Why Bayesians Needn’t Be Afraid of Observing Many Non-black Non-ravens

Abstract

According to Hempel’s raven paradox, the observation of one non-black non-raven confirms the hypothesis that all ravens are black. Bayesians such as Howson and Urbach (Scientific reasoning: the Bayesian approach, 2nd edn. Open Court, Chicago, 1993) claim that the raven paradox can be solved by spelling out the concept of confirmation in the sense of the relevance criterion. Siebel (J Gen Philos Sci 35:313–329, 2004) disputes the adequacy of this Bayesian solution. He claims that spelling out the concept of confirmation in the relevance sense lets the raven paradox reappear as soon as numerous non-black non-ravens are observed. It is shown in this paper that Siebel’s objection to the Bayesian solution is flawed. Nevertheless, the objection made by Siebel may give us an idea of how Bayesians can successfully handle situations in which we observe more than one non-black non-raven.

Appendices

Appendix 1

The probability of P(¬Ra_i & ¬Ba_i) can be determined in the following way: There are x non-black non-ravens and δ objects that are not non-black non-ravens so that we have all in all x + δ objects to sample from. Imagine that you draw i objects at random without replacement from the x + δ objects. The probability of drawing k non-black non-ravens while the ith object is a non-black non-raven as well (so that all in all k + 1 non-black non-ravens are drawn) can then be determined by dividing the number of favourable events by the number of possible events. For the first i − 1 draws there are x over k possibilities to draw a non-black non-raven while there are δ over i − 1 − k possibilities to draw an object which is not a non-black non-raven. Thus x over k and δ over i − 1 − k have to be multiplied to get the overall number of possibilities to obtain k non-black non-ravens in the first i − 1 draws. In the ith draw there still remain x − k possibilities to draw a non-black non-raven. Hence, x over k times δ over i − 1 − k needs to be multiplied by x − k to obtain the number of favourable events. The number of possible events is determined in a similar manner. There are x + δ over i − 1 possibilities to sample i − 1 objects while there remain x + δ − (i − 1) possibilities to draw the ith object. Thus we have x + δ over i − 1 times x + δ − (i − 1) for the number of possible events. Dividing favourable events by possible events yields

$$ \frac{{\left( {\begin{array}{*{20}c} {\text{x}} \\ {\text{k}} \\ \end{array} } \right)\left( {\begin{array}{*{20}c} \updelta \\ {{\text{i}} - 1 - {\text{k}}} \\ \end{array} } \right)({\text{x}} - {\text{k}})}}{{\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta } \\ {{\text{i}} - 1} \\ \end{array} } \right)({\text{x}} + \updelta - ({\text{i}} - 1))}}. $$

The number k of non-black non-ravens drawn in the first i − 1 draws may vary between 0 and k. The event of drawing m non-black non-ravens (0 ≤ m ≤ k) in the first (i − 1)th draws as well as a non-black non-raven in the ith draw excludes the event of drawing n non-black non-ravens (0 ≤ n ≤ k) in the first (i − 1)th draws as well as a non-black non-raven in the ith draw if m ≠ n. In order to obtain the probability for sampling a non-black non-raven on the ith draw we can thus simply sum up the probability for every k (0 ≤ k ≤ i − 1):

$$ {\text{P}}(\neg {\text{Ra}}_{\text{i}} \,\& \,\neg {\text{Ba}}_{\text{i}} ) = \sum\limits_{{{\text{k}} = 0}}^{{{\text{i}} - 1}} {\frac{{\left( {\begin{array}{*{20}c} {\text{x}} \\ {\text{k}} \\ \end{array} } \right)\left( {\begin{array}{*{20}c} \updelta \\ {{\text{i}} - 1 - {\text{k}}} \\ \end{array} } \right)({\text{x}} - {\text{k}})}}{{\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta } \\ {{\text{i}} - 1} \\ \end{array} } \right)({\text{x}} + \updelta - ({\text{i}} - 1))}}}. $$

