Alice has a $ 1/2 $ chance of winning the game on her first turn. If she doesn't, then the probability that she wins the game on her second turn is $ 1/8, $ since she must not win on her first flip ($ 1/2 $ chance), Bob must not win on his first flip ($ 1/2 $ chance), and then Alice must win on her second flip ($ 1/2 $ chance). The probability that she wins the game on her third turn is $ 1/32, $ and in general, the probability that she wins the game on her $ k^\text{th} $ turn is $ (1/2)^{2k-1} $. Thus, the probability that Alice wins is an infinite geometric series with first term $ 1/2 $ and common ratio $ 1/4 $. So, the probability that Alice wins the game is $$\frac{\frac12}{1-\frac14} = \boxed{\frac{2}{3}}.$$OR Note that the only difference between the odds of Alice or Bob winning is who goes first. Because Bob goes second, the odds of him winning on his $ k^\text{th} $ flip is half of the odds that Alice wins on her $ k^\text{th} $ flip, since Alice must first get a tails before Bob gets a chance to win. Thus, if $ a $ is Alice's chance of winning, and $ b $ is Bob's chance of winning, then $ a = 2b $. Also, since someone must win, $ a + b = 1 $. It follows that $ a = 2/3 $ and $ b = 1/3, $ so Alice has a $ \boxed{\frac{2}{3}} $ chance of winning the game.