Four coins with reflip problem? I came across the following problem today.

Flip four coins. For every head, you get $\$1$. You may reflip one coin after the four flips. Calculate the expected returns.

I know that the expected value without the extra flip is $\$2$. However, I am unsure of how to condition on the extra flips. I am tempted to claim that having the reflip simply adds $\$\frac{1}{2}$ to each case with tails since the only thing which affects the reflip is whether there are tails or not, but my gut tells me this is wrong. I am also told the correct returns is $\$\frac{79}{32}$ and I have no idea where this comes from.
 A: Your answer is $\frac {80}{32}$. The $\frac1 {32}$ is from where all four flips are heads.
A: 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 $\frac{79}{32}$.
A: 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
A: 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 
$$\mathbb{E}\left(\sum_{i=1}^5X_i-\prod_{i=1}^5X_i\right)=\left(\sum_{i=1}^5\mathbb{E}[X_i]-\prod_{i=1}^5\mathbb{E}[X_i]\right)=\frac52-\frac1{32}=\frac{79}{32}$$ 
A: Your gut is wrong, as pointed out already.
The possible outcomes from the initial flip are:
"4 heads" x 1
"3 heads" x 4
"2 heads" x 6
"1 heads" x 4
"0 heads" x 1
This gives an expected return of (4 + 12 + 12 + 4 + 0)/16 = 2
If you add 0.5 to each case except 4 heads, you get (4 + 14 + 15 + 6 + 0.5)/16 = 79/32
A: This particular problem can be solved simply by flipping five coins but winning a maximum of \$4.
That gives an expected value of: $$\frac{1\cdot 0 + 5\cdot 1 + 10\cdot 2 + 10\cdot 3 + (5+1)\cdot 4}{2^5}=\frac{79}{32}.$$
