We consider two versions of the problem:

  1. The coin is fair.
  2. The coin is biased, with probability \( p \) of heads.

Part 1: Fair Coin

Let \( X \) be the number of flips until the first head. Since the coin is fair:

\[ P(\text{Head}) = \frac{1}{2} \]

Then \( X \sim \text{Geometric}(p = \frac{1}{2}) \). The expectation is:

\[ \mathbb{E}[X] = \frac{1}{p} = \frac{1}{1/2} = \boxed{2} \]

Part 2: Biased Coin (Probability of Head = \( p \))

Now \( X \sim \text{Geometric}(p) \), where:

\[ P(X = k) = (1 - p)^{k - 1} p \]

Then the expected number of flips is:

\[ \mathbb{E}[X] = \frac{1}{p} \]

So the smaller \( p \) is, the longer you expect to wait for a head.

Conclusion

  • For a fair coin: \( \boxed{2} \) expected flips
  • For a biased coin with \( P(\text{Head}) = p \): \( \boxed{\frac{1}{p}} \) expected flips

Reference