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Warning: Most of my posts are riddles disguised as something else. For those who have enjoyed my contribution and are expecting more, DO NOT try to guess that the answer is a tugboat, geese, or the Winter Bonanza sale. This is NOT a riddle. You have been warned.

This is a question that has frustrated me to no end for the longest time. In a 1 on 1 on 1 game of skill and strategy, there are only two outcomes assuming roughly equal skill. Either (A) two players conflict and the victor (now weakened) is destroyed by the third player or (B) two players both focus their efforts on the same opponent to easily destroy them. The victor who contributed most (and was weakened because of it) is destroyed afterwards.

The argument can be made that all three fighting both opponents at the same time is possible, but it can't derive an outcome. Such game play can only result in an even stalemate until one player changes tactics, resulting in A or B.

It's simple and boring, with the outcome more dependent upon chance (or how disliked you are by the other players) than anything. Making the game itself complex can give allow players to turn the tables, but

What's the simplest possible 1 on 1 on 1 game that can be designed such that it discourages these outcomes and encourages something more interesting?

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    $\begingroup$ This is actually a good question, and I think you'll find people in boardgames.SE with answers as well. $\endgroup$ – generalcrispy Nov 21 '14 at 20:17
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    $\begingroup$ This question would probably be better off on Math.SE. It's not really about puzzles. $\endgroup$ – Joe Z. Nov 21 '14 at 20:27
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    $\begingroup$ @Travis_Kindred I think the answer is the abstract concept of Life. Right? $\endgroup$ – kaine Nov 21 '14 at 20:43
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    $\begingroup$ @kaine it's "geese." $\endgroup$ – hexparrot Nov 21 '14 at 20:46
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    $\begingroup$ This question appears to be off-topic because it is about game design and not puzzles $\endgroup$ – Ric Nov 21 '14 at 22:03
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If the game mechanics allow it, you could devise a situation where the victor gained certain benefits from their victory - experience points or whatever - sufficient to compensate for part or all of the damage they suffered.

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  • $\begingroup$ Bingo! We have a winner! I should have thought of that. It's done that way in most video games, so it would work. I'm thinking stratego-like numbered troops, where adjacent troops can attack together and the greatest sum wins. Ties would end in mutual destruction and victories would end with a power increase for the units engaged. $\endgroup$ – Travis Don Kindred Nov 24 '14 at 4:38
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Have some mechanics that give a player an advantage when two others are working together, and a disadvantage when two others are fighting. Balanced correctly, the whole thing can be a negative feedback loop that only clever strategy can overcome.

Maybe one way to do this is to have a circular advantage loop, like rock-paper-scissors. So if Rock wants to work with Paper to bring down Scissors, it needs to protect Paper against Scissors and also mount its own defense. Rock ends up losing resources on two fronts, faring no better than Scissors.

If Rock fights Scissors alone, make it so Rock will actually gain great strength by defeating Scissors (perhaps by commandeering resources), so Paper has an incentive to even things out, because even Rock, if built up enough, can defeat Paper.

****EDIT (because I don't have the rep to make comments) ****

Regarding chance ruining strategy, that's true only if the players are infinitely intelligent. Complex rules can make a system unpredictable for mere humans, which is exactly what noise does. The only difference between chance and complex rules is that chance is true noise.

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  • $\begingroup$ But if you did that perfectly, it would constantly bring the players back to equal. Stalemate would only be prevented by failure of the system. $\endgroup$ – Travis Don Kindred Nov 21 '14 at 21:00
  • $\begingroup$ ...or player error. It would depend on how strongly the system corrects itself. I would say this system encourages something more interesting, like you requested, but I see how there might be room for disagreement. $\endgroup$ – MackTuesday Nov 21 '14 at 21:22
  • $\begingroup$ If the system could determine the cause of a setback, we're in business. If it granted bonuses unless it detects poor choices, that could be cool. $\endgroup$ – Travis Don Kindred Nov 21 '14 at 21:27
  • $\begingroup$ Hm, good choices give benefits anyway. Giving bonuses to good choices would be the same as amplifying the benefits a player would get anyway, no? $\endgroup$ – MackTuesday Nov 21 '14 at 21:36
  • $\begingroup$ True. Perhaps it compares your overall standing/score with how well you've performed. It gives bonuses or hindrances based on the difference between the two. $\endgroup$ – Travis Don Kindred Nov 21 '14 at 21:42
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Two quick options:

  • 1 on 1 rounds: Three people want to play charades and don't know each other (simpler would be guess what number I am thinking). Adam goes to the front, acts out a clue, and lets Becky and Clara guess. Which ever gets it right wins a point. Becky goes up and does the same. Clara does it too. We then start again with Adam. As the competition for each round is 1 on 1; teaming up is not possible.
  • Three player Paper Scissors Rock: Some games can be designed such that no teamed up strategy offers an advantage for the team. We dictate that if all pick different or the same there is a tie and they go again. If two pick the same, they act as one move against the other player. The two remaining players play against each other. There is no advantage for both teammates whether they chose to play differently or the same. (Note: that this assumes that prizes for 1st, 2nd, and 3rd are chosen so a 50-50 chance of first and last is as valuable as a 100% chance of getting second.)
    For a strategy video game, this means that you cause teaming up to hurt yourself. One example follows like this: Adam has archers who can kill unarmored berzerkers; Becky has berzerkers who are great against the wussy armored cavaliers. Clara's cavaliers, however, can reach archers in time to cut them down. If Adam teams up with Clara he will lose just after Becky does. Becky won't team up with him though for the same reason. You will end up with an odd game, therefore, where everyone is trying to fortify themselves and the neighbor they have an advantage against while weakening their kryptonite. Teaming up, however, is out of the question until very end game.
    A more interesting option that is closer to RPS would be as follows: Adam picks auburn which is really red. Becky and Clara think he could have picked amber or aqua though (yellow or blue). He doesn't know their colors either. Red burns the yellow grain. Yellow drink the blue water. Blue douses the red flames. The strategy game is to collect your own color powerups which is what decides victory in a tie. If they guess what color you are, however, they take away your powerups easier than you can collect them. At end game, you colors are revealed and the purpose of your strategy becomes clear.

