Question: Player A and Player B compete in a ball game. Player A has a chance of p to earn a point, and Player B has a chance q=1−p to earn a point. The game ends whenever a player gets 10 points. Now Player A has 7 points and Player B has 5 points. What is the probability that Player A wins the game?
A tennis game
Answer: Player A has to win 3 more points. Before Player A wins the last point, Player B can only wins no more than 4 points.
Method 1. Let n be the number of further games. n can be 3,4,5,...,7. If n=3, then Player A wins all the games, the probability is p3. If n=4, then Player B may win one of the first 3 games, the probability is C31p3q. For general n, Player B may win any n−3 of the first n−1 games, and the probability is Cn−1n−3p3qn−3. Summing up, we have P(A)=(7∑n=3Cn−1n−3qn−3)p3=(1+3q+6q2+10q3+15q4)p3.
Method 2. If we use a to represent the event that Player A earns a point and we use b to represent the event that Player B earns a point, then (1+b+b2+…)a(1+b+b2+…)a(1+b+b2+…)a=(1+b+b2+b3+b4+…)3a3=(1+3b+6b2+10b3+15b4+…)a3
consists of all the cases that Player A earns 3 more points. It is easy to see that only up to b4 corresponding to that Player A wins the game. Therefore we have
P(A)=(1+3q+6q2+10q3+15q4)p3.
The two methods are essentially the same. In the second method, the counting is hidden behind the algebra and we do not need to consider case by case.
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