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[winner]

What I understand of game design says that to balance a multiplayer game all you have to do is make sure that no one thing is completely superseded by any other, people will take care of the rest.

To test this idea I took rock paper scissors and gave
a win by Rock a value of 3
a win by Scissors a value of 2
and a win by Paper a value of 1.

Obviously your choices no longer have an equal utility. But how often would you want to use each available choice?
rock is useful when your opponent uses scissors more than 1/3 of the time that the use paper because your opponent gains 1 point when they throw paper, and you gain 3 every time they throw scissors
this can be restated as (with '>' standing in for '>=')
 R = S>3P
the other two are
 S = 3P>2R

 P = 2R>S

if you notice the point that all these are true at is when for every 1 scissors you have 2 rocks and 3 paper.  This is the balance the game will end up at because this is the balance that you can't gain an advantage against. (Obviously if your opponent is using a different strategy then you can gain an advantage by using the symbol they are weak against)

..|Three P]Two R] S
P | 0| 0| 0]-1|-1] 2       sums to 0
R | 1| 1| 1] 0| 0]-3       sums to 0
S |-2|-2|-2] 3| 3] 0       sums to 0

you can clearly see that no matter how often they use any given sign there is no advantage they can gain on you (though in this case since neither the equation for paper nor that of scissors are true you gain an advantage by choosing only rock).

[zzzdude]

Now you say "you lose 1 point when your opponent throws paper". Now, your rules state that you gain a value if you win. Nothing at a loss. Are you saying that when you win with a rock, your value increases by 3, but your opponent decreases by 1 (value of paper)?

[winner]

What I'm saying is that paper beats rock. And your opponent's score is calibrated to 0, basically if they gain one point it looks the same as if you had lost one.  I'll modify it to try to make it clearer

[3]

This is why fighting games have tiers at tournament level. It's quite interesting when you think about it - for example, why is a high-tier character considered effective? Not only may or may not that character be blatantly more effective than other characters, but the character's effectiveness is also partially determined by its popularity - a greater popularity results in 1. a greater chance of a tournament win based purely on odds and 2. a greater chance of useful tactics being developed due to greater gross playing time.

Of course, it's possible for higher-tier characters to become even more popular due to their effectiveness, and thus become more effective... or not? More popularity equals a greater chance of other players developing tactics that work specifically against that character. This gets to the point that developers (specifically Capcom) actually stop testing characters for anything less than bugs and blatant overpowered-ness, and just let the community sort itself out.

It's really quite a web, and only applies heavily to certain types of games - those that involve the player making choices that affect the player character his/herself. The rule still applies to games that don't necessarily do this, however - take FPSs, wherein a combination of a certain weapon and a certain tactic might be extremely effective, forcing other players to find means of besting said tactic.

Personally, I both like and dislike this at the same time. In some ways it reminds me of games that are excessively chance-based - the concept of having to learn all kinds of "tricks of the trade" before I'm able to do anything whatsoever puts me off somewhat, but as long as those tactics evolve organically via simply playing, or are part of the gameplay instead of the basis of it, they can be a great design choice.

[eerr]

There is a calculation for this... but I can't remember it offhand.

[winner]

There is a calculation for this... but I can't remember it offhand.

do you know what it calculates or where I can find it?

[sonerohi]

My thought on seeing the title: How can I balance rock-paper?

[eerr]

There is a calculation for this... but I can't remember it offhand.

do you know what it calculates or where I can find it?

I think it calculates for the average thats most profitable, on average.
It also takes into account whats most profitable for your enemy.

Obviously if rock is overpowered people will use rock more and paper less. But you can't get away with using just one.

[olemars]

Reminds me of this Spring mod. There's probably a dl for it on jobjol.

[Innominate]

Let's say you know your opponents preferences for the game. They use rock r of the time, paper p and scissors s, with

Code: [Select]

r + p + s = 1.

Let's assume you also have a static preference system (but that theirs is independent of yours), let your preferences be R, P and S, with

Code: [Select]

R + P + S = 1.

Your expected return for this system is:

Code: [Select]

E = R(r*0 + p*-1 + s*3) + P(r*1 + p*0 + s*-2) + S(r*-3 + p*2 + s*0)
⇒ E = R(3s - p) + P(r - 2s) + S(2p - 3r)

The way to maximise this is to put all your eggs in one basket. Choose the one with the higher partial derivative, where

Code: [Select]

∂E/∂R = 3s - p
∂E/∂P = r - 2s
∂E/∂S = 2p - 3rImportantly, it is impossible for all three of these to be negative. That is, there is no possible static set of possibilities which cannot be beaten. Indeed, if we assume that r, s and p only apply to the next round, this holds true even for dynamic choices.

As it happens, the best combination is the one for which the equations are all equal to 0 (except swap the lower and upper case letters). This gives the entirely predictable optimal choice (assuming you don't know your opponent's decision-making):
R = 1/2, P = 1/6, S = 1/3.
In other words, you do rock one and a half times more often than scissors, and three times more often than paper. Basically, this does not overcome the problems of the normal version.

Note that, with this setup, your opponent's choice of weightings is entirely irrelevant to the expected value, which still comes out to be 0. That is to say, the best outcome you can hope for, not knowing your opponent's mind, is breaking even.

