Let's find the expected number of times you'll flip the $\color{#20A900}{second}$ coin once you pick it up.

$E_1 = \sum \text{(number of times it is flipped) × (probability that it's flipped that number of times)}$

$E_1 = 1 × (\frac{1}{2}) ^ 1 + 2 × (\frac{1}{2}) ^ 2 + 3 × (\frac{1}{2}) ^ 3 \ldots \infty$

$\color{#3D99F6}{\boxed{E_1 = 2}}$

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Let's find the expected number of times you'll flip the $\color{#20A900}{second}$ coin in total considering $\color{#D61F06}{first}$ coin also.

$E_2 = \sum \text{(number of times first coin is flipped) × (probability that it's flipped that number of times)} × E_1$

$E_1 = 1 × (\frac{1}{2}) ^ 1 × E_1+ 2 × (\frac{1}{2}) ^ 2 × E_1 + 3 × (\frac{1}{2}) ^ 3 × E_1\ldots \infty$

$E_2 = 2 × E_1$

$\color{#3D99F6}{\boxed{E_2 = 4}}$

Let $N$ be the number of times the first coin is tossed, and let $Y_j$ be the number of times the second coin is tossed after the $j$ th toss of the first coin. Thus the total number of tosses of the second coin is the random variable $Z \; = \; \sum_{j=1}^N Y_j$ Suppose that the probability of the first coin coming up tails is $p_1$ , and that the probability of the second coin coming up tails is $p_2$ . Then $N$ has the geometric distribution $\mathrm{Geo}(p_1)$ , while each of the random variables $Y_j$ have the geometric distribution $\mathrm{Geo}(p_2)$ , and the random variables $N,Y_1,Y_2,...$ are independent of each other.

If $N = n$ then $Z = Y_1 + Y_2 + \cdots + Y_n$ , and so $E[Z|N=n] \; = \; E\left[\sum_{j=1}^n Y_j\right] \; = \; \sum_{j=1}^n E[Y_j] \; = \; np_2^{-1}$ for all $n \ge 1$ . Thus we deduce that $E[Z|N] = p_2^{-1}N$ , and hence $E[Z] \; = \; E\big[E[Z|N]\big] \; = \; E[p_2^{-1}N] \; = \; p_2^{-1}E[N] \; = \; p_1^{-1}p_2^{-1}$ In the case $p_1=p_2=\tfrac12$ we obtain $E[Z] = \boxed{4}$ .