lean_workbook_plus_20857

For each integer $n\geq 0$ , let $p_n$ denote the probability that the $n^{\text{th}}$ flip after the first one is heads, with $p_0 = 1$ . Then by the Total Probability Formula, \n\begin{align*}p_{n+1} &= \operatorname{Pr}{(n+1)^{\text{st}}\text{ heads}\mid n^{\text{th}}\text{ heads}}\Pr{n^{\text{th}}\text{ heads}} &\hspace{10ex}+ \operatorname{Pr}{(n+1)^{\text{st}}\text{ heads}\mid n^{\text{th}}\text{ tails}}\Pr{n^{\text{th}}\text{ tails}}&= \frac23p_n + \frac13(1-p_n) = \frac13 + \frac13p_n.\end{align*} It is straightforward to derive $p_n = \tfrac{1+3^n}{2\cdot 3^n}$ , and now the desired answer is obtained by plugging in $n=2010$ .