Two fair coins are tossed until both turn up heads A penny and a dime are tossed together until both turn up heads, after which no more tosses are made. Find the expected number of times the penny comes up heads.
What I've tried:
Let $X$ and $Y$ be the number of times the penny and dime come up heads, respectively. Then, for $x = 1,2,3,\ldots$
$$P(X = x) = \sum_{y=1}^\infty P(X = x, Y = y) \\ = \sum_{y=1}^x P(X=x,Y=y) + \sum_{y=x+1}^\infty P(X=x, Y=y) \\ = \sum_{y=1}^x \sum_{n=x}^\infty \left( \frac{1}{4} \right) \binom{n-1}{x-1} \left( \frac{1}{2} \right)^{n-1} \binom{n-1}{y-1} \left( \frac{1}{2} \right)^{n-1} + \\ \sum_{y=x+1}^\infty \sum_{n=y}^\infty \left( \frac{1}{4} \right) \binom{n-1}{x-1} \left( \frac{1}{2} \right)^{n-1} \binom{n-1}{y-1} \left( \frac{1}{2} \right)^{n-1} \\ = \sum_{y=1}^x \sum_{n=x}^\infty \left( \frac{1}{4} \right)^n \binom{n-1}{x-1} \binom{n-1}{y-1} + \sum_{y=x+1}^\infty \sum_{n=y}^\infty \left( \frac{1}{4} \right)^n \binom{n-1}{x-1} \binom{n-1}{y-1}$$
Is there a way to simplify the above expression? Or is there an easier approach that I'm missing?

Imagine a clock that "ticks" only when the penny comes up heads. We need he expected number of ticks $T$ until the dime comes up heads. This follows a geometric distribution with success probability $\dfrac{1}{2}$. So $\mathbb{E}[T] = 2$.
