集邮者

我们把这个过程分成三个阶段：

1. 第 1 阶段：集邮者拿到第一种不同的邮票。
2. 第 2 阶段：集邮者拿到第二种不同的邮票。
3. 第 3 阶段：集邮者拿到第三种不同的邮票。

分别计算每个阶段所需邮票数的期望，再把它们相加。

第一张邮票一定是新种类，所以拿到第一种不同邮票所需的期望张数为：$E_1 = 1$

当集邮者已经有 1 种不同邮票后，下一次抽到新种类的概率是 $\frac{3}{4}$，因为 4 种邮票里还有 3 种尚未出现。抽到新种类所需的尝试次数服从成功概率为 $\frac{3}{4}$ 的几何分布，因此这一阶段的期望尝试次数为：

$$E_2 = \frac{1}{\frac{3}{4}} = \frac{4}{3}$$

现在集邮者已经有 2 种不同邮票。抽到第三种不同邮票的概率是 $\frac{2}{4} = \frac{1}{2}$，因为还有 2 种尚未出现。因此这一阶段的期望尝试次数为：

$$E_3 = \frac{1}{\frac{1}{2}} = 2$$

收集到 3 种不同邮票所需的总期望为三个阶段期望之和：

$E = E_1 + E_2 + E_3 = 1 + \frac{4}{3} + 2 = \frac{3}{3} + \frac{4}{3} + \frac{6}{3} = \frac{13}{3}$

因此，集邮者为了收集到 3 种不同邮票，期望需要抽到的邮票张数是 $\boxed{\frac{13}{3}}$。

---

Original Explanation

We'll break the process into three phases:

1. Phase 1: The collector finds the first unique stamp.
2. Phase 2: The collector finds the second unique stamp.
3. Phase 3: The collector finds the third unique stamp.

We will calculate the expected number of stamps needed in each phase and then sum them up.

The first stamp will always be a new type, so the expected number of stamps to find the first unique stamp is: $E_1 = 1$

Once the collector has one unique stamp, the probability of finding a new unique stamp on the next try is $\frac{3}{4}$ because 3 out of the 4 stamp types remain undiscovered. The expected number of trials to get a new stamp follows a geometric distribution with success probability $\frac{3}{4}$ . The expected number of trials to get a new type in this phase is:

$$E_2 = \frac{1}{\frac{3}{4}} = \frac{4}{3}$$

Now the collector has 2 unique stamps. The probability of finding a third unique stamp is $\frac{2}{4} = \frac{1}{2}$, since 2 out of the 4 types remained undiscovered. The expected number of trials to get a new type in this phase is

$$E_3 = \frac{1}{\frac{1}{2}} = 2$$

The total expected number of stamps to find three different types is the sum of the expectations for each phase:

$E = E_1 + E_2 + E_3 = 1 + \frac{4}{3} + 2 = \frac{3}{3} + \frac{4}{3} + \frac{6}{3} = \frac{13}{3}$

Thus, the expected number of stamps that the collector would have to find to get three different types is $\boxed{\frac{13}{3}}$.