10 weapon types, 2 proficiencies, rock-paper-scissors relationship?

I want to make a game where there's a total of 10 general weapon types. I want any given character to be able to be proficient with 2 of the weapon types. Looking at the fact that 10 is an even number, a perfect rock-paper-scissors relationship wouldn't be possible. "C(10,2)=45" however is an odd number, which should be workable into a rock-paper-scissors system.

Since the final number of total weapon pairings is an odd number, I reckon there should be an intuitive and systemically organized way to get a rock-paper-scissors relationship to exist where each weapon pairing is at an advantage against 22 of the other pairings while being at a disadvantage against the remaining 22 pairings. Assuming the 10 weapon types that a character can be proficient with 2 of are labeled A through to J, how should I organize it?

P.S.: A smaller example that runs into the same situation would be "C(6,2)=15"; weapons labeled A through F; if that's more convenient for explanation purposes. Please let me know if this question belongs under different tags or even over on the Mathematics section of SE instead of here; I wasn't sure where to put this question.

The simplest way to do it would be to lay out your weapon combinations in whatever order you see fit in a circular list. Each combination could have a bonus against the 22 combinations after it in the list, and a weakness against the 22 before it.

The player might appreciate a more straight-forward or easier-to-recognise logic to that order -- some sort of pattern.

With your 6 weapon example, you could have:

1. AB AC AF
2. BC BD BF
3. CD CE CF
4. DE DA DF
5. EA EB EF


Here you have a whole lot of rock-paper-scissors. Every combination of weapons has a "primary" and a "secondary" weapon according to a rock-paper-scissors between the first 5 weapon types, with the 6th weapon type (F) always being secondary (since, as you pointed out, we need an odd number for rock-paper-scissors).

We now have 5 super groups, and can have a simple 5-way rock-paper-scissors between them (group 1 beats groups 2 and 3, 2 beats 3 and 4, and so on).

If two characters are using the same primary weapon, we have a 3-way rock-paper-scissors. The way I've laid it out here, you could have AB beat AC, AC beat AF, and AF beat AB.

Expanding this to 10 weapons, you can divide the 45 combinations into 9 primary weapon groups of 5 combinations, with the 10th weapon always secondary:

1. AB AC AD AE AJ
2. BC BD BE BF BJ
3. CD CE CF CG CJ
4. DE DF DG DH DJ
5. EF EG EH EI EJ
6. FG FH FI FA FJ
7. GH GI GA GB GJ
8. HI HA HB HC HJ
9. IA IB IC ID IJ


While this kind of pattern may not be super easy to memorise, a well-designed UX could do a lot to make it clear which weapon is a character's primary weapon, which primary weapons beat which other primary weapons, and which secondary weapons beat which other secondary weapons (conveniently in the same order as primary weapons, except for the always-secondary weapon J).

I hope that makes sense!

With 11 weapon types, with 55 combinations, which, as you've pointed out, is divisible by the number of weapons, you don't need a special "secondary only" weapon:

01. AB AC AD AE AF
02. BC BD BE BF BG
03. CD CE CF CG CH
04. DE DF DG DH DI
05. EF EG EH EI EJ
06. FG FH FI FJ FK
07. GH GI GJ GK GA
08. HI HJ HK HA HB
09. IJ IK IA IB IC
10. JK JA JB JC JD
11. KA KB KC KD KE

• I had considered the possibility of relegating a weapon to some kind of "special" position to make an odd number of "normal" weapons, though if possible I'd like to avoid giving any one weapon special treatment. That format of laying it out does look good though; I'll probably use that format as a base to tinker with. Apr 28, 2016 at 3:09
• After trying to tinker with it a bit more, I'm pretty sure your system exactly as you presented it is probably indeed the single best possible way to go about it. I still wish that I could've given "weapon J" a normal role rather than the role of connecting point within it's primary's main group, but eh; if the math behind it simply can't be bent that way then I'll be better off spending my time working other things out. Apr 28, 2016 at 4:00
• Yeah, I guess part of the problem is that the number of combinations isn't divisible by the number of weapon types, so having all weapons have balanced positions isn't possible. Still, I understand it's not ideal to have one stand out so much from the others. Apr 28, 2016 at 4:25
• That'd work! :) I've expanded the answer to illustrate. Apr 28, 2016 at 5:27
• That should work, with one more condition: C(n,2)/n needs to be odd. For example, if n is 13, C(13,2) = 78, and 78/13 = 6. That means 6 secondary weapons for each primary weapon, which is even, so you wouldn't be able to have rock-paper-scissors between two characters with the same primary weapon. n = 15 should work, and I imagine you'd be able to shuffle them around amongst the old weapons such that the old advantage/disadvantage relationships aren't changed. Apr 28, 2016 at 7:42

While Jibb Smart already gave a good answer within the constraints of the question, I would like to step a bit outside the box and question: "Do you actually need perfect balance"? A perfectly imbalanced system can work too and often makes for far more interesting games.

For example, you can have one weapon combination which is slightly better than 43 others, but has a glaring weakness which can be easily exploited by the otherwise slightly underpowered combination 45. Everyone will play the first combination, until someone with 45 shows up and wipes the floor with them. Then players will look for a counter to 45 and try all kinds of different combinations.

Players will have far more fun with figuring out which combinations stand out of the crowd power-wise and then figure out which other combination can counter them than when you present them a perfectly balanced system where the counter for everything is obvious and in the end their weapon choice doesn't really matter because the ratio of players which are stronger or weaker than them will be 50/50 split in any case.

However, such a system is not easy to create. You really need to know and understand your game mechanics well to not accidentally create a combination which either has no viable counter or doesn't counter anything else.