# 用 Python 看條件機率

比較運算、述詞、蒙提霍爾問題，以及依證據更新，採 CS61A 的風格。

### Python 會顯示什麼？比較一個串列

*單選 · zh-hant*

Python 會顯示什麼？

```python
>>> child1 = [1, 2, 1]
>>> child1 == 1
```

- [ ] elementwise: `[True, False, True]`
- [ ] false: `False`
- [ ] error: `TypeError` 錯誤

你的答案: false

#### 提示

1. Python 的 `==` 會拿左邊的整個值和右邊的整個值比較。

2. 一個串列可能等於一個數字嗎？

#### 參考解答

`False`。Python 把整個串列 `[1, 2, 1]` 和數字 `1` 比較，兩者不相等。它不會逐元素比較；在 R 裡，`child1 == 1` 則會得到 `TRUE FALSE TRUE`。要得到 `[True, False, True]`，請寫 `[c == 1 for c in child1]`。

來源: [Adapted from Blitzstein & Hwang, Introduction to Probability, Chapter 2 and §2.10 (R), rewritten in Python](http://stat110.net) · Original exercise; the book's R examples are paraphrased, not copied

### Python 會顯示什麼？用 and 計數

*填空 · zh-hant*

Python 會顯示什麼？

```python
>>> pairs = [(1, 1), (1, 2), (2, 1), (1, 1)]
>>> sum([c1 == 1 and c2 == 1 for c1, c2 in pairs])
```

你的答案: 2 (允許的絕對誤差： 0)

#### 提示

1. 生成式的結果是 `[True, False, False, True]`。

2. `sum` 把每個 `True` 當作 1。

#### 參考解答

`2`。只有第一組和最後一組的兩個值都是 1，所以串列是 `[True, False, False, True]`，加總為 2。這就是 $A$＝兩個都是女孩時的 $n_{AB}$。

來源: [Adapted from Blitzstein & Hwang, Introduction to Probability, Chapter 2 and §2.10 (R), rewritten in Python](http://stat110.net) · Original exercise; the book's R examples are paraphrased, not copied

### 估計條件機率

*Python · zh-hant*

完成 `conditional_frequency(outcomes, a, b)`。`a` 和 `b` 是述詞，也就是接收一個結果、傳回 `True` 或 `False` 的函式。請傳回 $n_{AB}/n_B$：兩者都成立的結果數，除以 `b` 成立的結果數。如果 `b` 從未成立，傳回 `None`。

```python
def conditional_frequency(outcomes, a, b):
    """Estimate P(A | B) as n_AB / n_B over the list OUTCOMES.

    A and B are predicates: functions from an outcome to True or False.
    Return None when B never occurs.

    >>> families = [(1, 1), (1, 2), (2, 1), (2, 2)]
    >>> conditional_frequency(families, lambda f: f == (1, 1), lambda f: f[0] == 1)
    0.5
    """
    pass
```

#### 公開測試

Elder child is a girl

```pycon
>>> families = [(1, 1), (1, 2), (2, 1), (2, 2)]
>>> both_girls = lambda f: f == (1, 1)
>>> conditional_frequency(families, both_girls, lambda f: f[0] == 1)
0.5
```

At least one girl

```pycon
>>> families = [(1, 1), (1, 2), (2, 1), (2, 2)]
>>> both_girls = lambda f: f == (1, 1)
>>> round(conditional_frequency(families, both_girls, lambda f: 1 in f), 3)
0.333
```

B never happens

```pycon
>>> families = [(1, 1), (1, 2), (2, 1), (2, 2)]
>>> both_girls = lambda f: f == (1, 1)
>>> conditional_frequency(families, both_girls, lambda f: False) is None
True
```

Other outcomes

```pycon
>>> rolls = [1, 2, 3, 4, 5, 6]
>>> conditional_frequency(rolls, lambda r: r == 6, lambda r: r % 2 == 0)
0.3333333333333333
```

#### 提示

1. 用 `sum(1 for o in outcomes if b(o))` 數出 $n_B$。

2. 用同樣的方式數出 $n_{AB}$，條件改成 `a(o) and b(o)`。

3. 除法之前先檢查 $n_B = 0$ 的情況。

#### 參考解答

```python
def conditional_frequency(outcomes, a, b):
    n_b = sum(1 for o in outcomes if b(o))
    n_ab = sum(1 for o in outcomes if a(o) and b(o))
    return None if n_b == 0 else n_ab / n_b
```

這個函式對小孩或骰子一無所知，只會呼叫傳進來的述詞；正因如此，一個定義就能回答筆記中的每個問題。

來源: [Adapted from Blitzstein & Hwang, Introduction to Probability, Chapter 2 and §2.10 (R), rewritten in Python](http://stat110.net) · Original exercise; the book's R examples are paraphrased, not copied

