Stochastic Processes - Determining probability and expectation on tennis player question

Question

Two tennis players, $A$ and $B$, are rendered equal in a game. It takes two point lead for the winner. A player who has one point is said to benefit. Assuming that $A$ has a probability $p$ of winning each point and $B$ a probability $1-p$, independently of all other points, determine: $(a)$ the probability for $A$ to be declared the winner; $(b)$ the expected number of times that $A$ will benefit before the end of the game.

I have to prepare for an exam for the course of Stochastic Processes, and a part of the material bothers me enormously; essentially it is a problem resolution by conditioning method. Does someone could solve my problem and clearly explain each step? I tried, but I can not go very far in the problem. It should also say that my first course probability goes back more than three years; this may be part of why I have trouble on that kind of question.

Thanks!

Answer 1

After two points, either $A$ has won, $B$ has won, or the players are back where they started. These three events have probabilities $p^2,(1-p)^2$, and $2p(1-p)$, respectively.

Let $W$ be the event that $A$ wins, then from the above we have $$ \mathbb{P}(W)=p^2+2p(1-p)\mathbb{P}(W)$$ hence $$ \mathbb{P}(W)=\frac{p^2}{1-2p(1-p)}$$

A similar argument should work for the expectation.

Answer 2

There are 5 possible states for the Markov Chain.

State 1: A and B are equal

State 2 : A benefit

State 3 : B benefit

State 4 : A Win

State 5 : B Win

Let the probability that A wins starting from the state i be $a_i$. It's clear that $a_4 = 1$ and $a_5 = 0$. We have the following equations for the others (Using conditioning)

$$a_2 = a_1 (1-p) + a_4 p $$ $$a_3 = a_1 p + a_5 (1-p) $$ $$a_1 = a_2p + a_3 (1-p) $$

Solving the above equations, we get $a_1 = \frac{p^2}{2p^2-2p+1}$ (This is the required probability as the game begins in State 1)