Four coins with reflip problem?

Your temptation is right and your gut is wrong. You do get an extra $\frac12$ if you got tails at least once. The probability that you don't have a tail to reflip is $\frac1{16}$, so you get an extra $\frac12\left(1-\frac1{16}\right)=\frac{15}{32}$. This added to the base expectation of $2 = \frac{64}{32}$ gives $\frac{79}{32}$.

Expectation of first 4 flips is $2.

Expectation of the fifth flip, condition on there is a tail in the first 4 flips, is 0.5.

Expectation of the fifth flip, condition on there is no tail, is 0.

Probability there is a tail in the first 4 flips is 15/16.

Total expectation = 2 + 0.5*(15/16) + 0*(1/16) = 79/32

Let $X_i=1$ if $i$-th toss is head and $0$ otherwise.

The reward is $$\sum_{i=1}^4X_i + X_5\left( 1-\prod_{i=1}^4X_i\right)=\sum_{i=1}^5X_i-\prod_{i=1}^5X_i$$

Hence

$$\mathbb{E}\left(\sum_{i=1}^5X_i-\prod_{i=1}^5X_i\right)=\left(\sum_{i=1}^5\mathbb{E}[X_i]-\prod_{i=1}^5\mathbb{E}[X_i]\right)=\frac52-\frac1{32}=\frac{79}{32}$$