+0 Alice and Bob are playing a game. Alice starts first. On Alice's turn, she flips a coin. If she gets a heads, she wins. If not, it becomes Bob's turn. On Bob's turn, he flips a coin. If he gets a tails, he wins. If not, it becomes Alice's turn. What is the probability that Alice wins the game?
+394 The probability that Alice wins a game is the probability she will win on her first turn + the probability that she will win on her second turn + the probability she will win on her turn turn, and so on...
Alice has a \(\frac{1}{2}\) probability of winning on their first turn, finishing the game.
Since whenever someone wins, the game is finished Bob has a \(\frac{1}{2}\cdot\frac{1}{2}={(\frac{1}{2})}^{2}\) probability of winning on their first turn.
Alice has a \({(\frac{1}{2})}^{2}*\frac{1}{2}={(\frac{1}{2})}^{3}\) probability of winning on her their second turn.
Bob has a \({(\frac{1}{2})}^{3}*\frac{1}{2}={(\frac{1}{2})}^{4}\) probability of winning on their second turn.
Alice has a \({(\frac{1}{2})}^{4}*\frac{1}{2}={(\frac{1}{2})}^{5}\) probability of winning on her thrid turn.
We see Alice has a \({(\frac{1}{2})}^{1}+{(\frac{1}{2})}^{3}+{(\frac{1}{2})}^{5}+{(\frac{1}{2})}^{7}...\) chance of winning.
We can use the infinite geometric series fomula,\(\frac{\frac{1}{2}}{1-\frac{1}{4}}\), we get \(\frac{2}{3}\).
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