Rock, Paper, Scissors

For each round, Alice and Bob split two points, so after five rounds, Alice and Bob have 10 points between them. If Alice has 4 points, then Bob must have 6 points, so Alice did not earn more than Bob.

With respect to Alice's score, let $x$ denote the number of wins, $y$ the number of ties and $z$ the number of losses. If Alice's score is $2x + y + 0z = 2x + y$ , then Bob's score is $2z + y + 0x = 2z + y$ , which indicates that in a round:

• only one player can win the game, so another loses
• otherwise, both players earn a tie

The given situation can be presented as the following system of equations

$\begin{array}{rll} 2x + y &= 4&{\color{#3D99F6}\text{(Alice's score)}}\\ 2z + y& &{\color{#3D99F6}\text{(Bob's score)}}\\ x + y + z &= 5&{\color{#3D99F6}\text{(Alice's summary after five rounds)}} \end{array}$

where Bob's score value is not given.

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For this problem, suppose that Alice earns points strictly more than Bob. Then, $\begin{array}{rl} {\color{#3D99F6}\text{(Alice's score)}} &> {\color{#3D99F6}\text{(Bob's score)}}\\ 2x + y &> 2z + y \end{array}$ Since $x + y + z = 5$ , $\begin{array}{rl} 2x + y &> 2(5 - x - y) + y\\ 2x + y &> 10 - 2x - y\\ 2x + y + (2x + y) &> 10\\ 2(2x + y) &> 10\\ 2x + y &> 5 \end{array}$ However, by the given, $2x + y = 4$ , which contradicts the inequality above. Thus, it is impossible for Alice to earn more points than Bob.

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Here are all possibilities from the givens:

• Alice has $2$ wins and $3$ losses, whereas Bob has $3$ wins and $2$ losses. So Bob earns $6$ points.
• Alice has $1$ win, $2$ ties and $2$ losses, whereas Bob has $2$ wins, $2$ ties and $1$ loss. So Bob earns $6$ points.
• Alice has $4$ ties and $1$ loss, whereas Bob has $1$ win and $4$ ties. So Bob earns $6$ points.

So Bob ends up earning more points than Alice.