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Puzzling

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    $\begingroup$ @FirstNameLastName the "warm-up strategy" just demonstrates the concept that picking some S in that way enables everyone to coordinate an answer (in that case the correct one). In the case where it is not 0, they lose by some amount. The real strategy instead gets them all to lose on 0, while eking out a win on 1 and 2 $\endgroup$ Commented Sep 15 at 16:03
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    $\begingroup$ This strategy extends nicely to the next tiers at 26, 80 and so forth (n x 3 + 2), each time reducing the loss probability by a factor of 3. I wonder, is there any way to do better in between? Are odd numbered Ns salvageable? $\endgroup$ Commented Sep 15 at 17:08
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    $\begingroup$ One useful component for more general solutions might be that a generalization of the n=2 and n=6 strategies in the solution allows getting 1/3 chance of each of the total scores a/b/c for any numbers with a+b+c=0, using a number of players equal to the max(abs(a),abs(b),abs(c)). $\endgroup$ Commented Sep 15 at 20:04
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    $\begingroup$ @nitrodon In the 3 person case, If Bart always picks rock, then his wins and ties do nothing to improve the team's performance, while his losses kill the wins. This gives us a 4/9 chance of winning, which isn't great $\endgroup$ Commented Sep 15 at 21:43
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    $\begingroup$ @kagami: In the 3 person case, the two non-Bart players will choose their strategy based on Bart's hat. Ie, if Bart wins, they neutralize (p=1), if Bart ties, they play the normal two-player strategy (p=2/3), and if Bart loses, they shoot for two wins (p=1/3). End result is 2/3. $\endgroup$ Commented Sep 16 at 5:36