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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?

I don't know the answer.

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.

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  • $\begingroup$ 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. $\endgroup$ – TonyK Mar 9 at 21:04
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This is not a probability question. :) Look, Alice played rock 12 times, and Bob only played non-rock 12 times, and the question specified that there were no ties, so...

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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 $25$).

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Nothing to do with probability.

Alice played rock 12 times, since there are no ties, Bob played scissors 9 times and paper 3 times in those games (score 9 (A) : 3 (B) so far). The rest of the games (13) Bob played rock, so he won 6 times when Alice played scissors and lost 7 times when she played paper. The final score: 16 (A) : 9 (B)

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