Let us now determine the probability of P(¬Ra_i & ¬Ba_i| ¬Ra₁ & ¬Ba₁&...&¬Ra_i−1 & ¬Ba_i−1). There are (x − 0)· (x − 1) ·...· (x − (i − 1)) possibilities to draw only non-black non-ravens in i draws while there are (x + δ − 0)· (x + δ − 1) · … · (x + δ − (i − 1)) possibilities to draw from the overall x + δ objects. Thus we have

$$ {\text{P}}(\neg {\text{Ra}}_{\text{i}} \,\& \,\neg {\text{Ba}}_{\text{i}} |\neg {\text{Ra}}_{ 1} \,\& \,\neg {\text{Ba}}_{ 1} {\text{\& }} \ldots {\text{\& }} \neg {\text{Ra}}_{{{\text{i}} - 1}} \,\& \,\neg {\text{Ba}}_{{{\text{i}} - 1}} ) \, = \prod\limits_{{{\text{k}} = 0}}^{{{\text{i}} - 1}} {\frac{{{\text{x}} - {\text{k}}}}{{{\text{x}} + \updelta - {\text{k}}}}}. $$

Appendix 2

$$ \begin{aligned} \sum\limits_{{{\text{k}} = 0}}^{{{\text{i}} - 1}} {\frac{{\left( {\begin{array}{*{20}c} {\text{x}} \\ {\text{k}} \\ \end{array} } \right)\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta - {\text{x}}} \\ {{\text{i}} - 1 - {\text{k}}} \\ \end{array} } \right)({\text{x}} - {\text{k}})}}{{\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta } \\ {{\text{i}} - 1} \\ \end{array} } \right)({\text{x}} + \updelta - ({\text{i}} - 1))}}} & = \frac{1}{{\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta } \\ {{\text{i}} - 1} \\ \end{array} } \right)({\text{x}} + \updelta - ({\text{i}} - 1))}}\sum\limits_{{{\text{k}} = 0}}^{{{\text{i}} - 1}} {\left( {\begin{array}{*{20}c} x \\ {\text{k}} \\ \end{array} } \right)\left( {\begin{array}{*{20}c} \updelta \\ {{\text{i}} - 1 - {\text{k}}} \\ \end{array} } \right)(x - {\text{k}})} \\ & = \frac{1}{{\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta } \\ {{\text{i}} - 1} \\ \end{array} } \right)({\text{x}} + \updelta - ({\text{i}} - 1))}}\left( {{\text{x}}\sum\limits_{{{\text{k}} = 0}}^{{{\text{i}} - 1}} {\left( {\begin{array}{*{20}c} {\text{x}} \\ {\text{k}} \\ \end{array} } \right)\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta - {\text{x}}} \\ {{\text{i}} - 1 - {\text{k}}} \\ \end{array} } \right) - } \sum\limits_{{{\text{k}} = 0}}^{{{\text{i}} - 1}} {{\text{k}}\left( {\begin{array}{*{20}c} {\text{x}} \\ {\text{k}} \\ \end{array} } \right)\left( {\begin{array}{*{20}c} \updelta \\ {{\text{i}} - 1 - {\text{k}}} \\ \end{array} } \right)} } \right) \\ & = \frac{1}{{\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta } \\ {{\text{i}} - 1} \\ \end{array} } \right)({\text{x}} + \updelta - ({\text{i}} - 1))}}\left( {{\text{x}}\overbrace {{\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta } \\ {{\text{i}} - 1} \\ \end{array} } \right)}}^{{{\text{by}}\,{\text{Vandermonde}}'{\text{s}}\,{\text{Identity}}}} - {\text{x}}\sum\limits_{{{\text{k}} = 0}}^{{{\text{i}} - 1}} {\frac{\text{k}}{\text{k}}\left( {\begin{array}{*{20}c} {{\text{x}} - 1} \\ {{\text{k}} - 1} \\ \end{array} } \right)\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta - {\text{x}}} \\ {{\text{i}} - 1 - {\text{k}}} \\ \end{array} } \right)} } \right) \\ & = \frac{\text{x}}{{\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta } \\ {{\text{i}} - 1} \\ \end{array} } \right)({\text{x}} + \updelta - ({\text{i}} - 1))}}\left( {\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta } \\ {{\text{i}} - 1} \\ \end{array} } \right) - \sum\limits_{{{\text{k}} = 0}}^{{{\text{i}} - 1}} {\left( {\begin{array}{*{20}c} {{\text{x}} - 1} \\ {{\text{k}} - 1} \\ \end{array} } \right)\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta - 1 - ({\text{x}} - 1)} \\ {{\text{i}} - 1 - 1 - ({\text{k}} - 1)} \\ \end{array} } \right)} } \right) \\ & = \frac{\text{x}}{{\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta } \\ {{\text{i}} - 1} \\ \end{array} } \right)({\text{x}} + \updelta - ({\text{i}} - 1))}}\left( {\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta } \\ {{\text{i}} - 1} \\ \end{array} } \right) - \overbrace {{\frac{{({\text{i}} - 1)}}{{({\text{x}} + \updelta )}}\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta } \\ {{\text{i}} - 1} \\ \end{array} } \right)}}^{{{\text{by}}\,{\text{Vandermonde}}'{\text{s}}\,{\text{Identity}}}}} \right) \\ & = \frac{\text{x}}{{({\text{x}} + \updelta - ({\text{i}} - 1))}}\left( {1 - \frac{{({\text{i}} - 1)}}{{({\text{x}} + \updelta )}}} \right) = \frac{{{\text{x}}({\text{x}} + \updelta - ({\text{i}} - 1))}}{{({\text{x}} + \updelta )({\text{x}} + \updelta - ({\text{i}} - 1))}} = \frac{\text{x}}{{{\text{x}} + \updelta }} \\ \end{aligned} $$

