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There are $n\ge3$ players playing a game. In this game, one person will come out in first place, one in second, and so on. It's impossible to tie. The person in first place gets $n$ points, the person in second place gets $n-1$, and so on, so that the person in last place gets $1$ point. After playing this game some fixed number of times, the scores are tallied up and the winner is whoever has the highest score.

Can we fix a number of games (other than one game) in advance, so that a tie for first place is impossible? With two players this is easy - just play an odd number of games. But otherwise?

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Answer:

No, a tie is always possible.

Proof/construction:

If the number of rounds is even, have two players alternate being 1st and 2nd place respectively. The remaining players' scores can be anything.
If the number of rounds is odd, set up the first three games to end with tied scores by having one player be 1st, 1st, 3rd and the other be 2nd, 2nd, 1st. Then, have them alternate 1st and 2nd as before for the remaining games (if any). The other players can again have whatever scores; no matter how you do it, they will not be in the running for first.

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    $\begingroup$ Yup, this is basically the solution I used as well, although I expressed it as "if you can tie with $n$ and $m$ games, you can tie with $n+m$ games" and the fact that you can tie for 2 and 3 games. By the way, it might be a good idea to spoiler at least the first line of your answer, since knowing the answer might influence how a person thinks about the problem. $\endgroup$
    – Jack M
    Mar 14, 2015 at 1:49

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