Numbers game for 2017

Probability Level 3

The numbers $\frac{1}{1}, \frac{1}{2}, \frac{1}{3} , \ldots , \frac{1}{2017}$ are written on a blackboard. Andrew and Bob take turns, and Andrew goes first. On each turn, one of Andrew or Bob chooses two numbers $x$ and $y$ on the board, erases them and writes $x+y+xy$ . They do this until only one number is left on the board. Andrew wins if the number is odd, Bob wins if the number is even, and it is a draw if the number is not an integer.

Assuming Andrew and Bob play optimally, which player will win, or will it be a draw?

Andrew Draw Bob

1 solution

Brian Moehring
Feb 15, 2017

Define $x\diamond y = x+y+xy$ . Then $\begin{aligned} x\diamond (y\diamond z) &= x\diamond (y+z+yz) = x+(y+z+yz) + x(y+z+yz) = x+y+z + xy+xz+yz + xyz \\ (x\diamond y) \diamond z &= (x+y+xy)\diamond z = (x+y+xy) + z + (x+y+xy)z = x+y+z + xy+xz+yz + xyz \end{aligned}$ $x\diamond y = x+y+xy = y+x+yx = y\diamond x$ so that $\diamond$ is both associative and commutative. Therefore, the final number is independent is any strategies the two players use .

Therefore, we may choose the order to play the game.

Step 1: choose $1$ and $\frac{1}{2}$ . Then $1 \diamond \frac{1}{2} = 1 + \frac{1}{2} + 1\cdot \frac{1}{2} = 2$ .
Step 2: choose $2$ and $\frac{1}{3}$ . Then $2 \diamond \frac{1}{3} = 2 + \frac{1}{3} + 2\cdot \frac{1}{3} = 3$ .

and so on, the pattern continuing as

Step n: choose $n$ and $\frac{1}{n+1}$ . Then $n \diamond \frac{1}{n+1} = n + \frac{1}{n+1} + n\cdot \frac{1}{n+1} = n+1$ .

until

Step 2016: choose $2016$ and $\frac{1}{2017}$ . Then $2016 \diamond \frac{1}{2017} = 2017$ .

In summary, the final number will always be $2017$ , which is odd, so $\boxed{\text{Andrew wins}}$ .

Numbers game for 2017

1 solution

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