Two coin waits

1. Step 1 · Alice takes a flips and Bob takes more

   $\displaystyle \left(\frac{1}{2}\right)^{a} \times \left(\frac{1}{2}\right)^{a} = \left(\frac{1}{4}\right)^{a}$

   Alice’s first head comes on flip $a$ with chance $(1/2)^a$. Bob needs more than $a$ flips, so his first $a$ flips are all tails, also chance $(1/2)^a$.

2. Step 2 · Chance Bob needs more flips

   $\displaystyle \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots = \frac{1}{3}$

   Add $(1/4)^a$ over every $a$. This is what you’re told happened.

3. Step 3 · Given that, Alice took a flips

   $\displaystyle \left(\frac{1}{4}\right)^{a} \div \frac{1}{3} = \frac{3}{4},\, \frac{3}{16},\, \frac{3}{64},\, \dots$

   Divide by the chance of what you know. That’s a first-success wait with chance 3/4 each flip.

4. Step 4 · Average flips for Alice

   $\displaystyle 1 \div \frac{3}{4} = \frac{4}{3}$

   A chance of 3/4 each flip means $1 \div 3/4$ flips on average. Knowing Bob was slow makes Alice look quick.

Answer$\displaystyle \frac{4}{3} \approx 1.333$