### 蒙提會開哪一扇門？

*Python · zh-hant*

針對 1、2、3 號門，完成 `monty_opens(chosen, car)`。蒙提從不打開玩家選的門，也不打開有汽車的門。當有兩扇門可以開（玩家正好選中汽車）時，傳回編號較小的那一扇，讓結果可以預測。

```python
def monty_opens(chosen, car):
    """Return the door Monty opens among doors 1, 2, and 3.

    It is neither CHOSEN nor CAR. When two doors are allowed,
    return the lower-numbered one.

    >>> monty_opens(1, 2)
    3
    """
    pass
```

#### 公開測試

Only one door allowed

```pycon
>>> monty_opens(1, 2)
3
>>> monty_opens(2, 1)
3
>>> monty_opens(3, 1)
2
```

Player picked the car

```pycon
>>> monty_opens(1, 1)
2
>>> monty_opens(2, 2)
1
>>> monty_opens(3, 3)
1
```

#### 提示

1. 逐一檢查 `(1, 2, 3)` 中的每扇門，只留下既不是 `chosen` 也不是 `car` 的門。

2. 對允許的門取 `min`，就是編號較小的那一扇。

#### 參考解答

```python
def monty_opens(chosen, car):
    return min(door for door in (1, 2, 3) if door != chosen and door != car)
```

當 `chosen != car` 時，只有一扇門可以開；兩者相等時有兩扇，`min` 選編號較小的。實際遊戲中蒙提此時會隨機挑選，筆記中的模擬會把這個隨機性加回去。

來源: [Adapted from Blitzstein & Hwang, Introduction to Probability, Chapter 2 and §2.10 (R), rewritten in Python](http://stat110.net) · Original exercise; the book's R examples are paraphrased, not copied

### 換門獲勝的機率

*填空 · zh-hant*

你選 1 號門，蒙提打開一扇後面是山羊的門，而你永遠選擇換門。你贏得汽車的機率是多少？請以分數作答。

你的答案: 2/3 (允許的絕對誤差： 0)

#### 提示

1. 換門正好在你第一次選錯時會贏。

2. 第一次選錯的機率是 $2/3$。

#### 參考解答

$2/3$。如果第一次就選中汽車（機率 $1/3$），換門就會輸。如果第一次選到山羊（機率 $2/3$），蒙提只能打開另一扇山羊門，所以你換過去的那扇門後面一定是汽車。筆記中的模擬得到 $0.6690$。

來源: [Adapted from Blitzstein & Hwang, Introduction to Probability, Chapter 2 and §2.10 (R), rewritten in Python](http://stat110.net) · Original exercise; the book's R examples are paraphrased, not copied

### 依證據更新機率

*Python · zh-hant*

完成 `bayes_update(prior, p_evidence_if_a, p_evidence_if_not_a)`。已知 $P(A)$、$P(E\mid A)$ 與 $P(E\mid A^c)$，請用貝氏定理搭配全機率公式傳回 $P(A\mid E)$。測試傳入的是 `Fraction`，所以答案是精確值。假設 $P(E)>0$。

```python
from fractions import Fraction


def bayes_update(prior, p_evidence_if_a, p_evidence_if_not_a):
    """Return P(A | E) from the prior P(A), P(E | A), and P(E | not A).

    >>> bayes_update(Fraction(1, 4), 1, Fraction(2, 3))
    Fraction(1, 3)
    """
    pass
```

#### 公開測試

Monday: at least one girl

```pycon
>>> bayes_update(Fraction(1, 4), 1, Fraction(2, 3))
Fraction(1, 3)
```

Tuesday: the elder is a girl

```pycon
>>> bayes_update(Fraction(1, 3), 1, Fraction(1, 2))
Fraction(1, 2)
```

Evidence that rules out not-A

```pycon
>>> bayes_update(Fraction(1, 100), 1, 0)
Fraction(1, 1)
```

Evidence that says nothing

```pycon
>>> bayes_update(Fraction(1, 5), Fraction(1, 2), Fraction(1, 2))
Fraction(1, 5)
```

#### 提示

1. 分子是 $P(E\mid A)\,P(A)$。

2. 分母是在分子之外再加上 $P(E\mid A^c)\,(1-P(A))$。

#### 參考解答

```python
def bayes_update(prior, p_evidence_if_a, p_evidence_if_not_a):
    numerator = p_evidence_if_a * prior
    return numerator / (numerator + p_evidence_if_not_a * (1 - prior))
```

結果仍然是 $A$ 的機率，所以可以直接當成下一次的先驗再傳進去，這就是依序更新。在 $A$ 與非 $A$ 之下同樣可能出現的證據，不會改變先驗。

來源: [Adapted from Blitzstein & Hwang, Introduction to Probability, Chapter 2 and §2.10 (R), rewritten in Python](http://stat110.net) · Original exercise; the book's R examples are paraphrased, not copied