Expected Failures Before Success

The number of failures before the first success is geometric on $\{0,1,2,\dots \}$.

Its mean is $(1-p)/p$.

1. 1. Apply the shifted geometric mean

   For success probability $p=1/4$, the expected number of failures before the first success is $(1-p)/p$.

   $$E[\text{failures}]=\frac{1-1/4}{1/4}=3$$

Answer: 3