[Bricks]

This reminds me of the Guess-2/3-of-the-average game.  The best strategy is dependent entirely on the strategy of the other players, which generally isn't rational.  If it were, there would always be a clear choice, and it usually results in a universal tie.

The fighting game analogy is good, even though it seems like most fighting games don't have a true rock-paper-scissors trichotomy (Yeah, I sorta made that word up).  I remember PvP in WoW was often described as being a rock-paper-scissors system, but ultimately, it was balanced out by the fact that the unexpected, generally "weaker" classes/talent builds were uncommon enough that players didn't know how to handle them.

[winner]

Let's say you know your opponents preferences for the game. They use rock r of the time, paper p and scissors s, with

Code: [Select]

r + p + s = 1.

Let's assume you also have a static preference system (but that theirs is independent of yours), let your preferences be R, P and S, with

Code: [Select]

R + P + S = 1.

Your expected return for this system is:

Code: [Select]

E = R(r*0 + p*-1 + s*3) + P(r*1 + p*0 + s*-2) + S(r*-3 + p*2 + s*0)
⇒ E = R(3s - p) + P(r - 2s) + S(2p - 3r)

The way to maximise this is to put all your eggs in one basket. Choose the one with the higher partial derivative, where

Code: [Select]

∂E/∂R = 3s - p
∂E/∂P = r - 2s
∂E/∂S = 2p - 3rImportantly, it is impossible for all three of these to be negative. That is, there is no possible static set of possibilities which cannot be beaten. Indeed, if we assume that r, s and p only apply to the next round, this holds true even for dynamic choices.

As it happens, the best combination is the one for which the equations are all equal to 0 (except swap the lower and upper case letters). This gives the entirely predictable optimal choice (assuming you don't know your opponent's decision-making):
R = 1/2, P = 1/6, S = 1/3.
In other words, you do rock one and a half times more often than scissors, and three times more often than paper. Basically, this does not overcome the problems of the normal version.

Note that, with this setup, your opponent's choice of weightings is entirely irrelevant to the expected value, which still comes out to be 0. That is to say, the best outcome you can hope for, not knowing your opponent's mind, is breaking even.

yup exactly.

[Muz]

Uh, I think game theory works far better for this than derivatives. That derivative method is good if you have 100% certainty of the opponent's strategy. Since the opponent will be trying to fool you with his strategy, it's not a very realistic method.

Let's say it's a zero sum game, which means that if the winner gets +3 points, the loser gets -3 points (because his lagging behind the winner by 3 points). Ties count as 0 points.

	Rock	Paper	Scissors
Rock	0,0	-1,1	3,-3
Paper	1,-1	0,0	-2,2
Scissors	-3,3	2,-2	0,0

There's no Nash equilibrium here, which means that there's no strictly dominant strategy. Your game is already balanced because it's not a "no-brainer"!

Rock is a "safe" strategy, because it minimizes losses.. a good defensive one (and well-named?).
Paper minimizes wins, which makes it a benefit by being unexpected.
Scissors is a very risky strategy that puts you up for huge losses, only worth it if you know that the other player is going to play paper.

So, in short, you have rock as a rather default strategy. Paper is a poor strategy, specifically used for countering rock. Scissors is an even poorer strategy, and would normally be used along with a bluff to convince the opposite player that you're going to play rock.

By tripling the bonus for choosing rock, a person who plays rock somewhat than other strategies would be winning... making it a core strategy. The gameplay should then revolve around when you expect the other player to play rock, trying to bluff the other player into thinking you're going to play rock, and playing scissors when you think he's going to take the bluff.

I think the C&C games are a decent examples of this, where tanks are the rocks. Anti-tank infantry are like paper, and the anti-infantry infantry (who get run down by tanks) are like the scissors.

[winner]

The equations work just as well on expectations as they do on certainties (they will be exactly as useful as your expectations are accurate).  However I didn't intend them to be used for playing games because they ignore the dynamics of psychology so they say nothing about what your opponent is going to do. 
For me they were a cool tool for designing games because they allow you to fine tune the amount that you want different moves to be used.

[Muz]

The beautiful thing about Rock-Paper-Scissors is that it's always balanced. It makes things complex enough that you can't apply any theory to it right. You just can't powergame RPS with pencil, paper, calculator.

In fact, I just noticed that almost every RTS out there is an asymmetrical RPS game in some form. But most games exaggerate it a bit. Say, a horseman (rock) can kill 3 archers (scissors) before dying. If you upgrade the horseman, he might be able to kill 4-6 more. If you use a spearman (paper), he'd be able to take down the horseman.

If you look carefully, the easiest/most common method of balance is to increase the costs to suit the utility. One horseman would probably cost as much as 3 archers, and a spearman would cost as much as two of them. Something more versatile, like a horse archer that can beat both archers and spearmen, would probably cost even more than a typical horseman, just because of the versatility.

Gameplay of any RTS is essentially the same as asymmetrical RPS, but it adds the challenge of obtaining resources, reducing cost, and increasing utility.

If there's no cost, you really don't have to worry about balance.. any RPS design automatically balances itself.