检测结果阳性，你有多大可能真的生病？You test positive. How likely is it that you’re sick?

01预测Predict

02数一数阳性的人Count the positives

记下预测后，左边会给 1000 个人都做一次检测：带黑圈的是检测阳性的人，橙色是真的患病的人。只留下阳性的人，看看他们当中有多少是橙色的。

Once you save a prediction, all 1,000 people on the left take the test. A dark ring means a positive result; orange means the person really is sick. Keep only the positives, then see how many of them are orange.

03解释：问题被问反了Explain: the question got turned around

按顺序看图上的四处，每一步只亮起相关的部分。Look at four places in order; each step lights up only what it needs.

1. 先看点阵：1000 人里只有 10 个橙色的病人。这就是基础比率。Start with the grid: only 10 of the 1,000 people are orange. That is the base rate.
2. 看树的左支：10 个病人里有 9 个测出阳性。这是 90% 的来处。Follow the tree’s left branch: 9 of the 10 sick people test positive. That is where the 90% comes from.
3. 再看右支：990 个健康人的 9% 也是阳性，共 89 个误报。Now the right branch: 9% of the 990 healthy people also test positive, 89 false alarms.
4. 最后只看阳性的人：9 个病人对 89 个健康人，患病的只占 9.2%。Finally keep only the positives: 9 sick against 89 healthy, so only 9.2% are sick.

“90%”回答的是另一个问题：病人里有多少会测出阳性，即 P(阳性 | 患病)。你要的是 阳性的人里有多少真的患病，即 P(患病 | 阳性)。把两者弄混，叫作“条件倒置”。

“90%” answers a different question: of the sick, how many test positive, P(positive | sick). You want of the positives, how many are sick, P(sick | positive). Mixing them up is called the confusion of the inverse.

这两个数为什么差这么远？因为健康的人太多了。1000 人里只有 10 个病人，却有 990 个健康人；9% 的误报率作用在 990 人身上，就是 89 个假阳性，远多于 9 个真阳性。这就是基础比率：检测前就已知的患病比例。忽略它，叫作基础比率谬误。

Why are the two so far apart? Because healthy people vastly outnumber sick ones. Of 1,000 people only 10 are sick and 990 are healthy; a 9% false-alarm rate applied to 990 people makes 89 false positives, swamping the 9 true ones. That starting share is the base rate, and leaving it out is the base-rate fallacy.

同一件事，用比值来算The same thing, in odds

贝叶斯法则的比值形式：检测后的比值 = 检测前的比值 × 似然比。阳性的似然比 LR⁺ = 敏感度 ÷ 误报率，表示一个阳性结果把“患病对健康”的比值放大多少倍。

Bayes’ rule in odds form: posterior odds = prior odds × likelihood ratio. For a positive result, LR⁺ = sensitivity ÷ false-positive rate: how many times a positive multiplies the odds of sick to healthy.

为什么数人头更容易Why counting people helps

1978 年，研究者问哈佛医学院附属医院的医生和学生一道类似的题，最常见的答案是 95%，正确答案约 2%。Gigerenzer 与 Hoffrage 在 1995 年发现，把同样的题改写成“1000 人中有几个”的自然频数，答对的人大约多了两倍。左边的点阵就是自然频数。

In 1978 researchers gave a similar problem to physicians and students at Harvard-affiliated hospitals. The most common answer was 95%; the correct one was about 2%. Gigerenzer and Hoffrage found in 1995 that rewriting the same problems as natural frequencies (“so many out of 1,000 people”) roughly tripled the share of correct answers. The dot grid on the left is a natural-frequency picture.

04改变基础比率Change the base rate

保持检测不变，只把病变得更常见：同样一个阳性，意义完全不同。再试试提高敏感度和降低误报率，哪个对阳性结果的可信度影响更大？

Keep the test fixed and make the disease more common: the same positive now means something quite different. Then try raising sensitivity versus lowering the false-positive rate. Which moves the meaning of a positive more?

05再测一次Test again

把所有阳性的人再测一次。第一次检测后的概率，就成了第二次检测前的基础比率。每多一个独立的阳性，比值就再乘一次似然比。

Retest everyone who was positive. The probability after the first test becomes the base rate for the second. Each further independent positive multiplies the odds by the likelihood ratio once more.

这假设两次检测在“已知是否患病”的条件下相互独立。现实中，同一个原因（比如交叉反应）常会让同一个人反复误报，真实的提升会小得多。This assumes the two tests are independent given disease status. In practice the same cause (a cross-reacting antibody, say) often makes the same person test falsely positive twice, so the real gain is smaller.

06独立练习Independent practice

三道新题，提示一道比一道少：第一题只有点阵，没有数字；第二题只有空白的树；第三题没有图。

Three new cases, each with less support: the first gives you only the dots, no numbers; the second a blank tree; the third no figure at all.

07迁移：机场安检Transfer: airport screening