jmc

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 $$\frac{\frac{1}{2}}{1-\frac{1}{4}}=\boxed{\frac{2}{3}}.$$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.