Answer to Question #179022 in Statistics and Probability for eoa anoh

Question

In a tennis match, with the score at deuce, the game is won by the first player who gets a clear lead of two points.If the probability that given player wins a particular point isθ, and all points are played independently, what is the probability that player eventually wins the game

Answer 1

Solution:

Let us define

- sample space $\Omega$ to consist of all possible infinite sequences of points;

- event $W_{i}$ - nominated player wins the i th point;

- event $V_{i}$ - nominated player wins the game on the i th point;

- event V - nominated player wins the game.

We want P(V). The events $\left\{W_{1}, W_{1}^{\prime}\right\}$ partition $\Omega$ , and thus, by the Theorem of Total Probability

$P(V)=P\left(V \mid W_{1}\right) P\left(W_{1}\right)+P\left(V \mid W_{1}^{\prime}\right) P\left(W_{1}^{\prime}\right)$ ...(1)

Now $P\left(W_{1}\right)=\theta$ and $P\left(W_{1}^{\prime}\right)=1-\theta$ . To get $P\left(V \mid W_{1}\right)$ and $P\left(V \mid W_{1}^{\prime}\right)$ , we need to further condition on the result of the second point, and again use the Theorem: for example

$\begin{aligned} P\left(V \mid W_{1}\right) &=P\left(V \mid W_{1} \cap W_{2}\right) P\left(W_{2} \mid W_{1}\right)+P\left(V \mid W_{1} \cap W_{2}^{\prime}\right) P\left(W_{2}^{\prime} \mid W_{1}\right) \\ P\left(V \mid W_{1}^{\prime}\right) &=P\left(V \mid W_{1}^{\prime} \cap W_{2}\right) P\left(W_{2} \mid W_{1}^{\prime}\right)+P\left(V \mid W_{1}^{\prime} \cap W_{2}^{\prime}\right) P\left(W_{2}^{\prime} \mid W_{1}^{\prime}\right) \end{aligned}$

...(2)

where

$\begin{array}{ll} P\left(V \mid W_{1} \cap W_{2}\right)=1, & P\left(W_{2} \mid W_{1}\right)=P\left(W_{2}\right)=\theta \\ P\left(V \mid W_{1} \cap W_{2}^{\prime}\right)=P(V), & P\left(W_{2}^{\prime} \mid W_{1}\right)=P\left(W_{2}^{\prime}\right)=1-\theta, \\ P\left(V \mid W_{1}^{\prime} \cap W_{2}\right)=P(V), & P\left(W_{2} \mid W_{1}^{\prime}\right)=P\left(W_{2}\right)=\theta \\ P\left(V \mid W_{1}^{\prime} \cap W_{2}^{\prime}\right)=0, & P\left(W_{2}^{\prime} \mid W_{1}^{\prime}\right)=P\left(W_{2}^{\prime}\right)=1-\theta, \end{array}$

as,

given $W_{1} \cap W_{2}$ : the game is over, and the nominated player has won given $W_{1} \cap W_{2}^{\prime}$ : the game is back at deuce given $W_{1}^{\prime} \cap W_{2}$ : the game is back at deuce given $W_{1}^{\prime} \cap W_{2}^{\prime}$ } : the game is over, and the player has lost

and the results of successive points are independent. Thus

$\begin{aligned} P\left(V \mid W_{1}\right) &=(1 \times \theta)+(P(V) \times(1-\theta))=\theta+(1-\theta) P(V) \\ P\left(V \mid W_{1}^{\prime}\right) &=(P(V) \times \theta)+0 \times(1-\theta)=\theta P(V) \end{aligned}$

Hence, combining (1) and (2), we get

$\begin{array}{c} P(V)=(\theta+(1-\theta) P(V)) \theta+\theta P(V)(1-\theta)=\theta^{2}+2 \theta(1-\theta) P(V) \\ \end{array}$

$\Rightarrow P(V)=\frac{\theta^{2}}{1-2 \theta(1-\theta)}$