# Reply to Dr. Erasmus – Part 2: Straw Man and Invalid Inference

In this post I will reply to an objection that was raised by Dr. Jacobus Erasmus against my reasoning in one of my skeptical posts about the resurrection of Jesus.

**DR. ERASMUS COMMITS THE STRAW MAN FALLACY**

The most basic problem with the objection raised by Dr. Erasmus is that he commits the all-too-common STRAW MAN fallacy.

He does NOT understand my reasoning, and as a result he *mischaracterizes* my reasoning, and then he criticizes the *inaccurate and distorted* representation of my reasoning instead of criticizing my ACTUAL reasoning. I am being charitable in assuming that Dr. Erasmus did not intentionally mischaracterize and distort my reasoning. However, this charitable assumption leads me to the conclusion that Dr. Erasmus has *a poor understanding of probability*. He mischaracterized my reasoning because he does NOT have a good understanding of probability.

In the previous post where I began my reply to Dr. Erasmus, I have already pointed out the fundamental mistake that Dr. Erasmus made: there is no hint that he noticed that I was making use of CONDITIONAL PROBABILITY in my calculations. But for anyone who has *a basic understanding of probability*, it would have been obvious that I was making use of CONDITIONAL PROBABILITY. So, because Dr. Erasmus lacks a good understanding of probability, he failed to notice the obvious and important fact that I was making use of CONDITIONAL PROBABILITY. This resulted in his mischaracterizing my reasoning, and attacking a STRAW MAN, instead of pointing out a problem in my ACTUAL reasoning.

I could go into more detail in supporting my charge that Dr. Erasmus has committed a STRAW MAN fallacy against my skeptical post on the resurrection of Jesus, but I don’t see the point. He clearly misunderstood my reasoning, so his objection does NOT address my actual reasoning.

If Dr. Erasmus wants to make a *more serious effort* to understand my reasoning, then I would be happy to discuss any *new objections* that he might offer, but his first attempt at an objection completely misses the mark, and is unworthy of any more analysis and discussion by me.

### DR. ERASMUS MAKES AN INVALID INFERENCE

Not only does Dr. Erasmus take aim at a STRAW MAN by mischaracterizing my reasoning about the resurrection, he also FAILS to support his objection about that STRAW MAN. He attacks reasoning that is NOT mine, and his attack of that other reasoning is based on an INVALID INFERENCE, and thus it FAILS. His critique is *doubly wrong*. He aimed at *the wrong target* and he also *missed the target*!

In his attempted counterexample, Dr. Erasmus needs to establish two main things about his example, in order for his example to make the objection that he wants to make:

**The probabilities of (C1) and (C2) and (C3) are low.****The probability of (H2) is high.**

Dr. Erasmus FAILS to establish either one of these key points. I will skip over the problem with the first point, and focus on his FAILURE to establish the second point. Actually, I have already described the problem with the second point in my previous post where I began my reply. So, I will be very brief here.

Dr. Erasmus uses Bayes’ Theorem to infer that the UNCONDITIONAL probability of (H2) is 0.9. I agree that a probability of 0.9 is indeed “high”. The problem is that the instance of Bayes’ Theorem that he uses only yields a CONDITIONAL PROBABILITY of (H2), not an UNCONDITIONAL probability. The condition in this case is that (C1) and (C2) and (C3) are all true.

In other words, the probability of (H2) would be 0.9 IF *we knew for certain* that (C1) and (C2) and (C3) were *all true*. But if we knew for certain that (C1) and (C2) and (C3) were all true, then it cannot also be the case that the probability of each of those three claims is low.

In any case, Dr. Erasmus does NOT understand Bayes’ Theorem well enough to grasp the obvious fact that it is an INVALID INFERENCE to conclude that the UNCONDITIONAL probability of (H2) is 0.9 based on a calculation using his instance of Bayes’ Theorem. His formula implies only that the probability of (H2) *would be* 0.9 IF *we knew for certain* that (C1) and (C2) and (C3) were *all true.*

**CONCLUSION**

Dr. Erasmus does NOT have a good understanding of probability. As a result, *he distorts my reasoning* about the resurrection and attacks a STRAW MAN instead of pointing to a problem in my ACTUAL reasoning, and his attempt to attack that STRAW MAN itself FAILS because it is based on an INVALID INFERENCE, an inference that he would not have made if he had* a better understanding* of Bayes’ Theorem.

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### UPDATE on 1/7/19

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I took another look at Dr. Erasmus’ instance of Bayes’ Theorem (“the odds form of Bayes’ Theorem”) and discovered that instead of CALCULATING the probability of the hypothesis H2, he simply ASSIGNED a probability to H2!! So, it is clear that Dr. Erasmus does NOT understand Bayes’ Theorem, and *he is very confused* about what he was doing with the theorem.

