Snooker
A player has probability 0.4 of winning a single game. What is his probability of winning a 'best of 15 games' tournament?
Problem
In a particular snooker tournament, two players play "frames" against each other and the first to win 8 frames wins the match. This is known as a "best of 15" match. Assume that the probability of player A winning each frame is $p$, regardless of who starts. If A does not win the frame then his opponent does (there are no draws).
Suppose $p = 0.4$, then what is the probability that A wins the match? Now suppose that A is very slightly better than his opponent, say $p = 0.55$, what is the probability now of A winning the match? What is the probability of A winning the match when $p = 0.5$?Getting Started
You have to find the probability of the player winning 8 frames outright, or of winning 7 frames in the first 8 not winning the 8th and then winning the 9th or winning 7 of the first 8, losing the next two frames then winning the tenth or ... and so on.
Student Solutions
Thank you Sue Liu of Madras College, St Andrews for this solution.
This could be any 'best of 15 games' contest between two players where the object is to be the first to win 8 games (called frames in snooker) and the probability $p$ of winning a single game is constant.
We have to find the probability of player A winning the snooker match by adding the probabilities for all the possible outcomes. Player A can win in 8 frames (by winning the first 8 frames outright), or by winning any 7 of the first 8 then winning the ninth (when the match lasts 9 frames), or by winning any 7 of the first 9 then winning the tenth (when the match last 10 frames), or similarly player A can win a match which lasts for 11, 12 13 14 or 15 frames. Note that the last game, which decides the contest, must be won by A. Let $P(x)$ denote the probability of A winning a match with $x$ games in total.
$$\begin{eqnarray} P(8) &= {7\choose 7}p^7p \\ P(9) &= {8\choose 7}p^7(1-p)p\\ P(10)&= {9\choose 7}p^7(1-p)^2p \\ \ldots \\ P(15)&= {14\choose 7}p^7(1-p)^7p \end{eqnarray}$$
Let $f(p)$ denote the probability of A winning a 'best of 15' match when the probability of winning each frame is $p$. \[p = P(8) + P(9) + P(10) + P(11) + P(12) + P(13) + P(14) + P(15).\]
When $p = 0.4$ this gives the probability $f(0.4)$ of A winning a 'best of 15' match to be 0.2131.
When the probability of A winning each frame is $p = 0.55$ then the probability of the opponent winning a frame is $1-p = 0.45$ and the probability $f(0.55)$ of A winning the match is 0.6535.
When the probability of A winning each frame is $p = 0.5$ then the probability of A winning the match is 0.5 as expected.
Teachers' Resources
Why do this problem?
This problem gives a realistic and extended application of the binomial distribution, with an opportunity to consider the effects of the relative strength of opponents in a contest. It will allow students to consider how theoretical probability can be applied to a more complex situation.
Possible approach
Have students consider what the best of 15 means and what the possible outcomes are. How many matches might be played in total? It is helpful to consider the situation from the point of view of Player A and their wins and losses. What is significant about the final frame in any match? Before beginning any calculations, ask students to estimate the probability of Player A winning the match for different starting probabilities. How much might changing the probability of winning a single frame from $p=0.4$ to $p=0.55$ impact the probability of winning the match overall?
After minimal discussion to ensure a good understanding of the situation, students should be able to proceed with calculation in each of the scenarios described. Once the answers are found, compare with the initial estimates that the class made. Whose intuition was reliable? Students could discuss the interpretation of their results in terms of opponents of different relative strengths.