The question is taken from my math book and is supposed to be solved using generating functions without any tools like a calculator, but I don't know how.
Person A and person B is playing a game. In each round the winner gets 1 point, the loser gets 0 points and a draw gives both players a point. After an unknown number of rounds, person A has 4 points while person B has 2 points. In how many ordered combinations of rounds can this happen?
I first tried transferring the problem into an equation with three variables for person A win, person B win and draw which would have to equal 6 points in total. But failed.
Does anyone have a good solution to the problem? If there is a much easier way without generating functions, I would be happy to know that solution as well. The book says the answer is 41 combinations, but there is no explanation.
Answer is quite straight forward by enumerating cases. Observe that game must have at least 4 rounds (because in each round any player can gain almost 1 point), and cannot have more than 6 rounds (no matter the result of any round, at least one player gains 1 point in every round, for e.g. if there are 7 rounds, either A will have >4 points or B will have >2 points or both; note - points can only increase and never decrease).
4 rounds : Since A must gain point in every round, he can either Win or Draw. Since B cannot have >2 points, he must win 0 points in exactly 2 out of 4 rounds. The only way this can happen is : A wins 2 rounds, draws 2 rounds. Possibilities : 6
5 rounds : By similar analysis, you can prove that the only way to do this is when A wins exactly 3 rounds, draws 1 round & loses 1 round. Possibilities: 20
6 rounds : A must win exactly 4 rounds and lose exactly 2. Possibilities : 15
Total : 6+20+15 = 41