First is the worst, second is the best? A Markov chain analysis of the basketball game knockout

TL;DR

The paper addresses whether starting position affects winning probabilities in the playground game Knockout, and whether the adage "first is the worst" holds. It develops an absorbing Markov chain framework to model Knockout, solving the two-player case exactly and handling the $n$-player case numerically via a structured transition representation with $6n$ transient states per round and $n$ absorbing states. A key result is that the two-player win probability for the first shooter is $P_1$ wins = $P_2$ eliminated = $\dfrac{1}{3-p}$, with $p\in[0,1)$, placing $P_1$’s win probability in $[\tfrac{1}{3},\tfrac{1}{2})$; for $n>2$, the model uses a block-structured $T$ matrix and derives the per-elimination and total game-length expressions, showing the expected length per elimination is independent of $n$, while the total length scales with $(n-1)$. Numerical analyses reveal strong trends: the first position is almost always the worst, while even-numbered positions gain a noticeable bump and the last position is often optimal, with the magnitude of these effects depending on $p$ and $q$. The work further discusses extensions to unequally skilled players and outlines several questions for future research, including how superstars would influence optimal starting positions in mixed-skill groups.

Abstract

The game of Knockout is a classic playground game played with two basketballs. This paper uses a Markov process to analyze each player's probability of winning the game given their starting position in line and shooting percentages, assuming all players are equally skilled. The two-player case is solved in general for any probability of a long shot and short shot shooting percentage and the n-player case with n > 2 is solved numerically. In doing so, this paper answers the question of whether or not the playground wisdom of ``first is the worst, second is best'' is true. We also examine the average number of rounds it takes before the game ends, analyze trends in the data to recommend tips to win at Knockout, and provide questions in the case of players not being equally skilled.

Figures (5)

• Figure 1: Directed graph representing a 2-person game of Knockout
• Figure 2: Probability of winning by position when $n=7$, $p=0.4$ and $q = 0.9$
• Figure 3: Probability of winning by position when $n=100$, $p=0.4$ and $q = 0.9$
• Figure 4: Probability of winning by position when $n=10$, $p=0.4$ and $q = 0.9$
• Figure 5: Probability of winning by position when $n=10$, $q = 0.9$ and $p$ varies between $0.2, 0.5$, and $0.8$.

Theorems & Definitions (19)

• Theorem 2.1
• proof
• Theorem 2.2
• proof
• Theorem 3.1
• proof
• Corollary 3.2
• proof
• Remark 3.3
• Proposition 4.1