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Bernoulli Semantics and Ordinal Semantics for Conditionals

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Abstract

Conditionals with conditional constituents pose challenges for the Thesis, the idea that the probability of a conditional is the corresponding conditional probability. This note is concerned with two proposals for overcoming those challenges, both inspired by early work of van Fraassen: the Bernoulli Semantics associated with Stalnaker and Jeffrey, and augmented with a mechanism for obtaining “local probabilities” by Kaufmann; and a proposal by Bacon which I dub Ordinal Semantics. Despite differences in mathematical details and emphasis of presentation, both proposals lend themselves for use as a basis for a modal-theoretic interpretation of embedded conditionals.

The goal of this note is to compare the two frameworks by implementing a model for the interpretation of conditionals in each, based on the same underlying probability model for non-conditional sentences. I show that in the Ordinal model, certain sentences are assigned probabilities that do not accord with intuitions. This problem is familiar from the literature on Bernoulli models and can be addressed by introducing Kaufmann-style local probabilities into Ordinal models. I then show that Bernoulli Semantics has other limitations, in that it assigns probabilities in violation of the Thesis to certain very complex formulas. The upshot is that a fusion of the theories may be our best shot at getting the predictions right.

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Notes

  1. Khoo modifies the interpretation rule for right-nested conditionals to enforce the Import-Export Principle. See [18, 19] for relevant discussion.

  2. Much of Khoo’s discussion centers around the validity of the principle (CSO), which plays a role in Stalnaker’s result:

    $$ A \Rightarrow B, B \Rightarrow A, A \Rightarrow C \models B \Rightarrow C $$
    (CSO)

    Khoo seeks to uphold (CSO) because he finds it intuitively valid. I agree with the intuition, but in a probabilistic setting I take the relevant notion to be p-validity: roughly speaking, an inference is p-valid if the conclusion cannot be unlikely while all the premises are likely [1, 3]. And (CSO) is p-valid (see [6], for a recent proof).

  3. These sentences are cited verbatim from Khoo, including the disorienting oscillation between Past and Present tense. Given his wording, it is not clear whether the game is still ongoing or already completed. I will largely stay with Past tense in my own subsequent discussion.

  4. While Khoo does not give us Pr(C), some bounds are implied. First, note that \(\Pr (H)\) and \(\Pr (HWC)\) are fixed at .25 and .0475, respectively:

    $$ \begin{array}{@{}rcl@{}} \Pr(HWC) = \Pr(H)\Pr(W{\mid}H)\Pr(C{\mid}HW) = .25 \times .2 \times .96 = .0475 \end{array} $$
    (i)

    From this it follows that \(\Pr (W{\mid }HC)\) and \(\Pr (C{\mid }H)\) are inversely proportional:

    $$ \begin{array}{@{}rcl@{}} .0475 = .25 \times \Pr(C{\mid}H) \times \Pr(W{\mid}HC) \end{array} $$
    (ii)

    In other words, the more likely it is that Paul cheated if he had the weaker hand, the less likely it is that that cheating was successful. Now, Khoo does say that cheating with the weaker hand raises Paul’s chances of winning, but he does not say how high or by how much. Assuming that cheating with the weaker hand makes him more likely to win than to lose (i.e., \(\Pr (W{\mid }HC) > .5\)), we must conclude that Paul is rather unlikely to have cheated if he had the weaker hand (\(\Pr (C{\mid }H) < .38\)), and the prior probability that he cheated is less than .245:

    $$ \begin{array}{@{}rcl@{}} \Pr(C) = \Pr(H)\Pr(C{\mid}H) + \Pr(\overline{H} )\Pr(C{\mid} \overline{H} ) < .25 \times .38 + .75 \times .2 = .245 \end{array} $$
    (iii)
  5. It is worth noting in this connection that a link between conditionals of the form (AB) ⇒ C and the conditional probability \(\Pr (C{\mid }AB)\) has been suggested before by [9].

  6. Khoo’s definition says that the conditional is interpreted as \((A \supset B) \Rightarrow C\) at sequences at which AB is true. But that collapses into C by strong centering and since AB entails \(A \supset B\). Khoo does not seem to take this into account, as a result the probabilities he derives are slightly off.

