This is a simplification and slight modification of an interview question which I found online.
Quesiton: You flip 3 coins and if you get HHH or TTT you will win \$X. How much are you willing to pay for this game?
Solution: Since my chances of winning is $\frac{1}{4}$, I would pay \$$\frac{X}{4}$ to play this game.
Follow Up 1: What if you can have an option to re-flip any of the coins?
Solution: I still have $\frac{1}{4}$ chances of winning plus for any other outcomes I have $\frac{1}{2}$ chances of winning, so I am willing to pay \$$\frac{3*X}{4}$ to play this game.
Follow Up 2: What if you can re-flip a certain coin, let's say the last one
Solution: I still have $\frac{1}{4}$ chances plus if I get HHT or TTH I will re-flip. The chances I get HHT or TTH is $\frac{1}{4}$ and the chances that I win after re-flipping is $\frac{1}{8}$, so I am willing to pay \$$\frac{3*X}{8}$ to play this game.
Is my logic correct for all the above?
The first one and the last one are correct. The second is not correct.
For the second one, there is $\frac{1}{4}$ chance that we get $3H$ or $3T$ and $\frac{3}{4}$ chance that we get $2H, 1T$ or $2T,1H$. Given the option that we can re-flip a coin, we would re-flip the coin that we have one of. So if we had $2H,1T$ and we re-flip the coin that had $T$, there is $\frac12$ chance that we get $H$ and we win $ \$X$. Similarly for the case $2T, 1H$.
So expected win value $ = \displaystyle \left(\frac14 + \frac12 \cdot \frac34 \right) \cdot X = \frac{5X}{8}$.