停车计时器

Pay the Meter

专题: Probability / 概率
难度: L3
来源: OpenQuant

题目详情

你要向一个只接受真 25 美分硬币的停车计时器支付 1 美元。如果投进一枚假 25 美分硬币，计时器会直接重置为 $0.00$ 。设你有一个装得无限满的钱袋，其中 40% 的硬币是假 25 美分硬币。

计时器被付满之前，你期望总共要投入多少枚硬币？

You are attempting to pay 1 dollar for parking at a parking meter that only accepts real quarters. Upon inserting a fake quarter, the meter will reset back to $0.00. If you have an infinitely packed pouch where 40% of the coins are fake quarters.

How many coins do you expect to insert before the meter has been fully paid?

解析

你需要连续投入 4 枚真 25 美分硬币（每枚为真币的概率是 $p=0.6$ ）；一旦投进假币，金额就会重置为 $0$ 。因此，这个问题等价于在 $\text{Bernoulli}(p)$ 试验中等待连续 $r=4$ 次成功所需的期望时间。

一个已知结果是：

E[T_r] = \frac{1-p^r}{(1-p)p^r}

代入 $p=0.6$ ：

E[T_4] = \frac{(1-0.6^4)}{(1-0.6)0.6^4} = \frac{1-0.1296}{0.4 * 0.1296} \approx \boxed{16.79}

Original Explanation

You need four consecutive real quarters (each real with a probability of $p=0.6$ ); a fake resets the meter to $0$ . So this is the waiting time for $r=4$ consecutive successes in $\text{Bernoulli(p)}$ trials.

A known result:

E[T_r] = \frac{1-p^r}{(1-p)p^r}

Plugging, $p=0.6$ , $r=0.4$ :

E[T_4] = \frac{(1-0.6^4)}{(1-0.6)0.6^4} = \frac{1-0.1296}{0.4 * 0.1296} \approx \boxed{16.79}