  7. First, the sequences at which AB is true fall into two disjoint subsets, AB and \( \overline {A} (A \Rightarrow B)\). Thus:

    $$ \begin{array}{@{}rcl@{}} && \Pr(((A \Rightarrow^{K} B) \Rightarrow C) \wedge (A \Rightarrow B)) \\ && = \Pr(ABC) + \Pr(\overline{A} C) [ \Pr(AB) + \Pr(\overline{A} )\Pr(AB) + \Pr(\overline{A} )^{2}\Pr(AB) + {\ldots} ] \\ && = \Pr(ABC) + \Pr(\overline{A} C) \Pr(B{\mid}A) \end{array} $$
    (i)

    Second, at sequences at which AB is false, so is AB, thus the value of ABC is that of C at the first AB-world.

    $$ \begin{array}{@{}rcl@{}} && \Pr(((A \Rightarrow^{K} B) \Rightarrow C) \wedge (\overline{A \Rightarrow B} )) \\ && = \Pr(C{\mid}AB) [ \Pr(A \overline{B} ) + \Pr(\overline{A} )\Pr(A \overline{B} ) + \Pr(\overline{A} )^{2} \Pr(A \overline{B} ) + {\ldots} ] \\ && = \Pr(C{\mid}AB) \Pr(\overline{B} {\mid}A) \end{array} $$
    (ii)

    The probability in (A7) is the sum of (i) and (ii).

  8. Proof:

    $$ \begin{array}{@{}rcl@{}} \text{(A9)} &=& \Pr(C{\mid}AB)\Pr(A) + \Pr(\overline{A} C)\Pr(B{\mid}A) + \Pr(C{\mid}AB)\Pr(\overline{B} {\mid}A)\Pr(\overline{A} ) \\ &=& \Pr(C{\mid}AB)\Pr(A) + \Pr(\overline{A} C)\Pr(B{\mid}A) \\ &&+ \Pr(C{\mid}AB)\Pr(\overline{B} {\mid}A) - \Pr(C{\mid}AB)\Pr(\overline{B} {\mid}A)\Pr(A) \text{ since } \Pr(\overline{A} ) = 1 - \Pr(A) \\ &=& \Pr(C{\mid}AB)\Pr(A) + \Pr(\overline{A} C)\Pr(B{\mid}A) + \Pr(C{\mid}AB)\Pr(\overline{B} {\mid}A) \\ && - \Pr(C{\mid}AB)\Pr(A) + \Pr(C{\mid}AB)\Pr(B{\mid}A)\Pr(A) \text{ since } \Pr(\overline{B} {\mid}A) = 1 - \Pr(B{\mid}A) \\ &=& \Pr(\overline{A} C)\Pr(B{\mid}A) + \Pr(C{\mid}AB)\Pr(\overline{B} {\mid}A) + \Pr(ABC) = \text{(A7)} \end{array} $$
  9. Notice that Khoo’s probability will generally be close to \(\Pr (C{\mid }AB)\):

    $$ \begin{array}{@{}rcl@{}} \Pr(C{\mid}AB) = \Pr(C{\mid}AB)\Pr(A) + [ \Pr(C{\mid}AB)\Pr(B{\mid}A) + \Pr(C{\mid}AB)\Pr(\overline{B} {\mid}A)]\Pr(\overline{A} ) \end{array} $$

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Acknowledgments

I am grateful to the editor and the reviewer for very helpful comments on an earlier version. All remaining errors and misrepresentations are my own. This work was supported in part by the National Science Foundation (#2116972, “Research on conditional and modal language”).

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Appendix: Khoo on left-nested conditionals

Appendix: Khoo on left-nested conditionals

An anonymous reviewer suggested that I address a recent criticism of the Bernoulli approach to left-nested conditionals of the form (AB) ⇒ C. The criticism concerns cases in which none of A,B,C contain further embedded conditionals, thus the issue is not directly relevant to the comparison between Bernoulli Semantics and Ordinal Semantics since both agree on those conditionals. Nevertheless, I welcome this opportunity to discuss the criticism because it purports to identify an incorrect prediction of Bernoulli Semantics (hence presumably also of Ordinal Semantics). I will argue that the purported weakness is in fact an advantage.

The criticism is presented in [24] and [23]. I largely base my discussion on the latter. Khoo’s approach relies on sequences of worlds and bears some resemblances to Bernoulli semantics, but there are also important differences. For Khoo, the sequences represent Stalnakerian similarity orderings, rather than series of random picks. Thus a Khoo sequence contains each world exactly once, in contrast to Bernoulli sequences, in which worlds may occur repeatedly or not at all.

