# rock scissor paper probability

Alice and Bob played 25 games of rock-paper-scissors. Alice played rock 12 times, scissors 6 times, and paper 7 times. Bob played rock 13 times, scissors 9 times, and paper 3 times.

If there were no ties, who won the most games?

My thinking is:

• When Alice plays 12 rock, she can win only 9 times (since paper 3 times ob will make her loose) : 9/12
• When Alice plays, scissors 6 times, she could loose 6 times or win 3 times: : 3/9
• When Alice plays, paper 7 times, she could win 7 times or loose 6 times:7/13

Don't know if this is the right approach.

• The only reason that this problem is solvable is that Alice's $12$ rocks plus Bob's $13$ rocks account for all $25$ games. In general, you won't be able to work out what the final score is just from data like this. By the way, this is an extremely unlikely sequence of games: you might like to work out for yourself the probability that none of the $25$ games will be ties. – TonyK Mar 9 at 21:04

Of the $$12$$ times that Alice played rock, Bob played all of his scissors and paper winning $$3$$ times (for paper) and losing $$9$$ times (for scissors). In the other $$13$$ games, Bob played rock winning $$6$$ times (against Alice's scissors) and losing $$7$$ times (against Alice's paper). In total, Bob won $$3+6 = 9$$ times and lost $$9+7=16$$ times, meaning that Alice won the most games ($$16$$ out of $$25$$).