# probability that Janani will win the game?

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?

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}\$.
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 $\binom{3}{1} \frac{1}{2^3} + \binom{3}{2} \frac{1}{2^3} + \binom{3}{3} \frac{1}{2^3} = \frac{7}{8}$.