# 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%?

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}
\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}