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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 $p^3$. If $n=4$, then Player B may win one of the first 3 games, the probability is $C_1^3p^3q$. For general $n$, Player B may win any $n-3$ of the first $n-1$ games, and the probability is $C_{n-3}^{n-1}p^3q^{n-3}$. Summing up, we have $$P(A)=\left(\sum_{n=3}^7 C_{n-3}^{n-1} q^{n-3}\right)p^3=(1+3q+6q^2+10q^3+15q^4)p^3.$$

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 $$\begin{eqnarray*} &&(1+b+b^2+\ldots)a(1+b+b^2+\ldots)a(1+b+b^2+\ldots)a\\ &=&(1+b+b^2+b^3+b^4+\ldots)^3a^3\\ &=&(1+3b+6b^2+10b^3+15b^4+\ldots)a^3 \end{eqnarray*}$$ consists of all the cases that Player A earns 3 more points. It is easy to see that only up to $b^4$ corresponding to that Player A wins the game. Therefore we have $$P(A)=(1+3q+6q^2+10q^3+15q^4)p^3.$$

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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