Momentum-Switching Tennis Tournament

Probability · Hard · Free problem

Fix $0 < p < 1$. Alice and Bob play a best-of-$(2n-1)$ tennis tournament. Each round awards 1 point, and the first player to reach $n$ points wins.

Here is the twist: the trailing player always has the advantage. Specifically:

  • If one player has fewer points, that player wins the next round with probability $p$.
  • If one player has more points, that player wins the next round with probability
    - p$.
  • If they are tied, Alice wins the next round with probability $p$.

Find the probability that Alice wins the tournament when $n = 10$ and $p = 2/3$.

Hints

  1. Try computing the answer for small values of $n$ before attacking $n=10$. The pattern you find should make you suspicious that something deeper is going on.
  2. Condition on whether the score ever returns to a tie after the first round. Think about what the tournament looks like from any tied state -- does it remind you of a smaller version of the same problem?
  3. Set up a strong induction on $n$. At the first return to a tied score $(k,k)$, the remaining series is a best-of-$(2(n-k)-1)$ tournament starting from equal footing, which is exactly the inductive hypothesis.

Worked Solution

Setup: Best-of-$(2n-1)$ tournament where the trailing player wins each round with probability $p$, the leading player wins with probability

-p$, and ties go to Alice with probability $p$. We need $P(\text{Alice wins})$ for $n=10$, $p=2/3$.

Key Computation -- Small Cases:

For $n=1$: a single game from a tied state, so Alice wins with probability $p$.

For $n=2$ (best of 3), enumerate Alice's winning paths:

  • $WW$: Alice wins game 1 from a tie (prob $p$), then leads 1-0 and wins game 2 (prob
    -p$). Contribution: $p(1-p)$.
  • $WLW$: Win, lose, win. After $WL$ the score is tied 1-1, so game 3 has probability $p$. Contribution: $p \cdot p \cdot p = p^{3}$.
  • $LWW$: Lose, win, win. After $L$ Alice trails 0-1, so she wins game 2 with probability $p$. After $LW$ the score is tied 1-1, so game 3 has probability $p$. Contribution: $(1-p) \cdot p \cdot p = p^{2}(1-p)$.

Summing: $p(1-p) + p^{3} + p^{2}(1-p) = p[(1-p) + p^{2} + p(1-p)] = p[(1-p)(1+p) + p^{2}] = p[1 - p^{2} + p^{2}] = p$.

So the answer for $n=2$ is also $p$. This suggests the answer is $p$ for all $n$.

Proof by Strong Induction:

*Base case:* $n=1$. One game from a tie -- Alice wins with probability $p$.

*Inductive hypothesis:* For all

\leq k \leq n$, Alice wins a best-of-$(2k-1)$ tournament with probability $p$.

*Inductive step:* Show Alice wins the best-of-$(2n+1)$ tournament (first to $n+1$) with probability $p$. Condition on whether the score is ever tied again after the first round:

$P(\text{Alice wins}) = P(\text{Alice wins} \mid T)\,P(T) + P(\text{Alice wins} \mid T^{c})\,P(T^{c})$

*Case 1 ($T^{c}$ -- no subsequent tie):* Whoever wins the opening game (played from a tie, so Alice wins it with probability $p$) takes the lead and never relinquishes it. Therefore $P(\text{Alice wins} \mid T^{c}) = p$.

*Case 2 ($T$ -- a tie occurs):* At the first return to a tied score $(k,k)$ with

\leq k \leq n$, the remaining contest is first-to-$(n+1-k)$ from a tied state -- a best-of-$(2(n+1-k)-1)$ tournament. Since
\leq n+1-k \leq n$, the inductive hypothesis gives $P(\text{Alice wins} \mid T) = p$.

Combining:

$P(\text{Alice wins}) = p \cdot P(T) + p \cdot P(T^{c}) = p$

Result: The probability Alice wins is $p$ for every $n$. For $n=10$ and $p=2/3$:

$\boxed{\dfrac{2}{3}}$

Intuition

The key insight is that the momentum-switching rule is perfectly self-balancing. When a player falls behind, the rule hands them a higher win probability, which tends to pull the score back to a tie. When a player gets ahead, their win probability drops, making it harder to extend the lead. This tug-of-war means that the outcome of the entire tournament is determined entirely by what happens at tie states -- and at every tie state, Alice has the same probability $p$ of winning the next point. The beautiful consequence is that no matter how long the series is, all the complicated dynamics in between ties wash out, and the tournament probability collapses to $p$.

This is a great example of finding the right thing to condition on. The naive approach -- tracking the full Markov chain of scores -- works but obscures the structure. By conditioning on whether the score ever returns to a tie, you decompose the problem into pieces that are either trivially solved (no tie means the first game decides everything) or reducible to a smaller instance of the same problem (tie means restart a shorter tournament). This "condition on the first return to a nice state" trick shows up constantly in random walks, gambler's ruin problems, and Markov chain analysis. In interviews, it is the difference between grinding through a 19-state recursion and producing an elegant one-line answer.

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