Calvin's odd condition

Let $E$ be the expected value.

If Flip #1 = T, we're 1 flip in and exactly where we started.

If Flip #1 = H, then we have two possibilities:

• Flip #2 is H, then we're back to where we started, since we have an even number of heads.

• Flip #2 is T, then we're done.

Thus, $E = \frac{1}{2} \left(1+E\right) + \frac{1}{2}\left(\frac{1}{2}(2+E)+\frac{1}{2}(2)\right)$ which simplifies to $E = \frac{1}{2}(1+E) + \frac{1}{4}(2+E) + \frac{1}{4}(2) = \frac{3}{2} + \frac{3}{4}E$ so $E=6.$

This solution involves some probability theory.

Let c= E(X) where x is the event we want. Using conditional probability, we find that c=p(x/ tails )0.5 + p(x/heads)0.5 . Conditioning again gives us c= 0.5c + (2(0.5)+p(x/1tail and 1head)0.5)0.5 = 0.5c + 0.5 + (2+c)0.25. Rearranging for c gives us c= 6