The "every fail resets you" trick"一次反面就清零" 的小技巧

Let $E_n$ be the expected number of tosses to get $n$ heads in a row. Suppose you've already managed to get $n-1$ heads in a row (which took $E_{n-1}$ tosses on average). You toss one more coin:

设 $E_n$ 为得到 $n$ 个连续正面的期望抛掷次数. 假设你已经成功抛出 $n-1$ 个连续正面 (平均花费 $E_{n-1}$ 次). 再抛一次:

• With probability $\tfrac{1}{2}$: it's H → done in $E_{n-1}+1$ tosses.
• With probability $\tfrac{1}{2}$: it's T → you've used $E_{n-1}+1$ tosses and you're back at zero, so you need $E_n$ more.
• 概率 $\tfrac{1}{2}$: 正面 → 总共 $E_{n-1}+1$ 次完成.
• 概率 $\tfrac{1}{2}$: 反面 → 已用 $E_{n-1}+1$ 次, 但回到起点, 还需 $E_n$ 次.

With $E_0 = 0$:由 $E_0 = 0$:

Why this beats the full state machine为什么比完整状态机更直观

The key observation: a tails always sends you back to state 0, no matter where you are. So you only track "how many H in a row do I currently have," and the recurrence collapses to one line. No 4-state system, no linear algebra.

关键观察: 反面永远把你打回状态 0, 不管你已连了几个正. 所以只需跟踪"当前连了几个正", 递推关系一行就能写完. 不需要 4 状态系统, 也不需要线性代数.

Setup设置

Inscribe a quarter unit circle inside the unit square $[0,1]\times[0,1]$.

在单位正方形 $[0,1]\times[0,1]$ 内嵌一个四分之一单位圆.

• Area of square: $1$
• Area of quarter circle: $\dfrac{\pi}{4}$
• 正方形面积: $1$
• 四分之一圆面积: $\dfrac{\pi}{4}$

Random point & membership test随机点与判定

Draw $X \sim U(0,1)$ and $Y \sim U(0,1)$ independently. The point $(X,Y)$ lies inside the quarter circle iff its distance from the origin is $\leq 1$, i.e.

独立取 $X \sim U(0,1)$ 与 $Y \sim U(0,1)$. 点 $(X,Y)$ 在四分之一圆内当且仅当其到原点距离 $\leq 1$, 即

$$X^2 + Y^2 \leq 1$$

(No square root needed — squaring both sides keeps it cheap.)

(不需要开平方 — 两边平方后判断更省算力.)

Estimator估计量

Run $N$ trials. Let $K$ be the number of points satisfying $X^2+Y^2 \leq 1$. Then by LLN:

运行 $N$ 次试验. 设 $K$ 为满足 $X^2+Y^2 \leq 1$ 的点数. 由大数定律:

$$\frac{K}{N} \xrightarrow{N\to\infty} \frac{\pi}{4} \quad\Longrightarrow\quad \hat{\pi} = 4 \cdot \frac{K}{N}$$

Convergence rate (good follow-up to mention)收敛速度 (面试加分点)

Each trial is Bernoulli with $p = \pi/4$, so $K \sim \text{Bin}(N, \pi/4)$. The standard error is

每次试验是 $p=\pi/4$ 的伯努利, 故 $K \sim \text{Bin}(N, \pi/4)$. 标准误为

$$\text{SE}(\hat\pi) = 4 \sqrt{\frac{p(1-p)}{N}} \approx \frac{1.64}{\sqrt{N}}$$

Pseudocode伪代码

import random
N = 1_000_000
K = sum(1 for _ in range(N)
        if random.random()**2 + random.random()**2 <= 1)
pi_hat = 4 * K / N

Frame the distribution先写出分布

Let $X$ = number of heads. Then $X \sim \text{Bin}(400, 0.5)$ with

设 $X$ = 正面次数. 则 $X \sim \text{Bin}(400, 0.5)$:

$$\mu = np = 200, \quad \sigma^2 = np(1-p) = 100, \quad \sigma = 10$$

The exact $\binom{400}{220}/2^{400}$ is computable but useless in an interview. CLT to the rescue.

精确值 $\binom{400}{220}/2^{400}$ 算得出来但面试时没用. 这时该 CLT 上场.

Standardize标准化

$220$ is $\dfrac{220-200}{10} = 2$ standard deviations above the mean.

$220$ 距离均值 $\dfrac{220-200}{10} = 2$ 个标准差.

By 68-95-99.7, ~95% of mass is in $[-2, 2]$, so each tail has roughly $\dfrac{1-0.95}{2} = 2.5\%$.

由 68-95-99.7 法则, 约 95% 的质量在 $[-2, 2]$ 区间, 因此每条尾巴约 $\dfrac{1-0.95}{2} = 2.5\%$.

Interpretation A — "at least 220 heads"解读 A — "至少 220 次正面"

Interpretation B — "exactly 220 heads"解读 B — "恰好 220 次正面"

Continuity correction: $P(X=220) \approx P(219.5 < X < 220.5) = P(1.95 < Z < 2.05)$. Or use the local normal approximation $P(X=k) \approx \dfrac{1}{\sigma\sqrt{2\pi}} e^{-z^2/2}$ at $z=2$:

连续性修正: $P(X=220) \approx P(219.5 < X < 220.5) = P(1.95 < Z < 2.05)$. 或用局部正态近似 $P(X=k) \approx \dfrac{1}{\sigma\sqrt{2\pi}} e^{-z^2/2}$ ($z=2$):

$$P(X=220) \approx \frac{1}{10\sqrt{2\pi}} \cdot e^{-2} \approx \mathbf{0.54\%}$$

Why "55%" is way off为什么 "55%" 错得离谱

55% only makes sense for $P(X \geq 200)$ — about half. But 220 is meaningfully above the mean: 20 extra heads is twice the standard deviation, deep into the tail. Always anchor on $\mu$ and count $\sigma$ away from it.

55% 只在 $P(X \geq 200)$ 时合理 — 大约一半. 但 220 显著高于均值: 多出 20 次正是 2 倍标准差, 已经深入尾部. 永远先锚定 $\mu$, 再数离它有几个 $\sigma$.

Anchor on the break-even rate先锚定盈亏平衡利率

The bank's break-even rate must cover its expected loss plus costs:

银行的盈亏平衡利率必须覆盖 预期损失 加上各项成本:

$$r \approx r_f + \text{term premium} + \text{credit spread} + \text{op costs} + \text{margin}$$

Where credit spread is driven by

其中信用利差由如下公式驱动:

$$\text{Expected Loss} = \text{PD} \times \text{LGD} \times \text{EAD}$$

PD = probability of default, LGD = loss given default, EAD = exposure at default.

PD = 违约概率, LGD = 违约损失率, EAD = 违约风险敞口.

Side-by-side逐项对比