olymmath

HMMT 2006 C7

  • Combinatorics
  • Answer

Let nn be a positive integer, and let Pushover be a game played by two players, standing squarely facing each other, pushing each other, where the first person to lose balance loses. At the HMPT, 2n+12^{n+1} competitors, numbered 1 through 2n+12^{n+1} clockwise, stand in a circle. They are equals in Pushover: whenever two of them face off, each has a 50%50 \% probability of victory. The tournament unfolds in n+1n+1 rounds. In each round, the referee randomly chooses one of the surviving players, and the players pair off going clockwise, starting from the chosen one. Each pair faces off in Pushover, and the losers leave the circle. What is the probability that players 1 and 2n2^{n} face each other in the last round? Express your answer in terms of nn.

Dataset text NuminaMath olympiads. Quoted with source.

Show the answer

\frac{2^{n}-1}{8^{n}}

Solutions

No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

Add a solution

Locator hmmt/2006/c7 · Tag M7E364V018 · Edit this page on GitHub · Keys: j next, k previous