**Probabilism (initial formulation)**

- (Non-Negativity) Your credences should not be negative. In symbols: $c(A) \geq 0$, for all $A$ in $\mathcal{F}$.
- (Normalization I) Your credence in a necessarily false proposition should be 0. In symbols: $c(\bot) = 0$.
- (Normalization II) Your credence in a necessarily true proposition should be 1. In symbols: $c(\top) = 1$.
- (Finite Additivity) Your credence in the disjunction of two mutually exclusive propositions should be the sum of your credences in the disjuncts. In symbols: $c(A \vee B) = c(A) + c(B)$.

As is often pointed out, 0 and 1 are merely conventional choices. Like utilities, we can measure credences on different scales. But what are they conventional choices for? It seems to me that they must represent the lowest possible credence you can have and the highest possible credence you can have, respectively. After all, what we want Normalization I and II to say is that we should have lowest possible credence in necessary falsehoods and highest possible credence in necessary truths. It follows that Non-Negativity is not a normative constraint on your credences, which is how it is often presented. Rather, it follows immediately from the particular representation of our credences that we have chosen to. Suppose we chose a different representation, where -1 represents the lowest possible credence and 1 represents the highest. Then Normalization I and II would say that $c(\bot) = -1$ and $c(\top) = 1$, so Non-Negativity would be false.

One upshot of this is that Non-Negativity is superfluous once we have specified the representation of credences that we are using. But another is that Probabilism incorporates not only normative claims, such as Normalization I and II and Finite Additivity, but also a metaphysical claim, namely, that there is a lowest possible credence that you can have and a highest possible credence that you can have. Without that, we couldn't specify the representation of credences in such a way that we would want to sign up to Normalization I and II. Suppose that, for any credence you can have, there is a higher one than you could have. Then there is no credence that I would want to demand you have in a necessary truth--for any I demanded, it would be better for you to have one higher. So I either have to say that all credences in necessary falsehoods are rationally forbidden, or all are rationally permitted, or I pick some threshold above which any credence is rationally permitted. And the same goes, mutatis mutandis, for credences in necessary falsehoods. I'm not sure what the norm of credences would be if our credences were unbounded in one or other or both directions. But it certainly wouldn't be Probabilism.

So Non-Negativity is not a normative claim, but rather a trivial consequence of a metaphysical claim together with a conventional choice of representation. The metaphysical claim is that there is a minimal and a maximal credence; the representation choice is that 0 will represent the minimal credence and 1 will represent the maximal credence.

Next, suppose we make a different conventional choice. Suppose we pick real numbers $a$ and $b$, and we say that $a$ represents minimal credence and $b$ represents maximal credence. Then clearly Normalization I becomes $c(\bot) = a$ and Normalization II becomes $c(\top) = b$. But what of Finite Additivity? This looks problematic. After all, if $a = 10$ and $b = 30$, and $c(A) = 20 = c(\overline{A})$, then Finite Addivitity demands that $c(\top) = c(A \vee \overline{A}) = c(A) + c(\overline{A}) = 40$, which is greater than the maximal credence. So Finite Additivity makes an impossible demand on an agent who seems to have perfectly rational credences in $A$ and $\overline{A}$, given the representation.

The reason is that Finite Additivity, formulated as we formulated it above, is peculiar to very specific representations of credences, such as the standard one on which 0 stands for minimal credence and 1 stands for maximal credence. The correct formulation of Finite Additivity in general says: $c(A \vee B) = c(A) + c(B) - c(A\ \&\ B)$, for any propositions $A$, $B$ in $\mathcal{F}$. Thus, in the case we just gave above, if $c(A\ \&\ \overline{A}) = 10$, in keeping with the relevant version of Normalization I, we have $c(A \vee \overline{A}) = 20 + 20 - 10 = 30$, as required. So we see that it's wrong to say that Probabilism says that your credence in the disjunction of two mutually exclusive propositions should be the sum of your credences in the disjuncts--that's actually only true on some representation of your credences (namely, those for which 0 represents minimal credence).

Bringing all of this together, I propose the following formulation of Probabilism:

**Probabilism (revised formulation)**

- (Bounded credences) There is a lowest possible credence you can have; and there is a highest possible credence you can have.
- (Representation) We represent the lowest possible credence you have using $a$, and we represent the highest possible credence you can have using $b$.
- (Normalization I) Your credence in a necessarily false proposition should be the lowest possible credence you can have. In symbols: $c(\bot) = a$.
- (Normalization II) Your credence in a necessarily true proposition should be the highest possible credence you can have. In symbols: $c(\top) = b$.
- (Finite Additivity) $c(A \vee B) = c(A) + c(B) - c(A\ \&\ B)$, for any propositions $A$, $B$ in $\mathcal{F}$.

**Switching representations**

(i)

**Suppose $c(-)$ is a probability$_{a, b}$ function. Then $\frac{1}{b-a}c(-) - \frac{a}{b-a}$ is a probability function (or probability$_{0, 1}$ function).**

(ii) Suppose $c(-)$ is a probability function and $a, b$ are real numbers. Then $c(-)(b-a) + a$ is a probability$_{a, b}$ function.

**Dutch Book Argument**

The standard Dutch Book Argument for Probabilism assumes that, if you have credence $p$ in proposition $X$, then you will pay £$pS$ for a bet that pays £$S$ if $X$ and £$0$ if $\overline{X}$. But this assumes that you have credences between 0 and 1, inclusive. What is the corresponding assumption if you represent credences in a different scale? Shorn of its conventional choice of representation, the assumption is: (a) you will pay £$0$ for a bet on $X$ if you have minimal credence in $X$; (b) you will pay £$S$ for a bet on $X$ if you have maximal credence in $X$; (c) the price you will pay for a bet on $X$ increases linearly with your credence in $X$. Translated into a framework in which we measure credence on a scale from $a$ to $b$, the assumption is then: you will pay £$\frac{p-a}{b-a}S$ for a bet that pays £$S$ if $X$ and £$0$ if $\overline{X}$. And, with this assumption, we can find Dutch Books against any credence function that isn't a probability$_{a, b}$ function.

**Accuracy Dominance Argument**

The standard Accuracy Dominance Argument for Probabilism assumes that, for each world, the ideal or vindicated credence function at that world assigns 0 to all falsehoods and 1 to all truths. Of course, if we represent minimal credence by $a$ and maximal credence by $b$, then we'll want to change that assumption. We'll want to say instead that the ideal or vindicated credence function at a world assigns $a$ to falsehoods and $b$ to truths. Once we say that, for any credence function that isn't a probability$_{a, b}$ function, there is another credence function that is closer to the ideal credence function at all worlds.

So, the usual arguments for having a credence function that is a probability function when you represent your credences on a scale from 0 to 1 can be repurposed to argue that you should have a credence function that is a probability$_{a, b}$ function when you represent your credences on a scale from $a$ to $b$. And that gives us good reason to think that the second formulation of Probabilism above is correct.