© Zyanya Santuario, MATH+
Author: Mehmet Akif Yıldız
Challenge
The legendary Nutcracker Workshop is preparing for the grand Christmas Parade!
To be ready for any walnut emergency, they have 2025 walnuts — randomly shoved into three supply crates, none of them empty. By randomly we mean the following: if one lists the numbers a, b and c of walnuts contained in the three crates in increasing order, say a \leq b \leq c, then every triple (a,b,c) of positive integers with a+b+c=2025 is equally likely.
The Nutcracker Headquarters has just informed them that a huge walnut delivery is already on the way. The delivery elves have one very simple rule:
they will only unload into completely empty crates — and this shipment is so massive that they will need two of them!
If fewer than two crates are empty when they arrive, they will simply dump the entire shipment onto the floor, causing a rolling walnut avalanche across the Nutcracker Workshop.
Luckily, there are still a few hours left. So two clever workshop coordinators, Aria and Bram, decide to make a game out of clearing the crates:
The Game
Aria and Bram alternate turns. On a player’s turn, a legal move consists of the following:
- Choose two non-empty crates. Suppose they contain a and b walnuts with 1 \leq a \leq b.
- Choose a positive integer c with 1 \leq c \leq a.
- Transfer c walnuts from the crate containing less walnuts to the other crate. After the move, those two crates contain a - c and b + c walnuts.
A player loses if they cannot make a legal move anymore — that is, if there is only one non-empty crate left. Aria starts the game.
Assume Aria and Bram play optimally. Let p be the probability that Bram wins the game. Which of the options below is closest to p?
Possible Answers
- 0.1 %
- 0.2%
- 0.3%
- 0.5%
- 1%
- 2%
- 5%
- 10%
- 25%
- 50%