Seven heads and a flat prior

Idea. Average the chance of heads over what we now believe about $p$. That average is the mean of the posterior distribution, which here is a Beta distribution.

Step 1: the posterior. The prior density is 1 on $[0,1]$ and the likelihood of 7 heads and 3 tails is proportional to $p^7(1-p{)}^3$. So the posterior density is $$\pi (p)\propto p^7(1-p{)}^3,$$ the Beta(8, 4) distribution.

Step 2: the next toss. Given $p$, the next toss is heads with probability $p$, so $$\mathbb{P}(\text{heads})=\mathbb{E}[p\mid \text{data}]=\frac{{\int }_0^1{p}^8(1-p{)}^{3}dp}{{\int }_0^1{p}^7(1-p{)}^{3}dp}.$$ With the formula in the hint, $$\frac{8!3!/12!}{7!3!/11!}=\frac{8}{12}=\frac{2}{3}.$$

The rule of succession. After $k$ heads in $n$ tosses with a uniform prior, the next toss is heads with probability $$\frac{k+1}{n+2}.$$ It is as if you had seen one extra head and one extra tail before starting. Here $\frac{8}{12}=\frac{2}{3}$, a little below the observed frequency $0.7$: the prior pulls the estimate towards $\frac{1}{2}$, and the pull fades as $n$ grows.

The trap. Answering $0.7$, the maximum likelihood estimate. It ignores the prior and the uncertainty about $p$.