I'm working with the following problem:
In table tennis, a set is won by the first player to reach 11 points, unless the
score is $10–10$, in which case serves are alternated and the player who first
gets ahead by two points wins. Suppose that Ann wins a point as server with
probability $p_A$ and Bob wins a point as server with probability $p_B$. If the score
is $10–10$ and Ann serves, what is the probability that she wins the set?
Attempt:
For our sake, let $P(W)$ be the probability of Ann winning, where $W$ is the event of winning. We also let $A$ be the event of Ann winning a serve and $B$ the event of bob winning a serve.
We draw ourselves a diagram of the situation:
Using the law of probability, we have:
$$P(W) = P(W|A)P(A) + P(W| A \cap A)P(A \cap A) + 2P(W| A \cap B)P(A \cap B) + P(W| B \cap B)P(B \cap B)$$
The last term vanishes, since the probability of Ann winning becomes $0$ if Bob wins $2$ serves in a row.
Furthermore, we realize that $P(W) = P(W|A \cap B) := p$. Hence, we have:
$$ p = p_a^2 + 1 \cdot p_a^2 + 2p \cdot p_a \cdot p_b$$
Meaning we have $$ p = \frac{2p_a^2}{1-2p_a p_b}$$
but the answer sheet tells me that
$$ p = \frac{p_a(1-p_b)}{p_a+p_b-2p_a p_b}$$
I hope I can get some tips that'll help me solve this problem. I'm thinking that I get it wrong in the step where $P(W|A)P(A)$ is calculated, since it seems pretty off that this should be equal to the probability $P(W|A\cap A)P(A\cap A)$. Thanks.
Best Answer
Using the tip from my comment above, we have the events
We note that $A\cap B$ and $\overline A \cap \overline B$ both result in starting over. If the probability of winning is $p_W$, then we therefore have $$ P(W\mid A\cap B)=p_W\\ P(W\mid \overline A\cap \overline B)=p_W $$ The law of total probability yields $$ \begin{align} p_W={}&P(W\mid A\cap \overline B)P(A\cap\overline B)\\ &+P(W\mid A\cap B)P)A\cap B\\ &+P(W\mid\overline A\cap \overline B)P(\overline A\cap \overline B)\\ &+P(W\mid \overline A\cap B)P(\overline A\cap B)\\ ={}&1\cdot p_A(1-p_B)+p_W\cdot p_Ap_B\\ &+p_W\cdot(1-p_A)(1-p_B)+0\cdot (1-p_A)(1-p_B)\\ ={}&p_A(1-p_B)+p_W(p_Ap_B+(1-p_A)(1-p_B)) \end{align} $$ Solving $p_W=p_A(1-p_B)+p_W(p_Ap_B+(1-p_A)(1-p_B))$ for $p_W$ does indeed yield $$ p_W=\frac{p_A(1-p_B)}{1-(p_Ap_B+(1-p_A)(1-p_B))}\\ =\frac{p_A(1-p_B)}{p_A+p_B-2p_Ap_B} $$
Alternately, the "Starting over" case may be simplified away using $$ p_W=\frac{P(W\mid \text{not starting over})}{P(\text{not starting over})}\\ =\frac{p_A(1-p_B)}{p_A(1-p_B)+(1-p_A)p_B} $$