In Khoo’s framework, as in Bernoulli Semantics, first-order conditionals are interpreted by evaluating the consequent at the first world in the sequence at which the antecedent is true. However, Khoo introduces special interpretation rules for both right-nested and left-nested conditionals. The former need not concern us here.Footnote 1 The latter presents a challenge for his overall framework. While Khoo does not assume that conditionals are propositional – in his models, like in Bernoulli models, they can differ in truth value between sequences with the same base world – he does assume that their antecedents are propositional. That requirement is problematic for antecedents which themselves contain conditionals, as was first pointed out by [30]. I am not going to repeat that argument here (see also [10], Section 6.5). Bernoulli Semantics does not have the same problem because it does not require antecedents to be propositional. In Khoo’s framework, however, the upshot is that the Thesis cannot hold in general for left-nested conditionals.Footnote 2 Khoo argues that this is as it should be, on the grounds that the Thesis-compliant conditional probability is not always intuitively correct.

This last claim deserves some discussion because it has some prima facie plausibility. But it is ultimately on the wrong track (or so I argue). Khoo produces two examples in which left-nested conditionals receive probabilities that are different from what is predicted under the Bernoulli approach. As in Bacon’s vase example discussed in this paper, at issue is again the question whether such perceived violations are counterexamples to the semantic account, or rather evidence of an interaction between the standard semantics and contextually given information. One of Khoo’s examples is exactly parallel to one discussed at length in Kaufmann ([18], Section 5.6.1) and susceptible to the same analysis. Here I focus on the other example.

Poker Paul. Paul is playing poker against Nancy. Nancy has a weak hand, but it is still possible Paul’s is weaker. It is also possible for Paul to win even if he has the weaker hand, but since Nancy is a good player, that is unlikely. Cheating is a way to increase your success of winning with a weaker hand, and Paul is not opposed to cheating, though it is unlikely he cheated if he had the better hand. But, if Paul won with the weaker hand, it is overwhelmingly likeliy that he cheated. Now consider:

$$ \text{If Paul won if he had the weaker hand, he cheated.} $$
(A1)

Khoo argues that (A1) is “very likely” in this situation. For concreteness he supplies the following numbers, along with the corresponding paraphrases.Footnote 3

$$ \begin{array}{@{}rcl@{}} \Pr(H) & = .25 & \text{(Paul likely does not have the weaker hand)} \\ \Pr(W{\mid}H) & = .2 & \text{(Paul likely did not win if he had the weaker hand)} \\ \Pr(C{\mid} \overline{H} ) & = .2 & \text{(Paul likely did not cheat if he did not have the weaker hand)}\\ \Pr(C{\mid}HW) & = .95 & \text{(Paul likely cheated if he had the weaker hand and won)} \end{array} $$
(A2)

Based on these numbers, the conditional probability of C given HW is given in (A4). We already saw that this is also the probability that the conditional (A1) receives in a Bernoulli model with these probabilities.

$$ \begin{array}{@{}rcl@{}} \Pr(C{\mid}H \Rightarrow W) && = \Pr(C{\mid}HW)P(H) + \Pr(C \overline{H} )\\ && = \Pr(C{\mid}HW)P(H) + \Pr(\overline{H} )\Pr(C{\mid} \overline{H} )\\ && = .95 \times .25 + .75 \times .2 = .3875 \end{array} $$
(A3)

In Khoo’s judgment, this is too low for the conditional: he finds that (A1) is “very likely” in the given scenario. On the other hand, it is less clear (to me) what Khoo thinks of the conditional probability of C, given HW. He presents the example as evidence against the Thesis, thus he apparently accepts that the conditional probability is .3875. However, his justification for the claim that the conditional is very likely sounds much like a paraphrase of his reasoning about C after (hypothetically) learning HW: “Paul’s winning if he had the weaker hand most likely had to do with his cheating (since he was likely not good enough to beat Nancy with a weaker hand).” If this is Khoo’s reasoning in evaluating the conditional but not in calculating the conditional probability, I would like to know more about the difference between them. Absent that, I am not sure whether he thinks that both \(\Pr (C{\mid }H \Rightarrow W)\) and \(\Pr ((H \Rightarrow W) \Rightarrow C)\) should be high, or only the latter.

However that may be, my contention is that both of these probabilities are in fact .3875 – higher than the prior probability that Paul cheated, but not much higher.Footnote 4 Thus in this case, learning that Paul won if he had the weaker hand raises the probability that he cheated from less than .245 to .3875. I suspect that Khoo’s intuition that the probability ought to be higher is based on reasoning that involves a step not licensed by the scenario as given. Suppose the conditional antecedent is true. I concur with Khoo’s intuition that “Paul’s winning if he had the weaker hand most likely had to do with his cheating”, but we cannot conclude that he likely cheated. All we can conclude is that he likely cheated if he had the weaker hand. For the scenario explicitly stipulates that whether Paul cheated depended on his hand; thus the supposition that HW is true does not affect the (low) conditional probability that he cheated in the (likely) event that he had the stronger hand.

