The game can take an arbitrarily long time—and on average, it does take an arbitrarily long time.
First, the game can go on arbitrarily long: if you flip a coin $n$ times, you may get $n$ heads in a row. As a result, the game can last longer than any number $n$ of turns.
Next, the expected length of the game is also infinite. To see this, let $x$ be the expected length of the game assuming the first throw is heads. Evidently, $x$ is also the expected length of the game assuming the first throw is tails, and it’s also the expected length of the game overall.
We can get a formula for $x$ as follows. To play the game, you flip the coin once and get, say, heads. Now your score is +1. You flip the coin a second time. Either you get tails and win immediately (probability 1/2), or (probability 1/2) you get heads, your score goes up, and you must keep flipping coins until it gets back down to +1 again. The expected time it takes to get back to +1 starting from +1 is, of course, $x$.
Once you’ve returned to the score +1, you flip the coin again. You either get tails and win (probability 1/2), or your score goes up again, and you must keep flipping coins until it gets back down to +1 again, and ….
In formulas, the expected length of the game is:
$$x = underbrace{1}_{text{move right}} + underbrace{frac{1}{2}(1)}_{text{move left and win}} + frac{1}{2}left( underbrace{x}_{text{loop from +1 to +1}} + frac{1}{2}left[underbrace{1}_{text{left and win}} + x + frac{1}{2}(ldots)right]right)$$
$$x = 1 + frac{1}{2}left(1 + x + frac{1}{2}left[1+x+frac{1}{2}left(1 + x + ldotsright)right]right)$$
If we add $x$ to both sides, divide by two, and add one, we find that
$$1+x; =; 1 + frac{1}{2}left(1 + x + frac{1}{2}left[1+x+frac{1}{2}left(1 + x + ldotsright)right]right) qquad = x$$
so $x = x+1$. This problem occurs because the expected length is infinite— it exceeds all finite bounds.
We can see this result another way. The expected length of the game is:
$$x = sum_{text{winning sequence}} P(text{sequence})timestext{length}(text{sequence})$$
Of course, a winning sequence of throws is any sequence of heads and tails with the same number of heads and tails. If $n$ is the number of heads in such a sequence, the length of the sequence is $2n$ and the probability of the sequence is $frac{1}{2^{2n}}$. Also there are ${2n choose n}$ winning sequences containing $n$ heads.
Hence
$$x = sum_{n=1}^infty {2n choose n }frac{1}{2^{2n}} times 2n$$
which diverges to infinity.