Appendix 3

With “Appendix 1” in the background it is easy to see that

$$ {\text{P}}({\text{Ra}}_{\text{i}} \,\& \,\neg {\text{Ba}}_{\text{i}} ) = \sum\limits_{{{\text{k}} = 0}}^{{{\text{i}} - 1}} {\frac{{\left( {\begin{array}{*{20}c} \upbeta \\ {\text{k}} \\ \end{array} } \right)\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta - \upbeta } \\ {{\text{i}} - 1 - {\text{k}}} \\ \end{array} } \right)(\upbeta - {\text{k}})}}{{\left( {\begin{array}{*{20}c} {{\text{x}} + \updelta } \\ {{\text{i}} - 1} \\ \end{array} } \right)({\text{x}} + \updelta - ({\text{i}} - 1))}}}. $$

By systematically replacing the relevant occurrences of “x” with “β” in “Appendix 2” we obtain a proof that P(Ra_i & ¬Ba_i) = β/(x + δ) for all i. Thus we have:

$$ {\text{P}}(\neg {\text{Ra}}_{\text{i}} |\neg {\text{Ba}}_{\text{i}} ) = \frac{{{\text{P}}(\neg {\text{Ra}}_{\text{i}} \,\& \,\neg {\text{Ba}}_{\text{i}} )}}{{{\text{P}}({\text{Ra}}_{\text{i}} \,\& \,\neg {\text{Ba}}_{\text{i}} ) + {\text{P}}(\neg {\text{Ra}}_{\text{i}} \,\& \,\neg {\text{Ba}}_{\text{i}} )}} = \frac{{\frac{\text{x}}{{{\text{x}} + \updelta }}}}{{\frac{\upbeta }{{{\text{x}} + \updelta }} + \frac{\text{x}}{{{\text{x}} + \updelta }}}} = \frac{\text{x}}{{{\text{x}} + \upbeta }}. $$