One more option since I can:

  • Attacking one player does not work: Design a game where an "attack" damages both other players equally. This would require some thought to build but it is possible. The easiest seems to be either disguised one player games (you look like you fight each other but are really just trying to build your score faster) or a game with a shared resource pool. You win by taking what you need before the others can not buy killing them.
    Otherwise: Adam is amber; Becker, blue; Clara, Crimson. Adams economy relies on growing oranges and guar. Becky needs guar and plums. Clara needs plums and oranges. Without either they fail. Your best strategy to win the game is to tend your fields with your foes while offering goods to summon blights on you opponents shared fields.
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  • $\begingroup$ The only problem with Paper Scissors Rock is that saying there's strategy involved is only half right. $\endgroup$ – Travis Don Kindred Nov 21 '14 at 20:51
  • $\begingroup$ Sure sure but the simplist games (what you asked for) either are solvable (so only one person really can win) or rely on chance. This is the seed game on whose theory one could make a complicated game. $\endgroup$ – kaine Nov 21 '14 at 21:01
  • $\begingroup$ Do you play strategy video games? Zerg, Sprawl, Turtle (the three strategies), works exactly the same way as paper, rock, scissors. The complexities added only affect the tactics used, tactics which are irrelevant against the wrong strategy. If we started a game of stratego and a coin flip determined who got the higher number pieces and who got the lower numbers, strategy wouldn't matter. If you bet on red and black in roulette, chance doesn't matter. If both contribute evenly, neither matters. That's how I see it. Chance ruins strategy. $\endgroup$ – Travis Don Kindred Nov 21 '14 at 21:09
  • $\begingroup$ @Travis the easiest way to make a fun game is to make the player make calculated gambles that rely on chance or trying to predict an opponents moves. Nevertheless, I am not sure I understand your comment or what you apparently want. Strategy without any chance is tic tac toe. $\endgroup$ – kaine Nov 21 '14 at 21:26
  • $\begingroup$ I like this in Paper Scissors Rock. I like it because it takes two seconds to complete. Almost zero investment. A game like risk takes maybe 4-6 hours and can be ruined by two players ganging up or by a string of bad rolls. It's a huge letdown. In chess on the other hand, every failure is your fault. It feels good because you know you can learn from it and do better next time. You can never get better at flipping a coin, so you're going to get it about half the time. It's boring. That's what I'm wanting to avoid. $\endgroup$ – Travis Don Kindred Nov 21 '14 at 21:34
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Resurrection Mechanic:

A player must defeat both opponents in order to win. If he's the last standing and did not defeat both opponents, the opponent that he didn't personally defeat is resurrected with equal abilities / units / stats.

This can be reflected in real life in a situation where a regime has fallen, but there is still outlying individuals that believe in the cause and they are lying in wait for the perfect opportunity to strike.

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  • $\begingroup$ In situation A, that puts the weakened victor at a disadvantage and with a loss. In situation B, it wouldn't change anything. The winner already beat both others. $\endgroup$ – Travis Don Kindred Nov 21 '14 at 20:58
  • $\begingroup$ I disagree: In situation A the person on the sidelines wouldn't be the victor of the first confrontation, so he would eventually have to defeat a resurrected player of the same strength as himself. In situation B the resurrected player is again at the same strength as the second victor and it eliminates the "ganging up on an individual" problem. The main problem here is to present a mechanic that causes players to change targets frequently so that they use optimal resources. $\endgroup$ – Mathias Rechtzigel Nov 21 '14 at 21:08
  • $\begingroup$ Let's number our players for clarity. In (A), 1 and 2 attack each other. 3 sits and waits for 2 to beat 1, after which he beats 2 in turn. You suggest that 1 reenters the game with the same resources as 3. Assuming that 3 beats one also, let me ask you this. Who is the better player, 2 or 3 and why? In (B), let's say 1 and 2 gang up on three and three is eliminated. 3 has been beaten by both opponents now and is unable to gain reentry. $\endgroup$ – Travis Don Kindred Nov 21 '14 at 21:18
  • $\begingroup$ @TravisKindred 1 would have to beat 2 & 3 in a row in order to win. In (A) If 1 re-enters the game and loses, 3 would be the better player because he had a better strategy. If 3 loses, then 2 would re enter the game with the same resources as 1. In (B) When 3 is eliminated, only 1 or 2 would eliminate 3, not both. $\endgroup$ – Mathias Rechtzigel Nov 21 '14 at 21:33
  • $\begingroup$ What if 2 and 3 have the same strategy, to stay out of it. 1 attacks 2, and we're in condition (A). It doesn't determine comparative skill between 2 and 3. On (B), I understand what you're saying. That could possibly work. $\endgroup$ – Travis Don Kindred Nov 21 '14 at 21:39

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