Long-Run Win Proportion in a Serve-Switching Game

Expectation · Medium · Free problem

Alice and Bob play an infinite sequence of tennis points. One of them is chosen at random to serve the first point, and after that, the winner of each point serves the next one. Alice wins each point she serves with probability $p_1$, and she wins each point Bob serves with probability $p_2$.

Find the long-run proportion of points won by Alice as a function of $p_1$ and $p_2$. Report the numerical answer when $p_1 = 2/3$ and $p_2 = 1/3$.

Hints

  1. Think of "who is serving" as the state of a two-state Markov chain. What are the transition probabilities?
  2. Use a renewal argument: consider the expected number of points between consecutive Alice wins. Once Alice wins, she always serves next, so the renewal cycle starts from a known state.
  3. From state $A$ (Alice serving), if Bob wins the point (probability
    - p_1$), the number of subsequent points until Alice wins from state $B$ is $\text{Geometric}(p_2)$ with mean
    /p_2$. Combine these to get $E[T_A] = 1 + (1 - p_1)/p_2$.

Worked Solution

How to Think About It: This is a two-state Markov chain -- the state is "who is serving" and the transitions depend on who wins each point. The winner serves next, so winning begets serving advantage (or disadvantage, depending on the probabilities). To find Alice's long-run win proportion, you need two things: (1) how often Alice is serving in the long run, and (2) how often she wins conditional on who is serving. Before any math, notice something: if $p_1 + p_2 = 1$ (which is the case here with

/3 + 1/3 = 1$), there is a symmetry -- Alice's advantage on her own serve is exactly offset by her disadvantage on Bob's serve. So the answer should be
/2$. That is a good sanity check.

Quick Estimate: With $p_1 = 2/3$ and $p_2 = 1/3$, Alice wins two-thirds of her serves and one-third of Bob's serves. If she served half the time, her win rate would be $(1/2)(2/3) + (1/2)(1/3) = 1/2$. And since winning more leads to serving more, but serving more with $p_1 > 1/2$ means winning more -- there is a feedback loop. But the symmetry $p_1 + p_2 = 1$ means the feedback perfectly balances: she serves more than half the time, but her overall win rate stays at

/2$. Let's confirm with the exact calculation.

Approach: We use a renewal argument. Consider the inter-arrival times between consecutive points that Alice wins, then apply the renewal reward theorem.

Formal Solution:

Model this as a Markov chain with two states: $A$ (Alice serving) and $B$ (Bob serving). The transition rules are:

Let $E[T_A]$ be the expected number of points between consecutive Alice wins, starting from a point where Alice is serving. Condition on the outcome of the current point:

This gives:

$E[T_A] = p_1 \cdot 1 + (1 - p_1)\left(1 + \frac{1}{p_2}\right) = 1 + \frac{1 - p_1}{p_2}$

The key observation: once Alice wins a point, she is serving. So the renewal cycle always starts from state $A$. By the renewal reward theorem, the long-run fraction of points Alice wins is:

$\text{Proportion} = \frac{1}{E[T_A]} = \frac{1}{1 + \frac{1 - p_1}{p_2}} = \frac{p_2}{p_2 + 1 - p_1}$

We can verify this with the stationary distribution approach. Let $\pi_A$ be the long-run fraction of points where Alice is serving. The balance equation is:

$\pi_A = p_1 \pi_A + p_2(1 - \pi_A)$

Solving: $\pi_A(1 - p_1 + p_2) = p_2$, so $\pi_A = \frac{p_2}{1 - p_1 + p_2}$. The proportion of points Alice wins is:

$p_1 \pi_A + p_2(1 - \pi_A) = \frac{p_1 p_2 + p_2(1 - p_1)}{1 - p_1 + p_2} = \frac{p_2}{1 - p_1 + p_2}$

Both methods agree.

Answer: The long-run proportion of points won by Alice is

$\frac{p_2}{p_2 + 1 - p_1}$

With $p_1 = 2/3$ and $p_2 = 1/3$:

$\frac{1/3}{1/3 + 1/3} = \frac{1/3}{2/3} = \boxed{\frac{1}{2}}$

Intuition

This problem illustrates a clean application of renewal theory to Markov chains. The trick is recognizing that every time Alice wins a point, the system resets to the same state (Alice serving), so the gaps between her wins form i.i.d. renewal cycles. The long-run win fraction is just

$ divided by the expected cycle length -- no need to solve for the full stationary distribution, though both methods give the same answer.

The result $p_2 / (p_2 + 1 - p_1)$ has a nice interpretation: it depends on Alice's probability of winning Bob's serve ($p_2$) relative to the total "leakage" rate $p_2 + (1 - p_1)$. If Alice is dominant on both serves ($p_1$ close to

$, $p_2$ close to
$), the proportion approaches
$. The special case $p_1 + p_2 = 1$ always gives exactly
/2$ regardless of the individual values -- the serve advantage and disadvantage cancel perfectly. In practice, this kind of Markov chain renewal argument appears constantly in market microstructure (e.g., modeling sequences of buys and sells where the last trade influences the next) and in sports analytics.

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