The problem is as in the title, where assume optimal play in the second round to maximize number of heads (so in the second round we do not reflip coins with heads in the first round).
I know how to solve it by conditioning on the first outcome (the number of heads in the first round of the game). However, it ultimately leads to some binomial sums with binomials.
Intuitively, I expect 4 heads on first flip, then I have 4 coins left which I flip and expect 2 heads from in the second round. In total, I get 6 heads. Is it possible to justify this line of reasoning formally?
Best Answer
Instead of treating it as a "two turn" game, you can also think about flipping 8 different coins two times each. Let $\mathbf{1}_{n}$ be the indicator that coin $n$ revealed at least one head in the two flips. Then your score at the end of the game is $\sum_{n=1}^8 \mathbf{1}_n$. In expectation we get
$$\mathbb{E}\left[ \sum_{n=1}^8 \mathbf{1}_n \right] = \sum_{n = 1}^8 \Pr\left( \text{coin $n$ revealed at least one head} \right) = \sum_{n=1}^8 \frac{3}{4} = 6$$