Albert and Betty are playing the following game. There are $100$ blue balls in a red bowl and $100$ red balls in a blue bowl. In each turn a player must make one of the following moves: a) Take two red balls from the blue bowl and put them in the red bowl. b) Take two blue balls from the red bowl and put them in the blue bowl. c) Take two balls of different colors from one bowl and throw the balls away. They take alternate turns and Albert starts. The player who first takes the last red ball from the blue bowl or the last blue ball from the red bowl wins. Determine who has a winning strategy.
Problem
Source: Baltic Way 2014, Problem 8
Tags: combinatorics proposed, combinatorics
SCP
11.11.2014 18:40
Edited after remark of Rembetika: So when Albert did a , Betty does b . Albert does b => Betty does a Albert doesc and so will Betty do. After a turn of Betty, the number of blue balls in the red bowl is equal to the number of red balls in the blue bowl. As long as that number is at least 3, there is no problem. When it would be lower than 3, Albert made some bowl contains less than 3 balls of the opposite colour, so she can remove those and wins. ( the number of balls in always even in a bowl, by doing a or b she removes two of the balls of the oppositie colour, with c exactly one)
mavropnevma
11.11.2014 18:48
The way c) is stated, it can only be done by taking one red and one blue ball from a same bowl; so c) cannot be used as first move. It seems Betty might use some mirroring strategy ...
itslumi
03.07.2020 17:00
B can always ensure that One of the bowls will remain only with their balls(the red bowl with blue balls and blue bowl with red balls) and he can ensure that the number of balls in that bowl will remain always even,thus the game is finite implies that B will take the last two balls.