Fair Countable Lotteries and Reflection

Abstract

The main conclusion is this conditional: If the principle of reflection is a valid constraint on rational credences, then it is not rational to have a uniform credence distribution on a countable outcome space. The argument is a variation on some arguments that are already in the literature, but with crucial differences. The conditional can be used for either a modus ponens or a modus tollens; some reasons for thinking that the former is most reasonable are given.

Appendices

Appendix

This Appendix contains a longer and more precise version of the argument in Section 1.

The first thing to make more precise is the character of the functions. P₀ is stipulated to be a probability function on $\mathbb {N}^{2}$. Making the dependence of P₁ and P₂ on the outcomes of the lotteries explicit, these are not simply probability functions on $\mathbb {N}^{2}$, but rather functions from $\mathbb {N}^{2}$ into the space of probability functions on $\mathbb {N}^{2}$. For $n,m\in \mathbb {N}$, $s\subseteq \mathbb {N}^{2}$, and i = 1, 2, P_i(n, m)(s) represents the agent’s credence at t_i for the proposition that the ordered pair of outcomes belongs to s, if the ordered pair of outcomes is actually (n, m).

In continuity with the notation used above, I will abbreviate “{(N₁, N₂)∣ϕ}” as just “ϕ”. So, for instance, N₂ < l is the set {(N₁, N₂)∣N₂ < l}, i.e. $\mathbb {N}\times \{0,\ldots ,l-1\}$.

Premises

In the statements of the premises, all free variables are implicitly bound by initial universal quantifiers, restricted to

• ℕ for the variables “n”, “m”, “k”, and “l”,

• subsets of ℕ² for “s” and “t”,

• [0, 1] for “x”,

• {1, 2} for “i”, and

• the set of finite sets of mutually disjoint subsets of ℕ² for “S”.

The first four premises are uncontroversial principles of probability, including finite additivity:

• 1. $P_{i}(n,m)(\bigcup S)={\sum }_{s\in S}P_{i}(n,m)(s)$

• 2. $P_{i}(n,m)(\mathbb {N}^{2}\setminus s)=1-P_{i}(n,m)(s)$

• 3. P_i(n, m)(s) = 1 → P_i(n, m)(t) = P_i(n, m)(s ∩ t)

• 4. $P_{0}(\mathbb {N}^{2})=1$

The next group of premises encode the scenario stipulations concerning the events at t₁ and t₂:

• 5. P₁(n, m)(N₁ = n) = 1

• 6. P₁(n, m)(N₂ = l) = 0

• 7. P₂(n, m)(N₁ = k) = 0

• 8. P₂(n, m)(N₂ = m) = 1

The final premise comprises the relevant instances of the reflection principle:

• 9. P₀({(n, m)∣P_i(n, m)(s) = x}) = 1 → P₀(s) = x

In fact, this premise is a little more than the reflection principle. The premise does not say that if the t₀ credence of the t_i credence of s being x is 1, then the t₀ credence of s is x; but rather, that if the t₀ credence of an event that leads to the t_i credence of s being x is 1, then the t₀ credence of s is x. This is a strengthening of reflection that is justified when the agent knows which events will lead to which credences. Thus, this formal premise combines what is informally more easily thought of as several premises: the weak reflection principle, the ideal rationality of the agent, and that the agent knows at t₀ what will happen at t₁ and t₂ (except for the specific winning numbers).

Deduction

From 1 and 6:

• 10. $\forall n,m\in \mathbb {N}:P_{1}(n,m)(N_{2}\leq n)={\sum }_{i=0}^{n} P_{1}(n,m)(N_{2}=i)=0$

From 2 and 10:

• 11. $\forall n,m\in \mathbb {N}:P_{1}(n,m)(N_{2}>n)=1-P_{1}(n,m)(N_{2}\leq n)=1$

From 3:

• 12. $\forall n,m\in \mathbb {N}:P_{1}(n,m)(N_{1}=n)=1$→ P₁(n, m)(N₂ > N₁) = P₁(n, m)(N₂ > N₁ ∩ N₁ = n)

Because for all $n\in \mathbb {N}$, N₂ > N₁ ∩ N₁ = n is the same set as N₂ > n ∩ N₁ = n:

• 13. $\forall n,m\in \mathbb {N}:P_{1}(n,m)(N_{2}>N_{1}\mathrel \cap N_{1}=n)$= P₁(n, m)(N₂ > n ∩ N₁ = n)

From 3:

• 14. $\forall n,m\in \mathbb {N}:P_{1}(n,m)(N_{1}=n)=1$→ P₁(n, m)(N₂ > n ∩ N₁ = n) = P₁(n, m)(N₂ > n)

From 12, 13, and 14:

• 15. $\forall n,m\in \mathbb {N}:P_{1}(n,m)(N_{1}=n)=1$→ P₁(n, m)(N₂ > N₁) = P₁(n, m)(N₂ > n)

From 5, 11, and 15:

• 16. $\forall n,m\in \mathbb {N}:P_{1}(n,m)(N_{2}>N_{1})=1$

From 16:

• 17. $\{(n,m)\mid P_{1}(n,m)(N_{2}>N_{1})=1\}=\mathbb {N}^{2}$

From 4 and 17:

• 18. P₀({(n, m)∣P₁(n, m)(N₂ > N₁) = 1}) = 1

From 9 and 18:

• 20. P₀(N₂ > N₁) = 1

By analogous reasoning, using 7 and 8 instead of 5 and 6:

• 20. P₀(N₂ > N₁) = 0

From 19 and 20:

• 21. Contradiction