When Dr. Erasmus introduces his instance of Bayes’ Theorem, he makes this statement:

**We may now use Bayes’ Theorem to calculate the probability of H2.**

His aim, of course, is to use Bayes’ Theorem to SHOW that the probability of H2 is high, even when the probabilities of (C1), (C2), and (C3) are low. But in his example *he simply fills in the probability* for H2 on the right-hand side of the equation; he does NOT calculate that probability at all! Furthermore, he is assigns H2 the probability of .6, which is NOT what most people would consider to be a high probability! So, he shoots himself in *both feet*.

On the left-hand side of the equation is a single fraction with a CONDITIONAL probability in the numerator and a CONDITIONAL probability in the denominator:

**P(H2 | C1 & C2 & C3) / P(~H2| C1 & C2 & C3) **

This ratio of CONDITIONAL probabilities is apparently what Dr. Erasmus is attempting to calculate, because the values that he provides or assigns as inputs are *four fractions* that appear to each be *ratios of probabilities*, which is exactly what we find on the right-hand side of his instance of Bayes’ Theorem:

**[P(H2) / P(~H2)] x [P(C1|H2) / P(C1|~H2)] x [P(C2|H2) / P(C2|~H2)] x [P(C3|H2) / P(C3|~H2)]**

Here are the values that Dr. Erasmus assigns to the probabilities on the right-hand side of the equation:

**(0.6 / 0.4) x (0.7 / 0.5) x (0.8 / 0.5) x (0.9 / 0.3)**

Reading those assigned values back into the right-hand side of the equation, we see that he has assigned the following values to these probabilities:

**P(H2) = 0.6**

**P(~H2) = 0.4**

**P(C1|H2) = 0.7**

**P(C1|~H2) = 0.5**

**P(C2|H2) = 0.8**

**P(C2|~H2) = 0.5**

**P(C3|H2) = 0.9**

**P(C3|~H2) = 0.3**

When he assigned the value of 0.6 to P(H2), he was *assigning a moderate probability* to H2, the hypothesis that he was supposed to be using Bayes’ Theorem to SHOW that H2 has a high probability, even when (C1), (C2), and (C3) have low or moderate probabilities.

Instead of SHOWING that H2 has a high probability, Dr. Erasmus was SHOWING that he does NOT understand probability calculations, especially probability calculations that make use of Bayes’ Theorem.

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### UPDATE on 1/8/19

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Given the above probability values that Dr. Erasmus assigned to the elements of the right-hand side of his instance of Bayes’ Theorem, we can calculate the probabilities of of (C1), (C2), and (C3). When we do so, it turns out that NONE of them have a LOW probability, which means that his counterexample FAILS.

We know that the probability of (C1) is EQUAL to the SUM of the probability of (C1|H2) times the probability of (H2) and the probability of (C1|~H2) times the probability of (~H):

**P(C1) = [P(C1|H2) x P(H2)] + [P(C1|~H2) x P(~H2)] **

We can use the probabilities assigned by Dr. Erasmus to calculate the probability of (C1):

**P(C1) = (0.7 x 0.6) + (0.5 x 0.4) **

**P(C1) = 0.42 + 0.20**

**P(C1) = 0.62**

We can similarly infer the probabilty of (C2) using a similar formula to that used above:

**P(C2) = [P(C2|H2) x P(H2)] + [P(C2|~H2) x P(~H2)]**

We can use the probabilities assigned by Dr. Erasmus to calculate the probability of (C2):

**P(C2) = (0.8 x 0.6) + (0.5 x 0.4) **

**P(C2) = 0.48 + 0.20**

**P(C2) = 0.68**

We can also infer the probability of (C3) using a similar formula to that used above:

**P(C3) = [P(C3|H2) x P(H2)] + [P(C3|~H2) x P(~H2)] **

We can use the probabilities assigned by Dr. Erasmus to calculate the probability of (C3):

**P(C3) = (0.9 x 0.6) + (0.3 x 0.4) **

**P(C3) = 0.54 + 0.12**

**P(C3) = 0.66**

Based on the probabilities that Dr. Erasmus assigns to elements on the right-hand side of his instance of Bayes’ Theorem, the probabilities of (C1), (C2) and (C3) would be as follows:

**P(C1) = 0.62**

**P(C2) = 0.68**

**P(C3) = 0.66**

NONE of these probabilities is a LOW probability. A low probability would have to be less than .5, but all three of these probabilities are significantly greater than .5.

So, Dr. Erasmus FAILED to assign probabilities in such a way that (C1), (C2), and (C3) have low probabilities, and thus his counterexample FAILS. The point of his counterexample was to SHOW that the probability of the hypothesis H2 could be high even if the probabilities of (C1), (C2), and (C3) were low. But his assigned probability values FAIL to provide an example where the probabilities of (C1), (C2), and (C3) are all low. In his example NONE of these statements has a low probability.