This model is easy to analyze for 2 players. If player 1 chooses x≤1/2, player 2 chooses x+2ε, and the scores are x+ε and 1–x–ε respectively. If player 1 chooses x≥1/2, player 2 chooses x–2ε, and the scores will be 1–x+ε and x–ε respectively. Thus player 1 chooses 1/2 and player 2 chooses 1/2±2ε, and the scores will be 1/2+ε and 1/2–ε.

The 3 player case is more complicated. If player 1 chooses x≤1/4, player 2 chooses (1–x)/3, and player 3 chooses any number between them in the interval (x, (1–x)/3). In this case the average scores will be (5x+1)/6, (1–x)/3, and (1–x)/2 respectively. If player 1 chooses 1/4<x≤1/2, then player 2 chooses 1–x+2ε and player 3 chooses x–2ε. In this case the scores will be 1/2–x+2ε, 1/2-ε, and x–ε respectively. The cases where player 1 chooses x≥1/2 are similar by symmetry. Thus player 1 chooses 1/4, player 2 chooses 3/4, and player 3 chooses something between in the interval (1/4,3/4), giving average scores of 3/8, 3/8, and 1/4. What are the best strategies for more than 3 players?

This game is also interesting where players get multiple turns in a specified order. For example, for the order 112, player 1 chooses 1/4 and 3/4, and player 2 chooses any number between them in the interval (1/4, 3/4), and the scores will be 3/4 and 1/4 respectively. What are the best strategies for any order of players?

In fact, Dan Dima gives a convincing argument that for larger number of players n, the first (n–1) player will pick odd multiples of 1/(2n–2), with the first two players picking 1/(2n–2) and (2n–1)/(2n–2), leaving the last player to choose any number between them. This leads to scores (4n–9)/4(n–1)(n–2) for the first two players, (2n–5)/2(n–1)(n–2) for most of the players, and 1/2(n–1) for the last player.

If you can extend any of these results, please e-mail me. Click here to go back to Math Magic. Last updated 12/1/17.