1991).

  3. 3.

    The stipulation that Monty always opens a door with a goat behind it is sometimes erroneously omitted from the statement of the problem, as in vos Savant (1990–91) (vos Savant makes the assumption in her solution, but the original problem statement did not contain it). Without this stipulation, the correct probability is ½. That is, suppose we assume that Monty, rather than deliberately avoiding the door with the prize, simply chooses randomly which door to open, from the two doors that the contestant didn’t pick. Let h1 = [The car is behind door 1], h2 = [The car is behind door 2], h3 = [The car is behind door 3], and e = [Monty opens door 3 and there is a goat behind it]. After you have chosen door 1 but before Monty opens door 3, you should have the following credences: P(h1) = P(h2) = P(h3) = \( \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \); P(e|h1) = \( \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \); P(e|h2) = \( \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \); P(e|h3) = 0. Then the probability of door 1 having the prize behind it, given that Monty opens door 3 and reveals a goat, is given by Bayes’ Theorem as follows:

    $$ {\displaystyle \begin{array}{c}P\left({h}_1|e\right)=\frac{P\left({h}_1\right)P\left(e|{h}_1\right)}{P\left({h}_1\right)P\left(e|{h}_1\right)+P\left({h}_2\right)P\left(e|{h}_2\right)+P\left({h}_3\right)P\left(e|{h}_3\right)}\\ {}=\frac{\left(1/3\right)\left(1/2\right)}{\left(1/3\right)\left(1/2\right)+\left(1/3\right)\left(1/2\right)+\left(1/3\right)(0)}=\frac{1}{2}.\end{array}} $$

    The key is that in this version of the problem, P(e|h2) = ½. In the standard version (where Monty always avoids opening the door with the prize), P(e|h2) = 1. Substituting 1 for P(e|h2) in the above equation changes the final answer to \( \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right. \), the standard answer.

  4. 4.

    See the responses to Marilyn vos Savant’s famous column on the problem (vos Savant 1990–91).

  5. 5.

    This is essentially the approach recommended by Rescher (2001, ch. 2). But note that when discussing individual paradoxes, Rescher does not in fact rest content with rejecting the least plausible proposition. Rather, he offers explanations in each case of why the proposition to be rejected is false, explanations that would render the proposition much less plausible than it is initially.

  6. 6.

    But if you want the solution, see Huemer n.d. Hint: in order for the twins to be reunited, someone has to leave their initial, inertial reference frame.

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Huemer, M. (2018). Introduction. In: Paradox Lost. Palgrave Macmillan, Cham. https://doi.org/10.1007/978-3-319-90490-0_1

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