Question

There are $m$ blue marbles and $n$ red marbles on a table. Armaan and Babita play a game by taking turns. In each turn the player has to pick a marble of the colour of his/her choice. Armaan starts first, and the player who picks the last red marble wins. For how many choices of $(m, n)$ with $1 \leq m, n \leq 11$ can Armaan force a win?

Answer 1

66

Answer 2

The game is won by the player who picks the last red marble. This is a game where players can pick either a red or a blue marble. The presence of blue marbles allows a player to "pass" their turn of picking a red marble. The crucial factor is the parity of the number of red marbles, $n$.

1. If $n$ is odd: Armaan starts. He can always ensure he picks the $n$-th red marble. His strategy is to pick a red marble whenever it's his turn and red marbles are available. If $n$ is odd, Armaan picks the 1st, 3rd, ..., $n$-th red marble. Babita picks the 2nd, 4th, ..., $(n-1)$-th red marble. Blue marbles don't alter this outcome as players can use them to control whose turn it is when a certain number of red marbles remain. Armaan wins.

2. If $n$ is even: Armaan starts. If Armaan picks a red marble, $n-1$ (odd) red marbles remain for Babita. If Armaan picks a blue marble, $n$ (even) red marbles remain for Babita. In any case, Babita can ensure she is the one to pick the $n$-th red marble. For example, if $n$ is even, Armaan picks. If he picks red, $n-1$ red left. Babita picks red, $n-2$ red left. If Armaan picks blue, $n$ red left. Babita picks red, $n-1$ red left. Babita will always be the one to pick red marbles when the count of remaining red marbles is even, thus picking the $n$-th red marble. Babita wins, meaning Armaan cannot force a win.

Armaan forces a win if and only if $n$ is odd. Given $1 \leq m, n \leq 11$: The odd values for $n$ are $1, 3, 5, 7, 9, 11$ (6 values). For each of these 6 values of $n$, $m$ can be any of the 11 values from 1 to 11. Total number of winning pairs $(m, n)$ for Armaan = $6 \times 11 = 66$.