Thus while only Khoo knows for sure what is in Khoo’s mind, my best guess is that he implicitly interprets the conditional along the lines of (A4).

$$ \text{If Paul won if he had the weaker hand, he cheated}\ \textit{[if he had the weaker hand]}. $$
(A4)

The two conditional constituents of (A4) share the same antecedent. For this reason, the formula for calculating the probability of the sentence can be simplified as in (A6). The result is indeed a much higher probability.

$$ \begin{array}{@{}rcl@{}} && \Pr((H \Rightarrow W) \Rightarrow (H \Rightarrow C))\\ && = \frac{\Pr(HWHC) + \Pr(C{\mid}H)\Pr(HW \overline{H} ) + \Pr(W{\mid}H)\Pr(\overline{H} HC)} {\Pr(H \cup H)\Pr(W{\mid}H)} \\ && = \Pr(HWC) / \Pr(HW) = \Pr(C{\mid}HW) = .95 \end{array} $$
(A5)

From a linguistic perspective, it is not implausible that the conditional should be prone to reinterpretation in this way. It has long been observed that conditional antecedents, aside from their truth-conditional role within their sentences, tend to set up topics which participate in the structuring of the surrounding discourse context ([7, 13, 28], i.a.). In the case of (A4), the overarching discourse topic would be a certain property of propositions – specifically, the property of being true if Paul had the weaker hand. Within this context, the sentence states, in effect, that if Paul won has this property, then so does Paul cheated. The fact that the scenario stipulates that whether Paul cheated depended on his hand may facilitate this interpretation.

To be sure, embedded conditionals have not been discussed in the linguistic literature on topichood and discourse structure, and the above argument must remain speculative at this point. It certainly warrants further investigation.Footnote 5

Be that as it may, it is instructive to compare this hypothesis to Khoo’s own way around the problem with (A1). Khoo pursues two goals: to treat the conditional antecedent of (AB) ⇒ C as propositional, and to derive a high probability for (A1) in the scenario given above. His proposed solution is a radical kind of contextualism, according to which the antecedent denotes different propositions depending on the world sequence (not just the world) at which it is interpreted. Specifically, a conditional AB which occurs in a conditional antecedent is interpreted as either the material conditional \(A \supset B\) or the conjunction AB:Footnote 6

$$ \begin{array}{l} \textbf{Khoo's pointwise interpretation}\\ \quad(A \Rightarrow^{K} B) \Rightarrow C\ \text{is interpreted at}\ \sigma~ as: \left\{\begin{array}{lr} C & \text{ if } A \Rightarrow B \text{ is true at }\sigma\\ AB \Rightarrow C & \text{otherwise}{\kern48pt} \end{array}\right. \end{array} $$
(A6)

The expectation of these values is as in (A7); details are given in a footnote.Footnote 7

$$ \begin{array}{@{}rcl@{}} && \Pr((A \Rightarrow^{K} B) \Rightarrow C) \\ && = \Pr(ABC) + \Pr(\overline{A} C)\Pr(B{\mid}A) + \Pr(C{\mid}AB)\Pr(\overline{B} {\mid}A) \end{array} $$
(A7)

In the given scenario, this comes out to the probability in (A8).

$$ \begin{array}{@{}rcl@{}} && \Pr((H \Rightarrow W) \Rightarrow C) \\ && = \Pr(HWC) + \Pr(C \overline{H} )\Pr(W{\mid}H) + \Pr(C{\mid}HW)\Pr(\overline{W} {\mid}H) \\ && = .0475 + .15 \times .2 + .95 \times .8 = .8375 \end{array} $$
(A8)

Incidentally, (A7) is equivalent to (A9), which highlights the similarity to the \(\Pr (C{\mid }AB)\) I suggested above:Footnote 8,Footnote 9

$$ \begin{array}{@{}rcl@{}} \Pr(C{\mid}AB)\Pr(A) + [ \Pr(C{\mid} \overline{A} )\Pr(B{\mid}A) + \Pr(C{\mid}AB)\Pr(\overline{B} {\mid}A)]\Pr(\overline{A} ) \end{array} $$
(A9)

In sum, while I think Khoo has a point regarding the intuive probability of the sentence, this judgment is not a counterargument to Bernoulli Semantics. On the contrary, I maintain that Bernoulli Semantics is correct and the judgment is the result of an interaction with pragmatic factors. Khoo’s ad-hoc semantic redefinition does not do justice to either of those components.

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Kaufmann, S. Bernoulli Semantics and Ordinal Semantics for Conditionals. J Philos Logic 52, 199–220 (2023). https://doi.org/10.1007/s10992-022-09670-8

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