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# Help

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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?

Feb 18, 2024

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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 $$\frac{2}{3}$$.



Feb 18, 2024