Counting wins from rock paper scissors tallies

Two players play ten rounds of rock-paper-scissors, and no round is a draw. Over the ten rounds the first player threw rock $3$ times, paper once and scissors $6$ times; the second threw rock twice, paper $4$ times and scissors $4$ times. How many rounds did the first player win? Answer $-1$ if this cannot be determined.

$7$. With no draws, the first player's six scissors must meet non-scissors, i.e. all $2$ rocks and $4$ papers of the opponent, giving $4$ wins and $2$ losses. The opponent's four scissors then face the first player's $3$ rocks (wins) and $1$ paper (loss), for $4+3=7$ wins.