<p>I think you are doing the computation correctly, but Gigerenzer and Blank did not provide us with the full results of their experiment, preventing us from repeating their computations exactly: The data provided in columns 1 and 2 of the table are only the <strong>averages</strong>. The data in column 4 (Bayesian) is not a transformation of the average value using some formula (see below), but <strong>the original data transformed, and then averaged</strong>:</p>

<blockquote>
  <p>For each subject in the 70% base rate group we predicted, by using Bayes's theorem, how this subject would have responded in the 30% base rate group</p>
</blockquote>

<p>If you divide the first equation of the footnote by the second, and derive a formula for p30 as a function of p70, you get:</p>

<p>$$p_{30}(E|D) \approx \frac{p_{70}(E|D)}{5.44-4.44p_{70}(E|D)}$$</p>

<p>Notice that this function is strictly convex in the region [0,1], meaning that if you compute it on the average p70, you'll get a value that is lower than the value you would get by averaging the values for several data-points. </p>

<p>As an example, lets examine the Jack row: If you transform the mean p70 of 0.813, you get p30 of 0.44, which has a deviation of 0.27 from the observed mean p30 (as you probably found). Let's assume that the data in the Jack row, in the p70 column was obtained from two subjects, that responded 0.968 and 0.658 (yielding the average 0.813 that is reported in the table). Transform both of these values, and you'll get 0.848 and 0.261, respectively, yielding an average p30 of 0.553, which would give a Bayesian deviation of 0.161 exactly as reported in column 4.</p>

CC BY-SA 3.0
2012-08-07T22:32:58.110Z
2012-08-08T07:21:06.460Z
2012-08-08T07:21:06.460Z
Ofri Raviv
<p>I am having trouble understanding Table 1 of Gigerenzer, Hell, and Blank (1988, <a href="http://www.mpib-berlin.mpg.de/volltexte/institut/dok/full/gg/presenta/presenta.pdf" rel="noreferrer">PDF</a>, table on page 516):
<img src="https://i.sstatic.net/0bSxr.jpg" alt="table 1 gigerenzer"></p>

<p>Focusing on the Jack row, it is stated that the mean probabilities of Jack being an engineer were 71.4% in a low base rate condition and 81.3% in a high base rate condition. Thus participants partially took base rates into account. If they were completely ignoring base rates these values would be the same, rather than being separated by 9.9%. So I believe I understand what is going on in the Base Rate Neglect column.</p>

<p>However, I don't understand what is going on in the Bayesian column. There is an explanatory footnote on p516 but to me it was not illuminating. The footnote provides the following equations:</p>

<p>$$p_{70}(E|D)/(1-p_{70}(E|D)) = L\frac{0.7}{0.3}$$
$$p_{30}(E|D)/(1-p_{30}(E|D)) = L \frac{0.3}{0.7}$$</p>

<p>So, since the mean guess for the 70% condition was 81.3% we get </p>

<p>$$
\begin{align}
81.3/18.7 &amp; = L ~ 0.7/0.3 &amp;  \\
      &amp; = 81.3/18.7 &amp; = L ~2.33 \\
      &amp; = 4.35 &amp;= L ~2.33 \\
      &amp; = 1.86 &amp;= L \\
\end{align}
$$</p>

<p>However, then when I plug this $L$ value into the $p_{30}$ equation I get a value for $p_{30}(E|D)$ that deviates wildly both from the 'Bayesian' value and from what was actually observed. According to the table the average deviation from the Bayesian should be 16.1%, but according to my calculations it is much higher. It would be greatly appreciated if someone could tell me what I am doing wrong.</p>

<h3>References</h3>

<ul>
<li>Gigerenzer, G., Hell, W. &amp; Blank, H. (1988). Presentation and content: The use of base rates as a continuous variable.. Journal of Experimental Psychology: Human Perception and Performance, 14, 513. <a href="http://www.mpib-berlin.mpg.de/volltexte/institut/dok/full/gg/presenta/presenta.pdf" rel="noreferrer">PDF</a></li>
</ul>

