Rock, Paper, Scissors

[obi]

You die and go to hell. It turns out Satan was overthrown by Gauss. He amuses himself throughout eternity by making his inmates play mathematical games. He has decided that your game will be a game of Rock, Paper, Scissors.

In this game of RPS (Rock, Paper, Scissors), a win with rock gets you 10 dollars, a win with paper gets you 3, and a win with scissors gets you 1. You and your opponent start with 100 dollars each. It costs a dollar to play. In the event of a tie, both players get their dollar back.

The game only ends when one player runs out of money. You have all the time in the world to play this game, obviously, as you are dead. After all of the people assigned to play this bizarre game have completed, only the top 50% of the winners (based on how much money they have at the end) get released to go to heaven. The remainder rot in hell, where they will likely spend the rest of their time playing Chutes and Ladders with 1-sided dice, counting to infinity, and attempting to square the circle.

You obviously want to be in that upper 50% of the winners (so the upper quartile of the entirety of the players).

The question is: Do you ever use scissors?

What if it costs .50 to play? 1.5? What is the highest cost per play that you would be willing to pay to use scissors?

 

[DeLorean]

I'm assuming that you loose the money (e.g. 10 dollars loose when your opponent plays a paper to your rock), so I think this would be about prediction, much like pokemon. Something you could relate too.

Of course, you could not win every game, but you could win enough.

 

[obi]

No, it just costs a dollar to play. You don't lose anything extra for losing. This means that at a dollar per play, a win with scissors costs your opponent 1 dollar, and you gain / lose nothing.

 

[Captain Falcon]

I'm guessing that the most common play would be paper. It stops your opponent from winning a large amount of money with rock while still netting you 3$ if you win. Even if you lose, your opponent will not gain any money, and even if they use scissors 10 times, and you predict and use rock one of those times you will still have the upper hand. The more money you pay to play, the more you would see paper being used because its counter, scissors would be less acceptable as you would be losing money even if you won.

 

[Glen]

this is a really cool thread

i cant say i have an answer for you yet, but i was wondering, am i assuming my opponent is going to take the most logical choice as well? i havent thought that much about it yet but otherwise it seems to me that i would play differently depending upon my opponent's style of play.

 

[Dragontamer]

I'll play the Nash Equalibrium:

Rock == 1 / 14
Scissors == 3 / 14
Paper == 10/14

If my opponent chooses Rock / Paper on the average, we both gain points. If he chooses Scissors, we both will on the average lose points. Regardless, my expected win is tied with him, like Desteny Bond. In the end, I've forced a 50% chance of win and a 50% chance of losing.

Actually, thats if he wants to lose, we both can play the game forever if HE chooses rock. Paper == we both will lose on the average the most, and Scissors == we'll both lose gradually.

Considering that the first round is all about having more money than the other winners, I'll be sure to tell him about this scheme, while we both grow. He plays rock 100% of the time, I play the above odds and we both gain $1.14 on the average. After that, he decides when to let us fall and he'll start playing paper 100% of the time. After that, its a 50% / 50% chance. If it comes out that we both are not going to be the winner in the next round, we can start up that deal again.

What if it costs .50 to play? 1.5? What is the highest cost per play that you would be willing to pay to use scissors?

Irrelevant. We both will lose the same $$ on the average if I play at the equilibriuim point. I'd probably agree to playing if it is $5 or more, because if it is below $5 cooperating with my opponent can make the game drone on towards infinity.

 

[shadydentist]

Using dragontamer's numbers, lets take a look per hand:

Against rock:
1/14 tie (you 0, opponent 0)
3/14 lose(you 0, opponent 10)
10/14 win(you 3, opponent 0)
Expected outcome: opponent 30/14, you 30/14 or 2.14

Against paper:
1/14 lose (you 0, opponent 3)
3/14 win (you 1, opponent 0)
10/14 tie (you 0, opponent 0)
Average: you 3/14, opponent 3/14 or .21

Against scissors:
1/14 win (you 10, opponent 0)
3/14 tie (you 0, opponent 0)
10/14 lose (you 0, opponent 1)
Average: you 10/14, opponent 10/14 or .71

Against a player playing the same strategy, the odds would simply be
(1/14)(2.14) + (10/14)(.21) + (3/14)(.71) =.46 dollars per round.


Typically, nash equilibrium also assumes no communication between players. Plus, that makes this exercise more fun... if both players were completely rational, both would at first adopt this strategy.

Interesting to note that the expected outcome for one player all rock is greater than the nash equillibrium.

 

[TheMaskedNitpicker]

This does not appear to be a zero-sum game. Since the amount of money in the pool is likely to increase, the game is likely to last almost eternally. This really is Hell!

My question is this: if both players play with the Nash Equilibrium, what's the average number of games played before one of the players loses?

Also, the top 50% of winners can only be determined after all players have finished playing. With, say, 1,000,000 players, how long is the competition likely to take on average?

 

[Mekkah]

Haven't math'd this yet, but I think I will just make it easy for myself and spam paper, and if my opponent catches on I use some rocks in there which redeems losses quickly. Then again, lots of people will do that since it's an easy solution, so we'll all play eternally unless there's one idiot out there.

Also, the top 50% of winners can only be determined after all players have finished playing.

The game ends when one player runs out of money, so just wait for that idiot while keeping optimal gain, maybe with your opponent helping.

I'd say this is more interesting when the choice that nets you the most money loses to the one that nets you the least money.

 

[CaptKirby]

I would use scissors, because there is nothing wrong with not increasing your money to avoid losing it!

 

[Brain]

Actually, thats if he wants to lose, we both can play the game forever if HE chooses rock. Paper == we both will lose on the average the most, and Scissors == we'll both lose gradually.

