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Two people are playing a coin toss game with a fair penny. Manu gets a point if the penny lands on heads. Janani gets a point if the penny lands on tails. The score is Janani 9, Manu 7, in a game to 10 points. What is the probability that Janani will win the game?

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Hint: Janini will win if Manu loses. Manu wins only if the next three results are heads.

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Janini will win only if Manu loses. Probability that Manu wins the game: since Manu has 3 chances is $\frac{1}{8}\ $. Thus probability that Janini wins is 1- $\frac{1}{8}\ $ = $\frac{7}{8}\ $.

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The most number of tosses that can happen before someone wins for sure is 2 (if 2 heads are tossed). Thus after 3 tosses, it is impossible for the game to not yet be over. It is clear that Manu needs all of these tosses to be heads, which occurs with probability $\frac{1}{8}$. Thus, Janani wins with probability $\frac{7}{8}$.

An analogous way to think about this is to sum up the probabilities of getting 1, 2, and 3 tails in 3 tosses, all of which will lead to Janani winning. Even though Janani will win after just 1 tail, if we pretend that the coin is flipped a total of 3 times regardless, we can figure out the probability of this happening. This is $\binom{3}{1} \frac{1}{2^3} + \binom{3}{2} \frac{1}{2^3} + \binom{3}{3} \frac{1}{2^3} = \frac{7}{8}$.

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