Probability your name will be drawn out of a hat without replacement Let's say you put your name in a hat with 7 other names (n=8). Only 4 names will be drawn from said hat. Each name will be drawn one at a time, without replacement. What is the probability that your name will be drawn?
I've googled a bit and most of my results lead me to the secret santa problem. Which is similar, but not answering the same question me thinks.
Is the answer 1/8 * 1/7 * 1/6 * 1/5 ?
Am I over complicating this? Does none of this matter and the probability is 4/8 = 50%?
 A: In a sense you are over-complicating it, but it is still useful to be able to show that the intuitively correct answer corresponds to the correct answer under a draw-by-draw probability calculation.  Assuming simple random sampling without replacement, the probability that your name will be drawn is $4/8 = 0.5 = 50$%.
If you really want to see what happens when you break this probability calculation down onto a draw-by-draw basis, let $\mathcal{D}_i$ be the event that your name is drawn on the $i$th draw.  Under simple random sampling without replacement, the probability of being drawn in the first four draws is:
$$\begin{align}
\mathbb{P}(\mathcal{D}_{1:4})
&= 1 - \mathbb{P}(\bar{\mathcal{D}}_{1:4}) \\[16pt]
&= 1 - \mathbb{P}(\bar{\mathcal{D}}_1 \cap \bar{\mathcal{D}}_2 \cap \bar{\mathcal{D}}_3 \cap \bar{\mathcal{D}}_4) \\[12pt]
&= 1 - \prod_{i=1}^4 \mathbb{P}(\bar{\mathcal{D}}_i | \bar{\mathcal{D}}_1 \cap \cdots \cap \bar{\mathcal{D}}_{i-1}) \\[12pt]
&= 1 - \frac{7}{8} \cdot \frac{6}{7} \cdot \frac{5}{6} \cdot \frac{4}{5} \\[12pt]
&= 1 - \frac{4}{8} \\[12pt]
&= \frac{4}{8} \\[12pt]
&= \frac{1}{2} = 0.5, \\[6pt]
\end{align}$$
where the transition to the third line uses the chain rule of probability.  Or if you prefer to expand out based on probabilities of being drawn on the individual draws, you get:
$$\begin{align}
\mathbb{P}(\mathcal{D}_{1:4})
&= \ \ \ \mathbb{P}({\mathcal{D}}_1 \cap \bar{\mathcal{D}}_2 \cap \bar{\mathcal{D}}_3 \cap \bar{\mathcal{D}}_4) 
+ \mathbb{P}(\bar{\mathcal{D}}_1 \cap {\mathcal{D}}_2 \cap \bar{\mathcal{D}}_3 \cap \bar{\mathcal{D}}_4) \\[6pt]
&\quad + \mathbb{P}(\bar{\mathcal{D}}_1 \cap \bar{\mathcal{D}}_2 \cap {\mathcal{D}}_3 \cap \bar{\mathcal{D}}_4) 
+ \mathbb{P}(\bar{\mathcal{D}}_1 \cap \bar{\mathcal{D}}_2 \cap \bar{\mathcal{D}}_3 \cap {\mathcal{D}}_4) \\[12pt]
&= \ \ \ \frac{1}{8} \cdot \frac{7}{7} \cdot \frac{6}{6} \cdot \frac{5}{5}
+ \frac{7}{8} \cdot \frac{1}{7} \cdot \frac{6}{6} \cdot \frac{5}{5} \\[6pt]
&\quad + \frac{7}{8} \cdot \frac{6}{7} \cdot \frac{1}{6} \cdot \frac{5}{5}
+ \frac{7}{8} \cdot \frac{6}{7} \cdot \frac{5}{6} \cdot \frac{1}{5} \\[12pt]
&= \ \ \ \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \\[12pt]
&= \frac{4}{8} \\[12pt]
&= \frac{1}{2} = 0.5. \\[6pt]
\end{align}$$
