Flip 4 coins, for each head gain $1, what is EV if you are offered exactly 2 reflips? The question has been posted and answered before, but not with exactly two reflips, or none
For the question where you can reflip up to 2 coins, the expected value is 90/32 and I understand how it works
But I don’t understand how to approach the question when you have to reflip exactly 2 coins, or none at all.
My guess is just (75/32) expected value? Since there are 11 out of 16 cases of four flips, with at least two tails to reflip (with a half chance of getting those cases, ie 11/32) plus the base case of 2 dollars expected value, for a total of 75/32
Is this the right approach?
 A: First, we have to think about the optimal strategy when the choice is to only re-flip $0$ or $2$ coins.  Let $$X \sim \operatorname{Binomial}(n=4, p=1/2)$$ represent the random number of heads obtained on the first round of flips.  Obviously, if $X = 4$, there is no need to re-flip.  If $X \in \{0, 1, 2\}$, re-flips cannot worsen the outcome so it is optimal to choose two coins that are tails to re-flip.
If we re-flip exactly two coins that were tails, the expected number of additional heads we will get is of course $1$.  So if $Y$ is the total number of heads obtained after the re-flip, we have $$\operatorname{E}[Y \mid X] = X + 1, \quad X \in \{0, 1, 2\}.$$
So the only remaining case is when $X = 3$ and there is only one tail.  If we must re-flip two coins, the best choice of coins to re-flip includes the coin that shows tails; the other coin must be heads.  But re-flipping these two coins, on average, will not change the expected number of heads among those two coins, which will remain $1$, and is the same as if we were not to re-flip.  Hence in this case, we are indifferent to whether to re-flip or not.
This establishes the following:  if $Y$ is the random number of heads after the re-flipping strategy, then
$$\begin{align}
\operatorname{E}[Y \mid X = 4] &= 4 \\
\operatorname{E}[Y \mid X = 3] &= 3.
\end{align}$$
Now putting it all together,
$$\operatorname{E}[Y] = \sum_{x=0}^4 \operatorname{E}[Y \mid X = x]\Pr[X = x] = 1 \cdot \frac{1}{16} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{3}{8} + 3 \cdot \frac{1}{4} + 4 \cdot \frac{1}{16} = \frac{43}{16}.$$

As an exercise, what if the coins are biased?  Specifically, if the probability of obtaining heads on any given flip of a coin is $p$, what is the expectation as a function of $p$?
As a second exercise, what if we have $n$ coins, where $n \ge 4$?  What if the choice is to re-flip $m$ out of $n$ coins, or none?

Addendum.  Since the original question was not entirely clear, I will explain the experiment in more detail and show why my answer is consistent with this interpretation.
The original experiment is that you have four fair coins.  In Round 1, you flip all four coins, observe the result, and based on the number of heads you see, you are allowed to pick $0$, $1$, or $2$ of these coins to re-flip in Round 2, with the goal being to maximize the number of heads you have at the end of the second round.  The expected value of the number of heads at the end of the second round is $$1 \cdot \frac{1}{16} + 2 \cdot \frac{1}{4} + 3 \cdot \frac{3}{8} + \frac{7}{2} \cdot \frac{1}{4} + 4 \cdot \frac{1}{16} = \frac{45}{16}.$$  This agrees with the value provided in the original question, which also justifies the answer I provided above for the modified experiment in which only $0$ or $2$ coins can be re-flipped.
