Counting Paths to Deuce in a Tennis Game

Combinatorics · Medium · Free problem

Raj and Rachel are playing a simplified tennis game with the following rules:

  1. Scoring starts at $0$-$0$. When someone wins a point, their score increases by
    $.
  2. If a player reaches $4$ points before the other player reaches $4$, that player wins -- unless the score reaches $3$-$3$ (deuce). From deuce onward, a player must lead by two points to win.

How many distinct sequences of point outcomes lead to a score of $4$-$4$?

Hints

  1. What must be true about the score before either player can reach $4$ points without the game ending?
  2. Count the number of ways to interleave $3$ wins for each player over $6$ points -- this is a binomial coefficient.
  3. From $3$-$3$, the next two points must be split between the players. How many orderings are there for that?

Worked Solution

How to Think About It: The first thing to notice is that the score $4$-$4$ can only be reached if the game passes through $3$-$3$ first. Why? Because if either player reaches $4$ while the other has fewer than $3$ points (say $4$-

$, $4$-
$, or $4$-$0$), the game ends immediately. So the only route to $4$-$4$ goes through $3$-$3$. Once you see that, the problem splits into two independent counting steps: how many ways to reach $3$-$3$, and how many ways to go from $3$-$3$ to $4$-$4$.

Quick Estimate: To reach $3$-$3$, exactly $6$ points are played. We need to choose which $3$ of those $6$ points Raj wins (Rachel wins the other $3$). That is $\binom{6}{3} = 20$. From $3$-$3$, the next two points must be split one apiece -- either Raj then Rachel, or Rachel then Raj -- so

$ orderings. Rough answer: 0 \times 2 = 40$.

Approach: We count ordered sequences of point outcomes, splitting the problem at the deuce state.

Formal Solution:

Step 1 -- Reaching $3$-$3$: After exactly $6$ points, both players have $3$. The number of ways to interleave $3$ Raj-wins and $3$ Rachel-wins in $6$ slots is:

$\binom{6}{3} = 20$

Each such sequence is valid because no player reaches $4$ before the other reaches $3$ -- both arrive at $3$ simultaneously on the $6$th point.

Step 2 -- Going from $3$-$3$ to $4$-$4$: From deuce, the next two points must be won by different players (if the same player won both, the score would be $5$-$3$ and the game would have ended at $4$-$3$). The two valid orderings are:

That gives $ ways.

Step 3 -- Combine: The two stages are independent (the first $6$ points and the next $ points), so we multiply:

$\binom{6}{3} \times 2 = 20 \times 2 = 40$

Answer: There are $\binom{6}{3} \times 2 = 40$ distinct sequences of point outcomes that lead to a score of $4$-$4$.

Intuition

The key insight is that the game's stopping rule constrains which states are reachable. Since the game ends the moment either player hits $4$ while the other is below $3$, the score $4$-$4$ is only accessible through the bottleneck state $3$-$3$. This turns a potentially messy path-counting problem into a clean product: count the paths to the bottleneck, then count the paths from the bottleneck to the target.

This "bottleneck decomposition" pattern shows up constantly in combinatorics and probability -- whenever a process has a mandatory intermediate state, you can split the counting (or probability calculation) at that state and multiply. In finance, the same logic appears in barrier option pricing: the payoff depends on whether the price path passes through a critical level, and you decompose the analysis at that barrier.

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