CC BY-SA 3.0
2012-08-06T18:07:38.800Z
This would probably receive more attention on [Stats.SE](http://stats.stackexchange.com/).
eykanal
CC BY-SA 3.0
2012-08-08T08:27:21.100Z
@eykanal I think it would be on topic on both sites.  Questions at the interface of quantitative methods and psychology/cognitive science are most welcome here as seen by the upvotes.
Jeromy Anglim
<cognitive-psychology><decision-making><bayesian>
Problem understanding the calculation of normative (Bayesian) base rates
user1205901 - Слава Україні
# Problem understanding the calculation of normative (Bayesian) base rates

(Asked by: username_0 on 2012-08-06)

I am having trouble understanding Table 1 of Gigerenzer, Hell, and Blank (1988, [PDF](http://www.mpib-berlin.mpg.de/volltexte/institut/dok/full/gg/presenta/presenta.pdf), table on page 516):
![table 1 gigerenzer](https://i.sstatic.net/0bSxr.jpg)

Focusing on the Jack row, it is stated that the mean probabilities of Jack being an engineer were 71.4% in a low base rate condition and 81.3% in a high base rate condition. Thus participants partially took base rates into account. If they were completely ignoring base rates these values would be the same, rather than being separated by 9.9%. So I believe I understand what is going on in the Base Rate Neglect column.

However, I don't understand what is going on in the Bayesian column. There is an explanatory footnote on p516 but to me it was not illuminating. The footnote provides the following equations:

$$p\_{70}(E|D)/(1-p\_{70}(E|D)) = L\frac{0.7}{0.3}$$
$$p\_{30}(E|D)/(1-p\_{30}(E|D)) = L \frac{0.3}{0.7}$$

So, since the mean guess for the 70% condition was 81.3% we get

$$
\begin{align}
81.3/18.7 & = L ~ 0.7/0.3 & \\
& = 81.3/18.7 & = L ~2.33 \\
& = 4.35 &= L ~2.33 \\
& = 1.86 &= L \\
\end{align}
$$

However, then when I plug this $L$ value into the $p\_{30}$ equation I get a value for $p\_{30}(E|D)$ that deviates wildly both from the 'Bayesian' value and from what was actually observed. According to the table the average deviation from the Bayesian should be 16.1%, but according to my calculations it is much higher. It would be greatly appreciated if someone could tell me what I am doing wrong.

### References

* Gigerenzer, G., Hell, W. & Blank, H. (1988). Presentation and content: The use of base rates as a continuous variable.. Journal of Experimental Psychology: Human Perception and Performance, 14, 513. [PDF](http://www.mpib-berlin.mpg.de/volltexte/institut/dok/full/gg/presenta/presenta.pdf)

**Tags:** cognitive-psychology, decision-making, bayesian

> **Comments:**
> **username_1**: This would probably receive more attention on [Stats.SE](http://stats.stackexchange.com/).  
> **username_2**: @username_1 I think it would be on topic on both sites.  Questions at the interface of quantitative methods and psychology/cognitive science are most welcome here as seen by the upvotes.  

---

## 1 answers

### Answer by username_3 on 2012-08-07. Score: 5 | Accepted answer

I think you are doing the computation correctly, but Gigerenzer and Blank did not provide us with the full results of their experiment, preventing us from repeating their computations exactly: The data provided in columns 1 and 2 of the table are only the **averages**. The data in column 4 (Bayesian) is not a transformation of the average value using some formula (see below), but **the original data transformed, and then averaged**:

> For each subject in the 70% base rate group we predicted, by using Bayes's theorem, how this subject would have responded in the 30% base rate group

If you divide the first equation of the footnote by the second, and derive a formula for p30 as a function of p70, you get:

$$p\_{30}(E|D) \approx \frac{p\_{70}(E|D)}{5.44-4.44p\_{70}(E|D)}$$

Notice that this function is strictly convex in the region [0,1], meaning that if you compute it on the average p70, you'll get a value that is lower than the value you would get by averaging the values for several data-points.

As an example, lets examine the Jack row: If you transform the mean p70 of 0.813, you get p30 of 0.44, which has a deviation of 0.27 from the observed mean p30 (as you probably found). Let's assume that the data in the Jack row, in the p70 column was obtained from two subjects, that responded 0.968 and 0.658 (yielding the average 0.813 that is reported in the table). Transform both of these values, and you'll get 0.848 and 0.261, respectively, yielding an average p30 of 0.553, which would give a Bayesian deviation of 0.161 exactly as reported in column 4.


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