Considering that the first round is all about having more money than the other winners, I'll be sure to tell him about this scheme, while we both grow. He plays rock 100% of the time, I play the above odds and we both gain $1.14 on the average. After that, he decides when to let us fall and he'll start playing paper 100% of the time. After that, its a 50% / 50% chance. If it comes out that we both are not going to be the winner in the next round, we can start up that deal again.

If you want to grow, you should just both play rock/scissors in alternance, offset by 1. That way you both win 4 per turn on average. Moreover, the strategy is stable and you'll both grow equally.

Now, if you convince your foe to always play paper afterwards, he's an idiot. If you use the nash equilibrium against an idiot, you're also an idiot. The strategy to win against an idiot is to use him to gain an advantage. If your foe really does always uses paper, you have to skew the odds towards scissors not enough for him to notice it but enough to win safely. People are notoriously bad at evaluating randomness because they expect it to never streak: if you are careful to avoid them, you can probably pull off a 50% scissors rate against your opponent's paper without him noticing that you're not doing the nash equilibrium at all.

At first glance, what I would suggest is to use a non-uniform nash equilibrium. When you have a streak, skew the odds for the counter to your streak's counter. I'm fairly sure most opponents will instinctively (or even consciously) skew their odds towards scissors after seeing as little as two paper in a row because they project a third paper (and that will happen often with the nash equilibrium). Conversely, if your opponent tries to be random and he uses rock, he probably won't use rock next round because unlike randomness he thinks "I used rock last time so I should average it out". Use that to break him. Unless they use tools (like coins or die), your opponent _will_ have patterns and that is how you must destroy him. That might be difficult if you have no tools yourself, though :(

Also, if only 50% of all winners go to heaven, I would try to agree with my partner to boost both our scores so the current best player is always at 100. That way, we will have to play until randomness (or strategy) digs a 100 points gap between us That ought to be enough to beat the masses soundly. On the other hand I'm sure some fuckers will just play forever :(

 

[obi]

Of course, if you do strategize with your opponent, then you have to make sure to break the strategy before him, otherwise you will be at a disadvantage. "Offer them an open hand... then stab them in the back!"

Think about it. If you come with a strategy to raise both side's money until you reach some arbitrarily large amount for both, you'll be roughly equal at the time in which you agree to play to beat the opponent. Therefore, if you renege on your agreement just before reaching this very large amount of money for both sides, then you will probably have slightly less money, but their amount will be even lower, thus increasing your chances to win this encounter.

Obviously, your opponent will also be aware of his ability to turn on you. You just have to betray your opponent before they betray you. It's in both of your best interests to make the game last as long as possible before betrayal, but it's also in both of your best interests to be the one to initiate the betrayal, so you have two antagonistic forces determining the length of the game.

For another variation on this game (which in retrospect is probably more interesting), consider how things might change if a tie means neither side gets their dollar back. This allows the total pool of money to potentially shrink regardless of your throw (currently, paper vs. scissors means that paper loses a dollar, scissors gains / loses nothing; every other throw results in a gain for the winner greater than the loss for the loser).

 

[Dragontamer]

If I play at the Nash Equalibrium, I'll betray him the precise round he decides to betray me. It is the ultimate conservative strategy, playing exactly tit-for-tat. Every attempt for him to backstab me during the agreement hurts him as much as it hurts me by definition of that mixed stratetgy.

Its quite simple, I play the mixed strategy Nash Equalibrium point forever. He has the choice to kill both of us, or cooperate. I am impervious to backstabbing.

 

[Brain]

If I play at the Nash Equalibrium, I'll betray him the precise round he decides to betray me. It is the ultimate conservative strategy, playing exactly tit-for-tat. Every attempt for him to backstab me during the agreement hurts him as much as it hurts me by definition of that mixed stratetgy.

Its quite simple, I play the mixed strategy Nash Equalibrium point forever. He has the choice to kill both of us, or cooperate. I am impervious to backstabbing.

You may be impervious to backstabbing, your opponent might not. If your opponent's patterns can be abused, you should defect from the nash equilibrium to boost your own odds. 50% is few and your happiness is at stakes. That's also why a rational opponent would never cooperate: if I use rock all the time to boost us both, you can easily tell if I'm backing on the agreement but how can I tell that the lead you are building is because of luck and not you increasing the odds of using paper to make sure that you're gaining more than I am? I sure as fuck would do it. On the other hand, if I always use paper, you could easily precalculate a sequence of moves that makes you win while not straying far enough from the equilibrium to be statistically significant. I cannot risk that. Thus, if I was playing you, I would have to suggest the following strategy: we both use the nash equilibrium and whenever we both drop below 100 we do one paper/rock round and one rock/paper round, both gaining 1 point. Whoever wins the game will win with a good margin and I think the risk is worth the potential payoff: the person who is supposed to use paper then rock can only cheat by using paper then scissors (and gain an edge of 4), the other can only cheat by using scissors (and gain an edge of 1). Since the boosting phase will always happen as soon as one of us is at 99 and that the poorest would want to boost first, betrayal is pointless. If you refuse the deal we both use the nash equilibrium and that's that.

Also here's an interesting thing to know: what tools can you use in the game? As I said, humans are notoriously bad at being random, so if you want to play using the nash equilibrium, you'll need an unbiased source of randomness. If you can use computers, you'd have to be careful about your random number generator because if someone like me ever figures it out you are fucked. I guess you could just flip the coins used for the game :)

 

[Surgo]

You die and go to hell. It turns out Satan was overthrown by Gauss.

Sounds like a good time for some Gaussian elimination! Oh shit.

(I'll be here all night.)