Two Envelopes and Binding

ABSTRACT

This paper describes a way of defending a modification of Eckhardt's [2013] solution to the Two Envelopes Paradox. The defence is based on ideas from Arntzenius, Elga, and Hawthorne [2004].

KEYWORDS:

• two envelopes paradox
• binding
• decision theory

Notes

¹ Insert r = ³∕₄ in the result from Eckhardt [2013: 67].

² I observed 21 positive and 19 negative.

³ Using an empirical-statistical method is justified for the purpose of items I and II, because if, for each n ∈ ₀, we define the stochastic variable b – a_n to be equal to b – a if a = 2ⁿ and equal to 0 otherwise, and similarly for a – b_n, then the Law of Large Numbers holds for each b – a_n and a – b_n. This follows from the fact that these stochastic variables have well-defined, finite, expected values (see Feller [1968: 260]).

⁴ The symmetric ignorance principle is defined by Eckhardt [2013: 48] as follows: ‘an agent is symmetrically ignorant with respect to the choice of two options, if everything the agent knows about either option applies equally to both of them… The symmetrical ignorance principle states that symmetrical ignorance forestalls rational preference.’

⁵ In the interest of brevity, I based this summary on only one of Arntzenius et al.'s examples. Another that is particularly relevant for comparison with Two Envelopes is ‘Trouble in St. Petersburg’, but I leave that to the reader.

⁶ Proof. For any n > 0, the probability that the player ends up with 2ⁿ utils after having followed a strategy depends only on what that strategy prescribes regarding the situations where a is 2^n-1, 2ⁿ, or 2ⁿ⁺¹. For each of those three situations, the player can either swap or keep, which gives eight combinations. Calculating the probability of ending up with 2ⁿ utils for each of those combinations will show that all of them are different, except for the combination of swapping for all three and the combination of keeping for all three, which both yield a probability of 2^-1(3ⁿ·4^-n-1 + 3^n-1·4^-n). Something similar holds for n = 0, even though that case only gives rise to four combinations: only swapping for both a = 2⁰ and a = 2¹ and keeping for both give the same probability of ending up with 2⁰ utils. From this, it can be seen that the only two different strategies that have the same probability for all possible outcomes are the always-swap and the always-keep strategies; all other strategy-pairs will differ with respect to at least one possible outcome's probability.

⁷ This implies that, for every strategy N , there is a better strategy and there is a worse strategy: if N and n \N, then N{n} is better; if N and n , then {n} is better; if N and n , then N\{n} is worse; if N and n , then \{n} is worse.

⁸ I am grateful to Federico Luzzi and William Eckhardt for helpful discussion, and to the editor and reviewers for useful suggestions.